How does this program work?

Question

#include <stdio.h>  int main() {     float a = 1234.5f;     printf("%d\n", a);     return 0; } 

It displays a 0!! How is that possible? What is the reasoning?

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I have deliberately put a %d in the printf statement to study the behaviour of printf.

Answer 1

That's because %d expects an int but you've provided a float.

Use %e/%f/%g to print the float.

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On why 0 is printed: The floating point number is converted to double before sending to printf. The number 1234.5 in double representation in little endian is

00 00 00 00  00 4A 93 40 

A %d consumes a 32-bit integer, so a zero is printed. (As a test, you could printf("%d, %d\n", 1234.5f); You could get on output 0, 1083394560.)

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As for why the float is converted to double, as the prototype of printf is int printf(const char*, ...), from 6.5.2.2/7,

The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.

and from 6.5.2.2/6,

If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions.

(Thanks Alok for finding this out.)