英文:
Missing optimization: mov al, [mem] to bitfield-insert a new low byte into an integer
问题
我想替换整数中的最低字节。在x86上，这正好是 `mov al, [mem]`，但我似乎无法让编译器输出这个。我是不是漏掉了一个明显的被识别的代码模式，我是否理解错了什么，还是说这只是一个被忽视的优化？
GCC实际上使用了 `al`，但只是用于清零。
Clang几乎逐字地编译了这两者。
```asm
GCC:
mov eax, DWORD PTR [rdi]
movzx edx, BYTE PTR [rsi]
xor al, al
or eax, edx
ret
Clang:
mov ecx, -256
and ecx, dword ptr [rdi]
movzx eax, byte ptr [rsi]
or eax, ecx
ret
英文:
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
答案1
得分: 7
在x86上，这就是mov al, [mem]，但我似乎无法让编译器输出这个。
尝试这个，不涉及算术操作：
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
有点混乱，我知道，但编译器很擅长消除间接引用和对局部变量取地址的操作（虽然经过了几次尝试）。
使用联合体的另一种方法：
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;
}
请注意，这些解决方案是特定于端序的，但似乎符合您的需求，如有需要可以调整为其他端序。
英文:
> 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..).
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;
}
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.
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