Stack (data structure)
In computer science, a stack is a last in, first out (LIFO) abstract data type and data structure. A stack can have any abstract data type as an element, but is characterized by only three fundamental operations: push, pop and stack top. The push operation adds a new item to the top of the stack, or initializes the stack if it is empty. If the stack is full and does not contain enough space to accept the given item, the stack is then considered to be in an overflow state. The pop operation removes an item from the top of the stack. A pop either reveals previously concealed items, or results in an empty stack, but if the stack is empty then it goes into underflow state (It means no items are present in stack to be removed). The stack top operation gets the data from the top-most position and returns it to the user without deleting it. The same underflow state can also occur in stack top operation if stack is empty.
A stack is a restricted data structure, because only a small number of operations are performed on it. The nature of the pop and push operations also means that stack elements have a natural order. Elements are removed from the stack in the reverse order to the order of their addition: therefore, the lower elements are those that have been on the stack the longest.[1]
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History
The stack was first proposed in 1955, and then patented in 1957, by the German Friedrich L. Bauer.[2] The same concept was developed independently, at around the same time, by the Australian Charles Leonard Hamblin.[citation needed]
Abstract definition
A stack is a fundamental computer science data structure and can be defined in an abstract, implementation-free manner, or it can be generally defined as, Stack is a linear list of items in which all additions and deletion are restricted to one end that is Top.
This is a VDM (Vienna Development Method) description of a stack
Function signatures:
init: -> Stack push: N x Stack -> Stack top: Stack -> (N U ERROR) remove: Stack -> Stack isempty: Stack -> Boolean
(where N indicates an element (natural numbers in this case), and U indicates set union)
Semantics
top(init()) = ERROR top(push(i,s)) = i remove(init()) = init() remove(push(i, s)) = s isempty(init()) = true isempty(push(i, s)) = false
Reference [3]
Inessential operations
In modern computer languages, the stack is usually implemented with more operations than just "push","pop" and "Stack Top". Some implementations have a function which returns the current number of items on the stack. Alternatively, some implementations have a function that just returns if the stack is empty. Another typical helper operation stack top[4] (also known as peek) can return the current top element of the stack without removing it.
Software stacks
Implementation
In most high level languages, a stack can be easily implemented either through an array or a linked list. What identifies the data structure as a stack in either case is not the implementation but the interface: the user is only allowed to pop or push items onto the array or linked list, with few other helper operations. The following will demonstrate both implementations, using C.
Array
The array implementation aims to create an array where the first element (usually at the zero-offset) is the bottom. That is, array[0] is the first element pushed onto the stack and the last element popped off. The program must keep track of the size, or the length of the stack. The stack itself can therefore be effectively implemented as a two-element structure in C:
typedef struct { size_t size; int items[STACKSIZE]; } STACK;
The push() operation is used both to initialize the stack, and to store values to it. It is responsible for inserting (copying) the value into the ps->items[] array and for incrementing the element counter (ps->size). In a responsible C implementation, it is also necessary to check whether the array is already full to prevent an overrun.
void push(STACK *ps, int x) { if (ps->size == STACKSIZE) { fputs("Error: stack overflow\n", stderr); abort(); } else ps->items[ps->size++] = x; }
The pop() operation is responsible for removing a value from the stack, and decrementing the value of ps->size. A responsible C implementation will also need to check that the array is not already empty.
int pop(STACK *ps) { if (ps->size == 0){ fputs("Error: stack underflow\n", stderr); abort(); } else return ps->items[--ps->size]; }
If we use a dynamic array, then we can implement a stack that can grow or shrink as much as needed. The size of the stack is simply the size of the dynamic array. A dynamic array is a very efficient implementation of a stack, since adding items to or removing items from the end of a dynamic array is amortized O(1) time.
Linked list
The linked-list implementation is equally simple and straightforward. In fact, a simple singly linked list is sufficient to implement a stack—it only requires that the head node or element can be removed, or popped, and a node can only be inserted by becoming the new head node.
Unlike the array implementation, our structure typedef corresponds not to the entire stack structure, but to a single node:
typedef struct stack { int data; struct stack *next; } STACK;
Such a node is identical to a typical singly linked list node, at least to those that are implemented in C.
The push() operation both initializes an empty stack, and adds a new node to a non-empty one. It works by receiving a data value to push onto the stack, along with a target stack, creating a new node by allocating memory for it, and then inserting it into a linked list as the new head:
void push(STACK **head, int value) { STACK *node = malloc(sizeof(STACK)); /* create a new node */ if (node == NULL){ fputs("Error: no space available for node\n", stderr); abort(); } else { /* initialize node */ node->data = value; node->next = empty(*head) ? NULL : *head; /* insert new head if any */ *head = node; } }
A pop() operation removes the head from the linked list, and assigns the pointer to the head to the previous second node. It checks whether the list is empty before popping from it:
int pop(STACK **head) { if (empty(*head)) { /* stack is empty */ fputs("Error: stack underflow\n", stderr); abort(); } else { /* pop a node */ STACK *top = *head; int value = top->data; *head = top->next; free(top); return value; } }
Stacks and programming languages
Some languages, like LISP and Python, do not call for stack implementations, since push and pop functions are available for any list. All Forth-like languages (such as Adobe PostScript) are also designed around language-defined stacks that are directly visible to and manipulated by the programmer.
C++'s Standard Template Library provides a "stack" templated class which is restricted to only push/pop operations. Java's library contains a Stack class that is a specialization of Vector---this could be considered a design flaw, since the inherited get() method from Vector ignores the LIFO constraint of the Stack. PHP has an SplStack class.
Hardware stacks
A common use of stacks at the architecture level is as a means of allocating and accessing memory.
Basic architecture of a stack
A typical stack is an area of computer memory with a fixed origin and a variable size. Initially the size of the stack is zero. A stack pointer, usually in the form of a hardware register, points to the most recently referenced location on the stack; when the stack has a size of zero, the stack pointer points to the origin of the stack.
The two operations applicable to all stacks are:
- a push operation, in which a data item is placed at the location pointed to by the stack pointer, and the address in the stack pointer is adjusted by the size of the data item;
- a pop or pull operation: a data item at the current location pointed to by the stack pointer is removed, and the stack pointer is adjusted by the size of the data item.
There are many variations on the basic principle of stack operations. Every stack has a fixed location in memory at which it begins. As data items are added to the stack, the stack pointer is displaced to indicate the current extent of the stack, which expands away from the origin.
Stack pointers may point to the origin of a stack or to a limited range of addresses either above or below the origin (depending on the direction in which the stack grows); however, the stack pointer cannot cross the origin of the stack. In other words, if the origin of the stack is at address 1000 and the stack grows downwards (towards addresses 999, 998, and so on), the stack pointer must never be incremented beyond 1000 (to 1001, 1002, etc.). If a pop operation on the stack causes the stack pointer to move past the origin of the stack, a stack underflow occurs. If a push operation causes the stack pointer to increment or decrement beyond the maximum extent of the stack, a stack overflow occurs.
Some environments that rely heavily on stacks may provide additional operations, for example:
- Dup(licate): the top item is popped, and then pushed again (twice), so that an additional copy of the former top item is now on top, with the original below it.
- Peek: the topmost item is inspected (or returned), but the stack pointer is not changed, and the stack size does not change (meaning that the item remains on the stack). This is also called top operation in many articles.
- Swap or exchange: the two topmost items on the stack exchange places.
- Rotate (or Roll): the n topmost items are moved on the stack in a rotating fashion. For example, if n=3, items 1, 2, and 3 on the stack are moved to positions 2, 3, and 1 on the stack, respectively. Many variants of this operation are possible, with the most common being called left rotate and right rotate.
Stacks are either visualized growing from the bottom up (like real-world stacks), or, with the top of the stack in a fixed position (see image [note in the image, the top (28) is the stack 'bottom', since the stack 'top' is where items are pushed or popped from]), a coin holder, a Pez dispenser, or growing from left to right, so that "topmost" becomes "rightmost". This visualization may be independent of the actual structure of the stack in memory. This means that a right rotate will move the first element to the third position, the second to the first and the third to the second. Here are two equivalent visualizations of this process:
apple banana banana ===right rotate==> cucumber cucumber apple
cucumber apple banana ===left rotate==> cucumber apple banana
A stack is usually represented in computers by a block of memory cells, with the "bottom" at a fixed location, and the stack pointer holding the address of the current "top" cell in the stack. The top and bottom terminology are used irrespective of whether the stack actually grows towards lower memory addresses or towards higher memory addresses.
Pushing an item on to the stack adjusts the stack pointer by the size of the item (either decrementing or incrementing, depending on the direction in which the stack grows in memory), pointing it to the next cell, and copies the new top item to the stack area. Depending again on the exact implementation, at the end of a push operation, the stack pointer may point to the next unused location in the stack, or it may point to the topmost item in the stack. If the stack points to the current topmost item, the stack pointer will be updated before a new item is pushed onto the stack; if it points to the next available location in the stack, it will be updated after the new item is pushed onto the stack.
Popping the stack is simply the inverse of pushing. The topmost item in the stack is removed and the stack pointer is updated, in the opposite order of that used in the push operation.
Hardware support
Stack in main memory
Most CPUs have registers that can be used as stack pointers. Processor families like the x86, Z80, 6502, and many others have special instructions that implicitly use a dedicated (hardware) stack pointer to conserve opcode space. Some processors, like the PDP-11 and the 68000, also have special addressing modes for implementation of stacks, typically with a semi-dedicated stack pointer as well (such as A7 in the 68000). However, in most processors, several different registers may be used as additional stack pointers as needed (whether updated via addressing modes or via add/sub instructions).
Stack in registers or dedicated memory
The x87 floating point architecture is an example of a set of registers organised as a stack where direct access to individual registers (relative the current top) is also possible. As with stack-based machines in general, having the top-of-stack as an implicit argument allows for a small machine code footprint with a good usage of bus bandwidth and code caches, but it also prevents some types of optimizations possible on processors permitting random access to the register file for all (two or three) operands. A stack structure also makes superscalar implementations with register renaming (for speculative execution) somewhat more complex to implement, although it is still feasible, as exemplified by modern x87 implementations.
Sun SPARC, AMD Am29000, and Intel i960 are all examples of architectures using register windows within a register-stack as another strategy to avoid the use of slow main memory for function arguments and return values.
There are also a number of small microprocessors that implements a stack directly in hardware and some microcontrollers have a fixed-depth stack that is not directly accessible. Examples are the PIC microcontrollers, the Computer Cowboys MuP21, the Harris RTX line, and the Novix NC4016. Many stack-based microprocessors were used to implement the programming language Forth at the microcode level. Stacks were also used as a basis of a number of mainframes and mini computers. Such machines were called stack machines, the most famous being the Burroughs B5000.
Applications
Towers of Hanoi
One of the most interesting applications of stacks can be found in solving a puzzle called Tower of Hanoi. According to an old Brahmin story, the existence of the universe is calculated in terms of the time taken by a number of monks, who are working all the time, to move 64 disks from one pole to another. But there are some rules about how this should be done, which are:
- You can move only one disk at a time.
- For temporary storage, a third pole may be used.
- You cannot place a disk of larger diameter on a disk of smaller diameter.[5]
For algorithm of this puzzle see Tower of Hanoi. The C++ code for this solution can be implemented in two ways.
First Implementation (Without using Stacks)
Here we assume that A is first tower, B is second tower & C is third tower.
void TowersofHanoi(int n, int a, int b, int c) { //Move top n disks from tower a to tower b, use tower c for intermediate storage. if(n > 0) { TowersofHanoi(n-1, a, b, c); //recursion cout << " Move top disk from tower " << a << " to tower " << b << endl ; TowersofHanoi(n-1, a, b, c); //recursion } }
Output : (when there are 4 disks)
| Move disk | From peg | To peg |
|---|---|---|
| 1 | A | B |
| 2 | A | C |
| 1 | B | C |
| 3 | A | B |
| 1 | C | A |
| 2 | C | B |
| 1 | A | B |
| 4 | A | C |
| 1 | B | C |
| 2 | B | A |
| 1 | C | A |
| 3 | B | C |
| 1 | A | B |
| 2 | A | C |
| 1 | B | C |
Second Implementation (Using Stacks)
// Global variable , tower [1:3] are three towers arrayStack<int> tower[4]; void TowerofHanoi(int n) { // Preprocessor for moveAndShow. for (int d = n; d > 0; d--) //initialize tower[1].push(d); //add disk d to tower 1 moveAndShow(n, 1, 2, 3); /*move n disks from tower 1 to tower 3 using tower 2 as intermediate tower*/ } void moveAndShow(int n, int a, int b, int c) { // Move the top n disks from tower a to tower b showing states. // Use tower c for intermediate storage. if(n > 0) { moveAndShow(n-1, a, c, b); //recursion int d = tower[x].top(); //move a disc from top of tower x to top of tower[x].pop(); //tower y tower[y].push(d); showState(); //show state of 3 towers moveAndShow(n-1, c, b, a); //recursion } }
However Complexity for above written implementations is O(2^n). So it's obvious that problem can only be solved for small values of n (generally n <= 30). In case of the monks, the number of turns taken to transfer 64 disks, by following the above rules, will be 18,446,744,073,709,551,615; which will surely take a lot of time!!
Expression evaluation and syntax parsing
Calculators employing reverse Polish notation use a stack structure to hold values. Expressions can be represented in prefix, postfix or infix notations. Conversion from one form of the expression to another form may be accomplished using a stack. Many compilers use a stack for parsing the syntax of expressions, program blocks etc. before translating into low level code. Most of the programming languages are context-free languages allowing them to be parsed with stack based machines.
Evaluation of an Infix Expression that is Fully Parenthesized
Input: (((2 * 5) - (1 * 2)) / (11 - 9))
Output: 4
Analysis: Five types of input characters
* Opening bracket * Numbers * Operators * Closing bracket * New line character
Data structure requirement: A character stack
Algorithm
1. Read one input character
2. Actions at end of each input
Opening brackets (2.1) Push into stack and then Go to step (1)
Number (2.2) Push into stack and then Go to step (1)
Operator (2.3) Push into stack and then Go to step (1)
Closing brackets (2.4) Pop it from character stack
(2.4.1) if it is closing bracket, then discard it, Go to step (1)
(2.4.2) Pop is used three times
The first popped element is assigned to op2
The second popped element is assigned to op
The third popped element is assigned to op1
Evaluate op1 op op2
Convert the result into character and
push into the stack
Go to step (2.4)
New line character (2.5) Pop from stack and print the answer
STOP
Result: The evaluation of the fully parenthesized infix expression is printed on the monitor as follows:
Input String: (((2 * 5) - (1 * 2)) / (11 - 9))
| Input Symbol | Stack (from bottom to top) | Operation |
|---|---|---|
| ( | ( | |
| ( | ( ( | |
| ( | ( ( ( | |
| 2 | ( ( ( 2 | |
| * | ( ( ( 2 * | |
| 5 | ( ( ( 2 * 5 | |
| ) | ( ( 10 | 2 * 5 = 10 and push |
| - | ( ( 10 - | |
| ( | ( ( 10 - ( | |
| 1 | ( ( 10 - ( 1 | |
| * | ( ( 10 - ( 1 * | |
| 2 | ( ( 10 - ( 1 * 2 | |
| ) | ( ( 10 - 2 | 1 * 2 = 2 & Push |
| ) | ( 8 | 10 - 2 = 8 & Push |
| / | ( 8 / | |
| ( | ( 8 / ( | |
| 11 | ( 8 / ( 11 | |
| - | ( 8 / ( 11 - | |
| 9 | ( 8 / ( 11 - 9 | |
| ) | ( 8 / 2 | 11 - 9 = 2 & Push |
| ) | 4 | 8 / 2 = 4 & Push |
| New line | Empty | Pop & Print |
C Program
int main (int argc, char *argv[]) { struct ch *charactop; struct integer *integertop; char rd, op; int i = 0, op1, op2; charactop = cclearstack(); integertop = iclearstack(); while(1) { rd = argv[1][i++]; switch(rd) { case '+': case '-': case '/': case '*': case '(': charactop = cpush(charactop, rd); break; case ')': integertop = ipop (integertop, &op1); charactop = cpop (charactop, &op); while(op != '(') { integertop = ipush (integertop, eval(op, op1, op2); charactop = cpop (charactop, &op); if (op != '(') { integertop = ipop(integertop, &op2); integertop = ipop(integertop, &op1); } } break; case '\0': while (!= cemptystack(charactop)) { charactop = cpop(charactop, &op); integertop = ipop(integertop, &op2); integertop = ipop(integertop, &op1); integertop = ipush(integertop, eval(op, op1, op2); } integertop = ipop(integertop, &op1); printf("\n The fianl solution is: %d", op1); return 0; default: integertop = ipush(integertop, read - '0'); } } } int eval(char op, int op1, int op2) { switch (op) { case '+': return op1 + op2; case '-': return op1 - op2; case '/': return op1 / op2; case '*': return op1 * op2; } }
Output of the program:
Input entered at the command line: (((2 * 4) - (1 * 2) / (11 - 9)) [7]
Evaluation of Infix Expression which is not fully parenthesized
Input: (2 * 5 - 1 * 2) / (11 - 9)
Output: 4
Analysis: There are five types of input characters which are:
* Opening brackets
* Numbers
* Operators
* Closing brackets
* New line character (\n)
We do not know what to do if an operator is read as an input character. By implementing the priority rule for operators, we have a solution to this problem.
The Priority rule we should perform comparative priority check if an operator is read, and then push it. If the stack top contains an operator of prioirty higher than or equal to the priority of the input operator, then we pop it and print it. We keep on perforrming the prioirty check until the top of stack either contains an operator of lower priority or if it does not contain an operator.
Data Structure Requirement for this problem: A character stack and an integer stack
Algorithm:
1. Read an input character
2. Actions that will be performed at the end of each input
Opening brackets (2.1) Push it into stack and then Go to step (1)
Digit (2.2) Push into stack, Go to step (1)
Operator (2.3) Do the comparative priority check
(2.3.1) if the character stack's top contains an operator with equal
or higher priority, then pop it into op
Pop a number from integer stack into op2
Pop another number from integer stack into op1
Calculate op1 op op2 and push the result into the integer
stack
Closing brackets (2.4) Pop from the character stack
(2.4.1) if it is an opening bracket, then discard it and Go to
step (1)
(2.4.2) To op, assign the popped element
Pop a number from integer stack and assign it op2
Pop another number from integer stack and assign it
to op1
Calculate op1 op op2 and push the result into the integer
stack
Convert into character and push into stack
Go to the step (2.4)
New line character (2.5) Print the result after popping from the stack
STOP
Result: The evaluation of an infix expression that is not fully parenthesized is printed as follows:
Input String: (2 * 5 - 1 * 2) / (11 - 9)
| Input Symbol | Character Stack (from bottom to top) | Integer Stack (from bottom to top) | Operation performed |
|---|---|---|---|
| ( | ( | ||
| 2 | ( | 2 | |
| * | ( * | Push as * has higher priority | |
| 5 | ( * | 2 5 | |
| - | ( * | Since '-' has less priority, we do 2 * 5 = 10 | |
| ( - | 10 | We push 10 and then push '-' | |
| 1 | ( - | 10 1 | |
| * | ( - * | 10 1 | Push * as it has higher priority |
| 2 | ( - * | 10 1 2 | |
| ) | ( - | 10 2 | Perform 1 * 2 = 2 and push it |
| ( | 8 | Pop - and 10 - 2 = 8 and push, Pop ( | |
| / | / | 8 | |
| ( | / ( | 8 | |
| 11 | / ( | 8 11 | |
| - | / ( - | 8 11 | |
| 9 | / ( - | 8 11 9 | |
| ) | / | 8 2 | Perform 11 - 9 = 2 and push it |
| New line | 4 | Perform 8 / 2 = 4 and push it | |
| 4 | Print the output, which is 4 |
C Program
int main (int argc, char *argv[]) { struct ch *charactop; struct integer *integertop; char rd, op; int i = 0, op1, op2; charactop = cclearstack(); integertop = iclearstack(); while(1) { rd = argv[1][i++]; switch(rd) { case '+': case '-': case '/': case '*': while ((charactop->data != '(') && (!cemptystack(charactop))) { if(priority(rd) > (priority(charactop->data)) break; else { charactop = cpop(charactop, &op); integertop = ipop(integertop, &op2); integertop = ipop(integertop, &op1); integertop = ipush(integertop, eval(op, op1, op2); } } charactop = cpush(charactop, rd); break; case '(': charactop = cpush(charactop, rd); break; case ')': integertop = ipop (integertop, &op2); integertop = ipop (integertop, &op1); charactop = cpop (charactop, &op); while(op != '(') { integertop = ipush (integertop, eval(op, op1, op2); charactop = cpop (charactop, &op); if (op != '(') { integertop = ipop(integertop, &op2); integertop = ipop(integertop, &op1); } } break; case '\0': while (!= cemptystack(charactop)) { charactop = cpop(charactop, &op); integertop = ipop(integertop, &op2); integertop = ipop(integertop, &op1); integertop = ipush(integertop, eval(op, op1, op2); } integertop = ipop(integertop, &op1); printf("\n The final solution is: %d", op1); return 0; default: integertop = ipush(integertop, read - '0'); } } } int eval(char op, int op1, int op2) { switch (op) { case '+': return op1 + op2; case '-': return op1 - op2; case '/': return op1 / op2; case '*': return op1 * op2; } } int priority (char op) { switch(op) { case '^': case '$': return 3; case '*': case '/': return 2; case '+': case '-': return 1; } }
Output of the program:
Input entered at the command line: (2 * 5 - 1 * 2) / (11 - 9)
Output: 4 [7]
Evaluation of Prefix Expression
Input: / - 2 5 * 1 2 - 11 9
Output:' 4
Analysis: There are three types of input characters
* Numbers
* Operators
* New line character (\n)
Data structure requirement: A character stack and an integer stack
Algorithm:
1. Read one character input at a time and keep pushing it into the character stack until the new
line character is reached
2. Perform pop from the character stack. If the stack is empty, go to step (3)
Number (2.1) Push in to the integer stack and then go to step (1)
Operator (2.2) Assign the operator to op
Pop a number from integer stack and assign it to op1
Pop another number from integer stack
and assign it to op2
Calculate op1 op op2 and push the output into the integer
stack. Go to step (2)
3. Pop the result from the integer stack and display the result
Result: The evaluation of prefix expression is printed as follows:
Input String: / - * 2 5 * 1 2 - 11 9
| Input Symbol | Character Stack (from bottom to top) | Integer Stack (from bottom to top) | Operation performed |
|---|---|---|---|
| / | / | ||
| - | / | ||
| * | / - * | ||
| 2 | / - * 2 | ||
| 5 | / - * 2 5 | ||
| * | / - * 2 5 * | ||
| 1 | / - * 2 5 * 1 | ||
| 2 | / - * 2 5 * 1 2 | ||
| - | / - * 2 5 * 1 2 - | ||
| 11 | / - * 2 5 * 1 2 - 11 | ||
| 9 | / - * 2 5 * 1 2 - 11 9 | ||
| \n | / - * 2 5 * 1 2 - 11 | 7 | |
| / - * 2 5 * 1 2 - | 9 11 | ||
| / - * 2 5 * 1 2 | 2 | 11 - 9 = 2 | |
| / - * 2 5 * 1 | 2 2 | ||
| / - * 2 5 * | 2 2 1 | ||
| / - * 2 5 | 2 2 | 1 * 2 = 2 | |
| / - * 2 | 2 2 5 | ||
| / - * | 2 2 5 2 | ||
| / - | 2 2 10 | 5 * 2 = 10 | |
| / | 2 8 | 10 - 2 = 8 | |
| Stack is empty | 4 | 8 / 2 = 4 | |
| Stack is empty | Print 4 |
C Program
int main (int argc, char *argv[]) { struct ch *charactop = NULL; struct integer *integertop = NULL; char rd, op; int i = 0, op1, op2; charactop = cclearstack(); integertop = iclearstack(); rd = argv[1][i]; while(rd != '\0') { charactop = cpush(charactop, rd); rd = argv[1][i++]; } while(!emptystack(charactop)) { charactop = cpop(charactop, rd); switch(rd) { case '+': case '-': case '/': case '*': op = rd; integertop = ipop(integertop, &op2); integertop = ipop(integertop, &op1); integertop = ipush(integertop, eval(op, op1, op2)); break; default: integertop = ipush(integertop, read - '0'); } } } int eval(char op, int op1, int op2) { switch (op) { case '+': return op1 + op2; case '-': return op1 - op2; case '/': return op1 / op2; case '*': return op1 * op2; } } int priority (char op) { switch(op) { case '^': case '$': return 3; case '*': case '/': return 2; case '+': case '-': return 1; } }
Output of the program:
Input entered at the command line: / - * 2 5 * 1 2 - 11 9
Output: 4 [7]
Evaluation of postfix expression
The calculation: 1 + 2 * 4 + 3 can be written down like this in postfix notation with the advantage of no precedence rules and parentheses needed:
1 2 4 * + 3 +
The expression is evaluated from the left to right using a stack:
- when encountering an operand: push it
- when encountering an operator: pop two operands, evaluate the result and push it.
Like the following way (the Stack is displayed after Operation has taken place):
| Input | Operation | Stack (after op) |
|---|---|---|
| 1 | Push operand | 1 |
| 2 | Push operand | 2, 1 |
| 4 | Push operand | 4, 2, 1 |
| * | Multiply | 8, 1 |
| + | Add | 9 |
| 3 | Push operand | 3, 9 |
| + | Add | 12 |
The final result, 12, lies on the top of the stack at the end of the calculation.
Example in C
#include<stdio.h> int main() { int a[100], i; printf("To pop enter -1\n"); for(i = 0;;) { printf("Push "); scanf("%d", &a[i]); if(a[i] == -1) { if(i == 0) { printf("Underflow\n"); } else { printf("pop = %d\n", a[--i]); } } else { i++; } } }
Evaluation of postfix expression (Pascal)
This is an implementation in Pascal, using marked sequential file as data archives.
{ programmer : clx321 file : stack.pas unit : Pstack.tpu } program TestStack; {this program uses ADT of Stack, I will assume that the unit of ADT of Stack has already existed} uses PStack; {ADT of STACK} {dictionary} const mark = '.'; var data : stack; f : text; cc : char; ccInt, cc1, cc2 : integer; {functions} IsOperand (cc : char) : boolean; {JUST Prototype} {return TRUE if cc is operand} ChrToInt (cc : char) : integer; {JUST Prototype} {change char to integer} Operator (cc1, cc2 : integer) : integer; {JUST Prototype} {operate two operands} {algorithms} begin assign (f, cc); reset (f); read (f, cc); {first elmt} if (cc = mark) then begin writeln ('empty archives !'); end else begin repeat if (IsOperand (cc)) then begin ccInt := ChrToInt (cc); push (ccInt, data); end else begin pop (cc1, data); pop (cc2, data); push (data, Operator (cc2, cc1)); end; read (f, cc); {next elmt} until (cc = mark); end; close (f); end
}
Runtime memory management
A number of programming languages are stack-oriented, meaning they define most basic operations (adding two numbers, printing a character) as taking their arguments from the stack, and placing any return values back on the stack. For example, PostScript has a return stack and an operand stack, and also has a graphics state stack and a dictionary stack.
Forth uses two stacks, one for argument passing and one for subroutine return addresses. The use of a return stack is extremely commonplace, but the somewhat unusual use of an argument stack for a human-readable programming language is the reason Forth is referred to as a stack-based language.
Many virtual machines are also stack-oriented, including the p-code machine and the Java Virtual Machine.
Almost all calling conventions -- computer runtime memory environments—use a special stack (the "call stack") to hold information about procedure/function calling and nesting in order to switch to the context of the called function and restore to the caller function when the calling finishes. The functions follow a runtime protocol between caller and callee to save arguments and return value on the stack. Stacks are an important way of supporting nested or recursive function calls. This type of stack is used implicitly by the compiler to support CALL and RETURN statements (or their equivalents) and is not manipulated directly by the programmer.
Some programming languages use the stack to store data that is local to a procedure. Space for local data items is allocated from the stack when the procedure is entered, and is deallocated when the procedure exits. The C programming language is typically implemented in this way. Using the same stack for both data and procedure calls has important security implications (see below) of which a programmer must be aware in order to avoid introducing serious security bugs into a program.
Security
Some computing environments use stacks in ways that may make them vulnerable to security breaches and attacks. Programmers working in such environments must take special care to avoid the pitfalls of these implementations.
For example, some programming languages use a common stack to store both data local to a called procedure and the linking information that allows the procedure to return to its caller. This means that the program moves data into and out of the same stack that contains critical return addresses for the procedure calls. If data is moved to the wrong location on the stack, or an oversized data item is moved to a stack location that is not large enough to contain it, return information for procedure calls may be corrupted, causing the program to fail.
Malicious parties may attempt a stack smashing attack that takes advantage of this type of implementation by providing oversized data input to a program that does not check the length of input. Such a program may copy the data in its entirety to a location on the stack, and in so doing it may change the return addresses for procedures that have called it. An attacker can experiment to find a specific type of data that can be provided to such a program such that the return address of the current procedure is reset to point to an area within the stack itself (and within the data provided by the attacker), which in turn contains instructions that carry out unauthorized operations.
This type of attack is a variation on the buffer overflow attack and is an extremely frequent source of security breaches in software, mainly because some of the most popular programming languages (such as C) use a shared stack for both data and procedure calls, and do not verify the length of data items. Frequently programmers do not write code to verify the size of data items, either, and when an oversized or undersized data item is copied to the stack, a security breach may occur.
See also
References
- ^ http://www.cprogramming.com/tutorial/computersciencetheory/stack.html cprogramming.com
- ^ Dr. Friedrich Ludwig Bauer and Dr. Klaus Samelson (30. März 1957) (in german). Verfahren zur automatischen Verarbeitung von kodierten Daten und Rechenmaschine zur Ausübung des Verfahrens.. Deutsches Patentamt. http://v3.espacenet.com/origdoc?DB=EPODOC&IDX=DE1094019&F=0&QPN=DE1094019. Retrieved 2010-10-01.
- ^ Jones: "Systematic Software Development Using VDM"
- ^ Horowitz, Ellis: "Fundamentals of Data Structures in Pascal", page 67. Computer Science Press, 1984
- ^ a b Dromey, R.G. How to Solve it by Computer. Prentice Hall of India.
- ^ a b Data structures, Algorithms and Applications in C++ by Sartaj Sahni
- ^ a b c Gopal, Arpita. Magnifying Data Structures. PHI.
- Stack implementation on goodsoft.org.ua
- A Templated Stack Data Structure Example at AssignmentExpert.com
Further reading
- Donald Knuth. The Art of Computer Programming, Volume 1: Fundamental Algorithms, Third Edition.Addison-Wesley, 1997. ISBN 0-201-89683-4. Section 2.2.1: Stacks, Queues, and Deques, pp. 238–243.
- Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Section 10.1: Stacks and queues, pp. 200–204.
External links
- Stack Machines - the new wave
- Bounding stack depth
- Libsafe - Protecting Critical Elements of Stacks
- VBScript implementation of stack, queue, deque, and Red-Black Tree
- Stack Size Analysis for Interrupt-driven Programs (322 KB)
- Pointers to stack visualizations
- Paul E. Black, Bounded stack at the NIST Dictionary of Algorithms and Data Structures.
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