why is the output of the following code coming to be zero?

Question

main()
{
    float a=10;
    float c;
    float b=5.5;
    c=a+b;
    printf("%d",c);
}

The output of the above code is zero.Why is that? I am sorry if that is some really simple C concept, I am kind of a beginner.

It's undefined behaviour. You're lying to the compiler. (I once answered a [similar question](http://stackoverflow.com/a/7295097/596781).) — Kerrek SB, Aug 25 '13 at 17:42

C. K. Young · Accepted Answer · 2013-08-25T17:45:48.093

5

You need to use %f (or %e or %g, depending on your preferred format) instead of %d for floating-point numbers. Indeed, using %d for non-integers is "undefined behaviour".

printf("%f", c);

Alternatively, if you're trying to round the floating-point to an integer, you must cast it first.

printf("%d", (int) c);

edited Aug 25 '13 at 17:45

answered Aug 25 '13 at 17:40

C. K. Young

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Or in the unlikely case he's trying to round the number to an int: printf("%d", (int)c); – dcaswell Aug 25 '13 at 17:43
@user814064 Right, I'll add that to the answer. – C. K. Young Aug 25 '13 at 17:45

score 0 · Answer 2 · answered Aug 25 '13 at 17:48

0

See a+b=c will result c to become 15.5 when you try to printf c as a decimal "%d" it will be "undefined behaviour" as Chris said. If you do printf("%d",(int)c); your out put will become 15 and if you printf("%f",c); you will get 15.5

answered Aug 25 '13 at 17:48

Nxtmind

57
8

why is the output of the following code coming to be zero?

2 Answers2