What happens when float pointer is typecasted to char pointer?

Question

int main()
{
    float f = 12.2;
    char *p1;
    p1 = (char *)&f;
    printf ("%d", *p1);
}

This outputs 51.

Answer 1

1. You cast to (char) instead of (char*).
2. You print it out as integer.
3. Thus you get the least significant byte of f's address.

EDIT: According to the new markup you really truncate the float representation to its least significant byte (on little-endian machines)

Answer 2

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).

Answer 3

You're not casting a float to a char* (as stated in your original question), you're actually casting a float* to a char*.

That means, 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 bbbbbbbccccccccdddddddd

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 beacuase it's stored like this in memory (in a little-endian architecture):

00110011 00110011 01000011 01000001
dddddddd cccccccc bbbbbbbb aaaaaaaa

which means that the char* cast of the float* will get the dddddddd part.

On big-endian architectures, you would get the aaaaaaaa bit, 01000001, or 65 (0x41).

Answer 4

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.

Answer 5

The question is:

what happens when float is typecasted to char pointer

The precise answer is:

Undefined Behavior.