I am trying to understand union and here is a example code I wrote,

```
#include <stdio.h>
union god{
int body;
float spirit;
};
union god human;
int main(void)
{
float malice = 5.4;
human.spirit = malice;
printf("%ld %ld\n", sizeof(float), sizeof(int));
printf("%d %f\n", human.body, human.spirit);
return 0;
}
```

I expected 5 5.4 as output, because same memory is used for int and float types, I should be able to store both in same place.

However printing those values doesn't show my guess.

```
4 4
1085066445 5.400000
```

Output does not show both values as **5 5.4**.

int and float both have same size, so same memory is used for holding them.

1.But why the output is like this?

2.Am I missing some float, union, printf related basics?

c answered 3 months ago liliscent #1

For most of the implementation, `int`

and `float`

use essentially different representation in terms of bit. If you interpret the same bit sequence in `float`

as `int`

, the result is meaningless. You can't expect a `float`

representation of `5.4`

is also a `int`

representation of `5`

.

See here some typical implementation of `floating point`

, and here a typical representation of `int`

.

answered 3 months ago Eric Postpischil #2

The value in a float is stored in a special format. In the most common format, it is encoded with one bit for a sign, eight bits for an exponent, and 23 bits for the significand (a fraction). For 5.4, the sign bit is 0 (for positive), the exponent is 2 and is encoded as 129 (127 is added; the exponent is stored with a bias), and the fraction is 1.35000002384185791015625, using the bits 0x2ccccd. (Those are the fraction minus one, then multiplied by 2^{23}.) Put together, the sign bit, the exponent bits, and the significand bits form 0x40accccd. In decimal, that is 1085066445.

So, when you store 5.4 in the `float`

member of the union, the bits 0x40accccd are written to the storage. When you read the `int`

member, those bits are interpreted as 1085066445.