CODEWITHSUNDEEP
c++c++11type-conversion
Walter
Walter

Reputation: 45464

How to get the float with given bit-pattern (as int32_t) in C++?

I need a fast way to obtain the float with a given bit-pattern (provided as int32_t). Of course, the compiler should optimise the whole construction away. Simple conversion does a cast and reinterpret_cast<> is not allowed ...

Upvotes: 4

Views: 938

Answers (2)

Steve Jessop
Steve Jessop

Reputation: 279375

It's not reliable that the compiler will optimize this away, but it avoids UB provided that the value supplied really is a representation of a float (that is, it's the right size and its bit pattern doesn't hold a trap representation of float). GCC is at least sometimes capable of optimizing it away:

float convert(int32_t inputvalue) {
    float f;
    std::memcpy(&f, &inputvalue, sizeof(f));
    return f;
}

If the optimization is an important part of the question then the official answer is that there is no way to guarantee that an unknown compiler will make a given optimization. But this one is harder to optimize than most. It relies on the compiler "understanding" what memcpy does, which is a bigger ask than "understanding" what a pointer cast does.

Upvotes: 5

ecatmur
ecatmur

Reputation: 157444

The only fully portable method is to memcpy via a buffer:

static_assert(sizeof(float) == sizeof(int32_t), "!!");
char buf[sizeof(float)];
memcpy(buf, &i, sizeof(buf));
memcpy(&f, buf, sizeof(buf));

Usually the buffer can be elided:

static_assert(sizeof(float) == sizeof(int32_t), "!!");
memcpy(&f, &i, sizeof(float));

Upvotes: 2

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