Question

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What is printf’s behaviour when printing an int as float?

int main()
{
 int x=4;
 int y=987634;
 printf("%f %f",x,y);
}

On compiling this code i get an output as 0.000000 0.000000. Shouldn’t there be a type promotion of x and y to floating point numbers? Shouldn’t the O/P be 4.000000 and 987634.000000?
Can anyone help me with this. Thanx in Advance.

Answer 1

Conversions happen to arguments to functions with a prototype which includes the specific parameters. The prototype for printf() does not include the specific parameters after the first one

int printf(const char *format, ...);

so, no arguments after the 1st one get automatically converted except as defined by “default argument conversions” (basically any integer type with a rank lower than int to int and any floating-point type with a rank lower than double to double (thank you, Pascal Cuoq)). You need to convert them explicitly yourself with a cast operation

    printf("%f %f\n", (double)x, (double)y);

Ohhh … and you really, really, really should include the header that has the prototype in question (under penalty of Undefined Behaviour)

#include <stdio.h>