Missing optimization: mov al, [mem] to bitfield-insert a new low byte into an integer

Question

I want to replace the lowest byte in an integer. On x86 this is exactly mov al, [mem] but I can't seem to get compilers to output this. Am I missing an obvious code pattern that is recognized, am I misunderstanding something, or is this simply a missed optimization?

unsigned insert_1(const unsigned* a, const unsigned char* b)
{
    return (*a & ~255) | *b;
}
unsigned insert_2(const unsigned* a, const unsigned char* b)
{
    return *a >> 8 << 8 | *b;
}

GCC actually uses al but just for zeroing.

        mov     eax, DWORD PTR [rdi]
        movzx   edx, BYTE PTR [rsi]
        xor     al, al
        or      eax, edx
        ret

Clang compiles both practically verbatim

        mov     ecx, -256
        and     ecx, dword ptr [rdi]
        movzx   eax, byte ptr [rsi]
        or      eax, ecx
        ret

Answer 1

On x86 this is exactly mov al, [mem] but I can't seem to get compilers to output this.

Try this one, arithmetic-free:

unsigned insert_4(const unsigned* a, const unsigned char* b)
{
    unsigned int t = *a;
    unsigned char *tcp = (unsigned char *) & t;
    tcp[0] = *b;
    return t;
}

---

insert_4(unsigned int const*, unsigned char const*):
        mov     eax, DWORD PTR [rdi]
        mov     al, BYTE PTR [rsi]
        ret

A bit screwy, I know but the compilers are good at removing indirection and address taken for local variables (took a couple of tries though..).

godbolt x86-64 gcc 13.1 -O3

---

An alternative using union:

unsigned insert_5(const unsigned* a, const unsigned char* b)
{
    union {
        unsigned int ui;
        unsigned char uc;
    } u;
    u.ui = *a;
    u.uc = *b;
    return u.ui;
}

godbolt x86-64 gcc 13.1 -O3

---

Note, these solutions are endian-specific, though it seems like what you're looking for, and, as needed can be adjusted for the other endian.