Consider the following program: #include <stdio.h> #include <stdlib.h>…

2007

Consider the following program: 

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

#define EOF -1

#define MAX_STACK_SIZE 100
int s[MAX_STACK_SIZE];
int top = -1;

void push(int value) { if (top < MAX_STACK_SIZE - 1) s[++top] = value; } /* push the argument on the stack */
int pop(void) { if (top == -1) return EOF; return s[top--]; }  /* pop the top of the stack */
void flagError() { fprintf(stderr, "Error!\n"); exit(1); }

int main() {
    int c, m, n, r;
    while ((c = getchar()) != EOF) {
        if (isdigit(c))
            push(c - '0');
        else if ((c == '+') || (c == '*')) {
            m = pop();
            n = pop();
            r = (c == '+') ? n + m : n * m;
            push(r);
        } else if (c != ' ')
            flagError();
    }
    printf("%d\n", pop());
}

What is the output of the program for the following input ? 5 2 * 3 3 2 + * +

Answer: B. 25 — Solution: evaluate the postfix expression using a stack: push digits, and on an operator pop the top two values (first popped is the right operand), apply the…

  1. A.

    15

  2. B.

    25

  3. C.

    30

  4. D.

    150

Attempted by 298 students.

Show answer & explanation

Correct answer: B

Solution: evaluate the postfix expression using a stack: push digits, and on an operator pop the top two values (first popped is the right operand), apply the operator, and push the result.

  • Token "5": push 5 → stack = [5]

  • Token "2": push 2 → stack = [5, 2]

  • Token "*": pop 2 and 5, compute 5 * 2 = 10, push 10 → stack = [10]

  • Token "3": push 3 → stack = [10, 3]

  • Token "3": push 3 → stack = [10, 3, 3]

  • Token "2": push 2 → stack = [10, 3, 3, 2]

  • Token "+": pop 2 and 3, compute 3 + 2 = 5, push 5 → stack = [10, 3, 5]

  • Token "*": pop 5 and 3, compute 3 * 5 = 15, push 15 → stack = [10, 15]

  • Token "+": pop 15 and 10, compute 10 + 15 = 25, push 25 → stack = [25]

Final result: the stack contains 25, so the program prints 25.

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