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Home/ Questions/Q 8274007
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T07:37:07+00:00 2026-06-08T07:37:07+00:00

I have the following code, and the only thing I can physically see it

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I have the following code, and the only thing I can physically see it evaluating is line 18 which is the call to printf(). It doesn’t go any further.

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

int main (void) {

    int cylNum;
    double disp, pi, stroke, radius;
        pi = 3.14159;

    printf("Welcome to the Engine Displacement Calculator!\n");

    cylNum = scanf("Enter number of cylinders (then press enter): \n");

    stroke = scanf("Enter stroke: \n");

    radius = scanf("Enter radius: \n");

    disp = radius * radius * pi * stroke * cylNum;

    printf("Displacement is: %f", disp);

    getchar();
    printf("Press any key to exit!");
    return 0;
}
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  1. Editorial Team
    Editorial Team
    2026-06-08T07:37:08+00:00Added an answer on June 8, 2026 at 7:37 am

    the variables you’re trying to read should be parameters to “scanf()”, not the result of scanf():

    printf("Enter number of cylinders (then press enter): ");
    scanf("%d", &cylNum);
    ...
    
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