打印出浮子的内六角

Question

So I thought that I could print the internal hex of a float like this:

const auto foo = 13.0F;
const auto bar = reinterpret_cast<const unsigned char*>(&foo);

printf("0x%02hhX%02hhX%02hhX%02hhX\n", bar[0], bar[1], bar[2], bar[3]);

This outputs:

0x00005041

But when I look at the debugger, the hex that it is reporting for foo is:

0x003EF830

Can someone help me understand why this didn't work and what I'd need to do to make it work?

Answer 1

The hex value for the floating point value 13.0f is:

0x41500000

Confirmation:

Take 0x41500000 which is in binary:

0100,0001,0101,0000,0000,0000,0000,0000

which split into 1 bit sign, 8 bit exponent and 23bit mantissa is:

0 : 10000010 : 10100000000000000000000

And we can interpret this using the standard floating evaluation of:

-1^sx 1.<mantissa> x 2^(exponent-127)

as

(-1)^0 x 1.101 x 2^(10000010-1111111)

which is

1.101bx (10b)^11b

which is

1.625 x 8

which is

13.0f

Which means your output is correct (assuming a little endian format)

How you are reading it in the debugger is incorrect

Answer 2

You almost got it. I'm not sure what the debugger is displaying but it's not your float. Maybe it's the address of it?

The actual float value is 0x41500000 . You can check it here: IEEE-754 Floating Point Converter . If the link doesn't work you'll have to find the online floating point analyzer/description by yourself.

You did it the right way but you forgot that Intel x86 CPUs (I suppose that your CPU is Intel x86) are little endian . That means that higher significance bytes are on higher memory addresses. So you should print out bytes in the reverse order than you're printing them, like this:

printf("0x%02hhX%02hhX%02hhX%02hhX\n", bar[3], bar[2], bar[1], bar[0]);