Casting to (int *) of (&num) when num is float in C
c, casting, pointers
Solution
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`.
Problem
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.