CODEWITHSUNDEEP
cfloating-pointnumber-formatting
James Raitsev
James Raitsev

Reputation: 96481

Integer representation as float, clarification needed

Take 

int x = 5;
float y = x;

//"I know it's a float .. you know it's a float .. but take it's address
// and pretend you're looking at an integer and then dereference it"

printf("%d\n", *(int*)&y); //1084227584

Why am i seeing this number?

Can be represented as:

[sign bit]                              - 0
[8 bits worth of exp] - 129 (129-127=2) - 1000|0001
[23 bits of .xxxxxxx] - 25              - 1100|1

Put together, i have 

[sign bit][8 bits worth of exp][23 bits worth of .xxx]
0         10000001             11001000000000 //2126336

What am i missing please?

Upvotes: 3

Views: 193

Answers (2)

Brett Hale
Brett Hale

Reputation: 22348

Others have pointed out it's not portable... but you know this already, and you've specified 64-bit OS X. Basically, you have the mantissa wrong. 1.25 is represented with an implicit leading bit for 1.0. The first explicit bit of the mantissa represents 0.5 and the second bit 0.25. So the mantissa is actually: 01000000000000000000000.

The sign bit 0, and biased exponent 10000001, followed by the mantissa gives: 0x40a00000 which is 1084227584 decimal.

Upvotes: 3

John3136
John3136

Reputation: 29266

Why am i seeing this number? Because you are printing a float as an int.

I know, I know, you clearly already know this, but the bottom line is the behaviour is undefined. On your system are ints and floats the same size? Have you looked up the standards your compiler uses to store floating points?

Upvotes: 0

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