Casting to (int *) of (&num) when num is float in C

Question

Why is the ouput of the following code : " i is: 1075838976"?

#include <stdio.h>

int main(){
  int i = 2;
  float num = 2.5;

  i = *((int *)& num);
  printf("i is: %d\n", i);
}

Isn't it equivalent to :

#include <stdio.h>

int main(){
  int i = 2;
  float num = 2.5;

  i = num;
  printf("i is: %d\n", i);
}

Which outputs : "i is 2"? Thanks.

No, it's (obviously) not equivalent. The one reinterprets the bit-pattern (invoking undefined behaviour on the way), the other does a value-conversion. — Daniel Fischer, Jul 23 '13 at 15:23
Highly related, possibly even a duplicate: http://stackoverflow.com/q/3452439/319403 — cHao, Jul 23 '13 at 15:33

score 11 · Accepted Answer · answered Jul 23 '13 at 15:25

No, it's not equivalent at all.

This:

i = num;

converts the value of num (which is 2.5) from float to int. The conversion truncates the value to 2.

This:

i = *((int *)& num);

takes a pointer to the float object num, converts it to int*, and dereferences the resulting pointer.

If you're "lucky", this takes the bits making up the representation of num, pretends that they're the representation of an int, and gives you that results.

If you're not "lucky", then int and float may be of different sizes, a float object might be incorrectly aligned to be treated as an int object, or the result could even be a "trap representation" (though the latter is rare).

(I put "lucky" in quotation marks because, really, the best thing this code can do is blow up in your face, which immediately lets you know you're doing something questionable. The behavior is undefined, which means it's an error, but the implementation isn't required to warn you about it, either at compile time or at run time.)

The particular value you're getting, 1075838976, can be represented in hexadecimal as 0x40200000. If you look up the way float values are represented on your system, you can probably figure out how that bit pattern (0100 0000 0010 0000 0000 0000 0000 0000) makes up the the sign, mantissa, and exponent values that represent the value 2.5.

score 3 · Answer 2 · answered Jul 23 '13 at 15:30

In C "cast" is sort of "overloaded" and does two entirely different things:

-- Cause code to be generated to convert a "scalar" value between, say, int and float.

-- Coerce a pointer to a different type. This does not actually generate any code and does not convert anything.

((int *)& num) is coercing a pointer from float* to int*, and does not do any data conversion.

score 0 · Answer 3 · answered Jul 23 '13 at 15:24

0

You are telling the machine to interpret the binary number explicitly as an int, without going through a cast.

answered Jul 23 '13 at 15:24

Daniel A. White

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Casting to (int *) of (&num) when num is float in C

3 Answers3