bitwise casting uint32_t to float in C/C++

Question

I'm receiving a buffer from a network which was converted to an array of 32-bit words. I have one word which is defined as an ieee float by my interface document. I need to extract this word from the buffer. It's tough to cast from one type to another without invoking a conversion. The bits are already adhere to the ieee float standard, I don't want to re-arrange any bits.

float ieee_float(uint32_t f)
{
    return *((float*)((void*)(&f)));
}

Answer 1

In C++, the right way ™ is:

float ieee_float(uint32_t f)
{
    static_assert(sizeof(float) == sizeof f, "`float` has a weird size.");
    float ret;
    std::memcpy(&ret, &f, sizeof(float));
    return ret;
}

Both GCC and Clang at -O1 and above generate the same assembly for this code and a naive reinterpret_cast<float &>(f) .

Answer 2

You have several options, as stated here :

• Use the union solution: since C11 it is explicitly allowed (as said in the other answer).
• Instead of using an array of 32 bit words, use an array of 8 bit words (uint8_t), since char types can be aliased to any type.

Answer 3

There's no C\/C++ language. They're different languages with different rules. The valid way in C is to use a union, but that's not allowed in C++. See

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  Even reinterpret_cast<\/code> for type punning In C++20 a new cast type called std::bit_cast<\/code><\/a> was created exactly for this purpose

    
          
   
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