Expression in C is not giving required output It keeps giving Zero [duplicate]

Question

void main()
{
    int a = 5;
    int b = 10;
    int c;
    c = (a / b) * 100;

    //5/2=0.5
    //0.5*100=50

    printf("C value is : %d", c);
}

Why the output is Zero when the output should be 50. This is a question which I have been not caring for a long time. I wanna know what's happening behind that expression.

Answer 1

The expression a / b has integers for both operands, so integer division is performed which truncates any fractional portion of the result.

You'll need to cast one of the operands to a floating point type so that floating point division is carried out.

c = ((double)a / b) * 100;

Also, while in this particular case you should get the exact result you want, floating point arithmetic is not exact and rounding errors may occur.

Answer 2

Because the int value of 5 / 10 is 0 (0.5 gets rounded down). You need to use doubles or floats for this

void main()
{
    float a = 5;
    float b = 10;
    float c;
    c = (a / b) * 100;

    //5/2=0.5
    //0.5*100=50

    printf("C value is : %f", c);
}

outputs 50.0

Answer 3

following the comments above, float should be the best answer.

void main()
{
    int a = 5;
    int b = 10;
    int c;
    c = a * 100 / b;

    printf("C value is : %d", c);
}