Most efficient standard-compliant way of reinterpreting int as float

Question

Assume I have guarantees that float is IEEE 754 binary32. Given a bit pattern that corresponds to a valid float, stored in std::uint32_t, how does

Answer 1

Afaik, there are only two approaches that are compliant with strict aliasing rules: memcpy() and cast to char* with copying. All others read a float from memory that belongs to an uint32_t, and the compiler is allowed to perform the read before the write to that memory location. It might even optimize away the write altogether as it can prove that the stored value will never be used according to strict aliasing rules, resulting in a garbage return value.

It really depends on the compiler/optimizes whether memcpy() or char* copy is faster. In both cases, an intelligent compiler might be able to figure out that it can just load and copy an uint32_t, but I would not trust any compiler to do so before I have seen it in the resulting assembler code.

Edit:
After some testing with gcc 4.8.1, I can say that the memcpy() approach is the best for this particulare compiler, see below for details.

---

Compiling

#include 

float foo(uint32_t a) {
    float b;
    char* aPointer = (char*)&a, *bPointer = (char*)&b;
    for( int i = sizeof(a); i--; ) bPointer[i] = aPointer[i];
    return b;
}

with gcc -S -std=gnu11 -O3 foo.c yields this assemble code:

movl    %edi, %ecx
movl    %edi, %edx
movl    %edi, %eax
shrl    $24, %ecx
shrl    $16, %edx
shrw    $8, %ax
movb    %cl, -1(%rsp)
movb    %dl, -2(%rsp)
movb    %al, -3(%rsp)
movb    %dil, -4(%rsp)
movss   -4(%rsp), %xmm0
ret

This is not optimal.

Doing the same with

#include 
#include 

float foo(uint32_t a) {
    float b;
    char* aPointer = (char*)&a, *bPointer = (char*)&b;
    memcpy(bPointer, aPointer, sizeof(a));
    return b;
}

yields (with all optimization levels except -O0):

movl    %edi, -4(%rsp)
movss   -4(%rsp), %xmm0
ret

This is optimal.