Question: I'm trying to extract the byte value of a double from the computer's memory. The reason for this is accuracy, as the final value that is represented to the user has some rounding performed.

Ideally, I want to be able to extract the sign, exponent and mantissa from the number (IEEE 754 standard)

From what I understand, casting to an unsigned char is the way to go. I have the following code snippet (stolen elsewhere), but I don't trust the results - there's no byte output for any integer values:

```
double d = 2;
unsigned char *p = (unsigned char*)&d;

```

Can anyone guide me in the right direction in order to extract byte representation of double numbers accurately, or give any advice/comments on how to proceed?

Answer: 
You're doing it right. You can treat `p` as an array `unsigned char[sizeof(double)]` now.

For example:

```
for (int i = 0; i != sizeof(double); ++i)  printf("%02X ", p[i]);

```

For `d = 0.125` this prints `00 00 00 00 00 00 C0 3F`. Reversing endianness and decomposing into parts, this is:

```none
00111111 11000000 0 0 0 0 0 0
   3F       C0    0 0 0 0 0 0

0   01111111100   0...<52 times>...0
S     Exponent         Mantissa
+     1023 - 3           1.0

```

The exponent field's value is 1020, which gives an exponent of −3 after correcting by the bias of 1023, and the mantissa is 1.0, after including the implicit leading 1. So the value is 2−3 = 1/8.
If your motivation is only accuracy, use `printf("%a\n", d);`

It displays the same information as going to the binary representation would, but it displays it in a format that's almost human-readable. Here (that is, for `d=2;`), it displays `0x1p+1`. The number after `0x` is the mantissa, normalized between 1 and 2, and in hexadecimal. The number after `p` is the exponent. It is displayed in decimal but it represents a power of two.