int main()
{
float f = 12.2;
char *p1;
p1 = (char *)&f;
printf (\"%d\", *p1);
}
This outputs 51.
The question is:
what happens when float is typecasted to char pointer
The precise answer is:
Undefined Behavior.
There is slight deviation to your question I am off the topic,I usually use in such cases the snpritf Example:to fisrt format and then you can play the formatted buffer.
EDIT: According to the new markup you really truncate the float representation to its least significant byte (on little-endian machines)
Mostly what EFraim says, except that you did cast to char*, only the stackoverflow markup was wrong.
So you get the least significant byte of f's internal representation (in IEEE-754).
Assuming the other issues mentioned are fixed, you are getting the integer version of whatever the bit pattern happens to be for that float. Floats have a fairly complicated encoding, so it will be nothing obviously related to the number you put in the float.
You can cast a float*
to a char*
just fine, it's the using of such a beast that may be problematic.
When you de-reference it, you'll simply get the char
representation of the first part (but see below to understand what this really means, it's not as clear as you may think) of the float.
If you're talking about IEE754 floats, 12.2 in IEEE754 float is (abcd are the octets):
S EEEEEEEE MMMMMMMMMMMMMMMMMMMMMMM (sign, exponent, mantissa).
0 10000010 10000110011001100110011
a aaaaaaab bbbbbbbccccccccffffdffffffffd
The 00110011
at the end is the 51 (0x33) that you're seeing. The reason you're seeing the last bit of the float is because it's stored like this in memory (in a little-endian architecture):
00110011 00110011 01000011 01000001
ffffdffffffffd cccccccc bbbbbbbb aaaaaaaa
which means that the char*
cast of the float*
will point at the ffffdffffffffd
part.
On big-endian architectures, you would get the aaaaaaaa
bit, 01000001
, or 65 (0x41).