Printing floating point using integer type specifier

Question

I have the below C code.

int main (void)
{
    printf("%d\n",5/(float)4.7);
    printf("Size of int = %d",sizeof(int));
    return(1);
}

But the result is 1610612736 I was expecting the result to be 1 since we are tying to print an integer. Why this is the case? How float value is getting converted to a large value in printf statement? I am using codeblock + mingw

@MartinJames the question is not a duplicate. The OP didn't realize that `5/(float)4.7` leads to a float value. But he wants to print an integer (and correctly has `%d` format specifier). — Paolo, Feb 23 '16 at 15:57
[What happens when I use the wrong format specifier?](http://stackoverflow.com/q/16864552/995714), [What can happen if printf is called with a wrong format string?](http://stackoverflow.com/q/14504148/995714) — phuclv, Feb 23 '16 at 16:08
The `printf("Size of int = %d",sizeof(int));` also has a mismatch between the format specifier and the type of the value it corresponds to. It should be `printf("Size of int = %zd",sizeof(int));` or `printf("Size of int = %d",(int)sizeof(int));`, although it would be better to use `printf("Size of int = %zu",sizeof(int));` or `printf("Size of int = %u",(unsigned int)sizeof(int));`. — Ian Abbott, Feb 23 '16 at 18:00

Paolo · Accepted Answer · 2016-02-23T16:00:36.427

4

as you're trying to print an integer you have to cast the result to int

printf("%d\n",(int)(5/(float)4.7));

otherwise 5/(float)4.7 is a float

You see 1610612736 because the float value is "interpreted" as and int leading to and odd result.

edited Feb 23 '16 at 16:00

answered Feb 23 '16 at 15:48

Paolo

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Printing floating point using integer type specifier

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