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Explain the evaluation of expressions of stacks in C language
Stack is a linear data structure, where data is inserted and removed only at one end.
Algorithms
Given below is an algorithm for Push ( ) −
- Check for stack overflow.
if (top = = n-1)
printf("stack over flow");- Otherwise, insert an element into the stack.
top ++ a[top] = item
Given below is an algorithm for Pop ( ) −
- Check for stack underflow.
if ( top = = -1) printf( "stack under flow");
Otherwise, delete an element from the stack.
item = a[top] top --
Given below is an algorithm for Display ( ) −
if (top == -1)
printf ("stack is empty");Otherwise, follow the below mentioned algorithm.
for (i=0; i<top; i++)
printf ("%d", a[i]);Application of stacks
Let us understand the conversions of expressions of stacks in C language.
Expression − It is a legal combination of operands and operations.
Types of expressions
There are three types of expressions in C language on which the conversions and valuation can be carried out. They are explained below −
Infix expression − Operator is in between the operands. For example, A+B
Prefix expression − Operator is before the operands. For example, +AB
Postfix expression − Operator is after the operands. For example, AB+
Evaluation of postfix expression
Algorithm
Scan the input string from left to right.
For each input symbol,
If it is a digit then, push it on to the stack.
If it is an operator then, pop out the top most two contents from the stack and apply the operator on them. Later on, push the result on to stack.
If the input symbol is ‘\0’, empty the stack.
Program
Following is the C program for an evaluation of postfix expression −
#include<stdio.h>
int top = -1, stack [100];
main ( ){
char a[50], ch;
int i,op1,op2,res,x;
void push (int);
int pop( );
int eval (char, int, int);
printf("enter a postfix expression:");
gets (a);
for(i=0; a[i]!='\0'; i++){
ch = a[i];
if (ch>='0' && ch<='9')
push('0');
else{
op2 = pop ( );
op1 = pop ( );
res = eval (ch, op1, op2);
push (res);
}
}
x = pop ( );
printf("evaluated value = %d", x);
getch ( );
}
void push (int n){
top++;
stack [top] = n;
}
int pop ( ){
int res ;
res = stack [top];
top--;
return res;
}
int eval (char ch, int op1, int op2){
switch (ch){
case '+' : return (op1+op2);
case '-' : return (op1-op2);
case '*' : return (op1*op2);
case '/' : return (op1/op2);
}
}Output
When the above program is executed, it produces the following result −
Run 1: enter a postfix expression:45+ evaluated value = 9 Run 2: enter a postfix expression: 3 5 2 * + evaluated value = 13
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