cast float to unsigned int in C with gcc

Question

I am using gcc to test some simple casts between float to unsigned int.

The following piece of code gives the result 0.

const float maxFloat = 4294967295.0;
unsigned int a = (unsigned int) maxFloat;
printf("%u\n", a);

0 is printed (which I belive is very strange).

On the other hand the following piece of code:

const float maxFloat = 4294967295.0;
unsigned int a = (unsigned int) (signed int) maxFloat;
printf("%u\n", a);

prints 2147483648 which I belive is the correct results.

What happens that I get 2 different results?

Answer 1

If you first do this:

printf("%f\n", maxFloat);

The output you'll get is this:

4294967296.000000

Assuming a float is implemented as an IEEE754 single precision floating point type, the value 4294967295.0 cannot be represented exactly by this type because there's aren't enough bits of precision. The closest value it can store is 4294967296.0.

Assuming an int (and likewise unsigned int) is 32 bits, the value 4294967296.0 is outside the range of both of these types. Converting a floating point type to an integer type when the value cannot be represented in the given integer type invokes undefined behavior.

This is detailed in section 6.3.1.4 of the C standard which dictates conversion from floating point types to integer types:

1 When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined.⁶¹⁾

...

61) The remaindering operation performed when a value of integer type is converted to unsigned type need not be performed when a value of real floating type is converted to unsigned type. Thus, the range of portable real floating values is (−1, Utype_MAX+1).

The footnote in the above passage is referencing section 6.3.1.3, which details integer to integer conversions:

1 When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.

2 Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.

3 Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

The behavior you see in the first code snippet is consistent with an out-of-range conversion to an unsigned type when the value in question is an integer, however because the value being converted has a floating point type it is undefined behavior.

Just because one implementation does this doesn't mean that all will. In fact, gcc gives a different result if you change the optimization settings.

For example, on my machine using gcc 5.4.0, given this code:

float n = 4294967296;
printf("n=%f\n", n);
unsigned int a = (unsigned int) n;
int b = (signed int) n;
unsigned int c = (unsigned int) (signed int) n;
printf("a=%u\n", a);
printf("b=%d\n", b);
printf("c=%u\n", c);

I get the following results with -O0:

n=4294967296.000000
a=0
b=-2147483648
c=2147483648

And this with -O1:

n=4294967296.000000
a=4294967295
b=2147483647
c=2147483647

If on the other hand n is defined as long or long long, you would always get this output:

n=4294967296
a=0
b=0
c=0

The conversion to unsigned is well defined by the C standard as sited above, and the conversion to signed is implementation defined, which gcc defines as follows:

The result of, or the signal raised by, converting an integer to a signed integer type when the value cannot be represented in an object of that type (C90 6.2.1.2, C99 and C11 6.3.1.3).

For conversion to a type of width N, the value is reduced modulo 2^N to be within range of the type; no signal is raised.

Answer 2

Assuming IEEE 754 floating point numbers, the number 4294967295.0 can't be stored exactly in a float. It will be stored as 4294967296.0 instead (which is 2³²).

Further assuming your unsigned int has 32 value bits, this is just by one too large to fit in an unsigned int, so the result of the conversion is undefined according to the C standard -- 0 is a "reasonable" outcome.

In your second case, you have undefined behavior as well, and I have no theory what's happening here on the representation level. Fact is, the number is much too large for a 32 bit signed int (still assuming this is what your machine uses).

---

From this remark in your question:

prints 2147483648 which I belive is the correct results.

I assume you wanted to see the representation of your float in memory. Casting will convert the value, so that's not the way to see the representation. The following code would do:

int main(void) {
    const float maxFloat = 4294967295.0;
    unsigned char *floatBytes = &maxFloat;
    for (int i=0; i < sizeof maxFloat; ++i)
    {
        printf("0x%02x ", floatBytes[i]);
    }
    puts("");
}

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