访问 float 的 4 个字节是否会破坏 C++ 别名规则

Question

I need to read the binary content of a file and turn the extracted bytes into single precision floating point numbers. How to do this has already been asked here . That question does have proper answers but I'm wondering whether a particular answer is actually valid C++ code.

That answer gives the following code:

float bytesToFloat(uint8_t *bytes, bool big_endian) {
    float f;
    uint8_t *f_ptr = (uint8_t *) &f;
    if (big_endian) {
        f_ptr[3] = bytes[0];
        f_ptr[2] = bytes[1];
        f_ptr[1] = bytes[2];
        f_ptr[0] = bytes[3];
    } else {
        f_ptr[3] = bytes[3];
        f_ptr[2] = bytes[2];
        f_ptr[1] = bytes[1];
        f_ptr[0] = bytes[0];
    }
    return f;
}

Is this actually valid C++ code? I'm not sure whether it violates any aliasing rules.

Note that I'm targeting platforms with big endian where a float is guaranteed to be at least 32 bits long.

Answer 1

Is this actually valid C++ code?

Potentially yes. It has some pre-conditions:

• std::uint8_t must be an alias of unsigned char
• sizeof(float) must be 4
• bytes + 3 mustn't overflow a buffer.

You can add a checks to ensure safe failure to compile if the first two don't hold:

static_assert(std::is_same_v<unsigned char, std::uint8_t>);
static_assert(sizeof(float) == 4);

I'm not sure whether it violates any aliasing rules.

unsigned char is excempted of such restrictions. std::uint8_t , if it is defined, is in practice an alias of unsigned char , in which case the shown program is well defined. Technically that's not guaranteed by the rules, but the above check will handle the theoretical case where that doesn't apply.

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float is guaranteed to be at least 32 bits long.

It must be exactly 32 bits long for the code to work. It must also have exactly the same bit-level format as was on the system where the float was serialised. If it's standard IEE-754 single precision on both ends then you're good; otherwise all bets are off.