What is the output of the following program? #include <stdio.h> int funcf (int…

2004

What is the output of the following program? 

#include <stdio.h>
int funcf (int x);
int funcg (int y);

main()
{
    int x = 5, y = 10, count;
    for (count = 1; count <= 2; ++count)
    {
        y += funcf(x) + funcg(x);
        printf ("%d ", y);
    }
}

funcf(int x)
{
    int y;
    y = funcg(x);
    return (y);
}

funcg(int x)
{
    static int y = 10;
    y += 1;
    return (y+x);
}

Answer: A. 43 80 — Key idea: funcg has a static variable that persists across calls and is incremented each time funcg runs. funcf simply calls funcg and returns its result, so…

  1. A.

    43 80

  2. B.

    42 74

  3. C.

    33 37

  4. D.

    32 32

Attempted by 33 students.

Show answer & explanation

Correct answer: A

Key idea: funcg has a static variable that persists across calls and is incremented each time funcg runs. funcf simply calls funcg and returns its result, so each loop iteration calls funcg twice.

  • Initial values: main's x = 5, main's y = 10. static y inside funcg is initialized to 10.

  • First loop iteration:

    • funcf(x) calls funcg: static y increments 10 -> 11, funcg returns 11 + 5 = 16.

    • The direct call funcg(x): static y increments 11 -> 12, funcg returns 12 + 5 = 17.

    • Sum added to main's y = 16 + 17 = 33, so main's y becomes 10 + 33 = 43. Program prints 43.

  • Second loop iteration:

    • funcf(x) calls funcg: static y increments 12 -> 13, funcg returns 13 + 5 = 18.

    • The direct call funcg(x): static y increments 13 -> 14, funcg returns 14 + 5 = 19.

    • Sum added to main's y = 18 + 19 = 37, so main's y becomes 43 + 37 = 80. Program prints 80.

Final output: 43 80

Note: The result depends on the order in which the two function calls are executed because funcg modifies a static variable. The trace above shows the sequence where funcf (which calls funcg) runs before the separate funcg call, producing the shown output.

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