EOF output using% f

Question

#include<stdio.h>
int main()
{
    printf("%d",EOF);
}

generates -1 which is perfectly fine, but

#include<stdio.h>
int main()
{
    printf("%f",EOF);
}

produces 0.000. How can someone explain this when the expected result is -1.000?

+3

Gaurav jain 09 June 15 at 12:19

2 answers

EOF

is of type int

( signed

). You must not use the wrong format specifier for printing int

, or it will cause undefined behavior.

+5

haccks 09 June 15 at 12:21

Natasha Dutta · Accepted Answer · 2015-06-09T12:19:52+0000

Using an invalid format specifier for any particular argument in printf()

causes undefined behavior .

EOF

has a type int

. You can only use %d

for variable type int

.

FWIW, if you want floating point representation int

you need a cast

variable (but I personally recommend avoiding this)

printf("%f",(float)EOF);