有趣的C，有趣的编译器 [英] Amusing C, amusing compiler
问题描述
一个函数
int foo（struct T * x）
{
return（x + 1）-x;
}
应始终返回1，无论T是如何定义的。 （对于你的学生来说，int可以用ptrdiff_t取代
。）
首先，看看我的编译器（着名的IAR
EWARM for ARM）跳过箍来得到这个答案。
[对于sizeof（struct T）== 6，
00000000 061080E2 ADD R1，R0，＃+ 6
00000004 18209FE5 LDR R2，?? foo_0 ;; 0xaaaaaaab
00000008 92318CE0 UMULL R3，R12，R2，R1
0000000C 2CC1A0E1 LSR R12，R12，＃+ 2
00000010 0210A0E1 MOV R1， R2
00000014 912083E0 UMULL R2，R3，R1，R0
00000018 2331A0E1 LSR R3，R3，＃+ 2
0000001C 03004CE0 SUB R0， R12，R3
00000020 0EF0A0E1 MOV PC，LR ;;返回
?? foo_0：
00000024 ABAAAAAA DC32 0xaaaaaaab
]
你的编译器怎么样？票价？
[MSVC说得对：
mov eax，1
ret 0
]
另一件事是逻辑上，因为实际类型没关系，
它可能是一个不完整的类型。但是，如果我只是在foo的正文之前说
struct T;
，则编译失败。是好还是有道理？
- 方舟
A function
int foo(struct T *x)
{
return (x+1)-x;
}
should always return a 1, no matter how T is defined. (And int could be
replaced with ptrdiff_t for you pedants.)
For one thing, it was amusing to watch how my compiler (the famed IAR
EWARM for ARM) jumps through hoops to arrive at this answer.
[ For sizeof(struct T)==6,
00000000 061080E2 ADD R1,R0,#+6
00000004 18209FE5 LDR R2,??foo_0 ;; 0xaaaaaaab
00000008 92318CE0 UMULL R3,R12,R2,R1
0000000C 2CC1A0E1 LSR R12,R12,#+2
00000010 0210A0E1 MOV R1,R2
00000014 912083E0 UMULL R2,R3,R1,R0
00000018 2331A0E1 LSR R3,R3,#+2
0000001C 03004CE0 SUB R0,R12,R3
00000020 0EF0A0E1 MOV PC,LR ;; return
??foo_0:
00000024 ABAAAAAA DC32 0xaaaaaaab
]
How does your compiler fare?
[MSVC gets it right:
mov eax, 1
ret 0
]
Another thing is that, logically, since the actual type doesn''t matter,
it could be an incomplete type. However, if I just say
struct T;
before the foo''s body, compilation fails. Is it good and/or justified?
- Ark
推荐答案
Ark写道：
Ark wrote:
一个函数
int foo（struct T * x）
{
return（x +1）-x;
}
应始终返回1，无论T是如何定义的。 （对于你的学生来说，int可以用ptrdiff_t取代
。）
首先，看看我的编译器（着名的IAR
EWARM for ARM）跳过箍来得到这个答案。
[对于sizeof（struct T）== 6，
00000000 061080E2 ADD R1，R0，＃+ 6
00000004 18209FE5 LDR R2，?? foo_0 ;; 0xaaaaaaab
00000008 92318CE0 UMULL R3，R12，R2，R1
0000000C 2CC1A0E1 LSR R12，R12，＃+ 2
00000010 0210A0E1 MOV R1， R2
00000014 912083E0 UMULL R2，R3，R1，R0
00000018 2331A0E1 LSR R3，R3，＃+ 2
0000001C 03004CE0 SUB R0， R12，R3
00000020 0EF0A0E1 MOV PC，LR ;;返回
?? foo_0：
00000024 ABAAAAAA DC32 0xaaaaaaab
]
你的编译器怎么样？票价？
[MSVC说得对：
mov eax，1
ret 0
]
另一件事是逻辑上，因为实际类型没关系，
它可能是一个不完整的类型。但是，如果我只是在foo的正文之前说
struct T;
，则编译失败。它是好的和/或合理的吗？
- Ark
A function
int foo(struct T *x)
{
return (x+1)-x;
}
should always return a 1, no matter how T is defined. (And int could be
replaced with ptrdiff_t for you pedants.)
For one thing, it was amusing to watch how my compiler (the famed IAR
EWARM for ARM) jumps through hoops to arrive at this answer.
[ For sizeof(struct T)==6,
00000000 061080E2 ADD R1,R0,#+6
00000004 18209FE5 LDR R2,??foo_0 ;; 0xaaaaaaab
00000008 92318CE0 UMULL R3,R12,R2,R1
0000000C 2CC1A0E1 LSR R12,R12,#+2
00000010 0210A0E1 MOV R1,R2
00000014 912083E0 UMULL R2,R3,R1,R0
00000018 2331A0E1 LSR R3,R3,#+2
0000001C 03004CE0 SUB R0,R12,R3
00000020 0EF0A0E1 MOV PC,LR ;; return
??foo_0:
00000024 ABAAAAAA DC32 0xaaaaaaab
]
How does your compiler fare?
[MSVC gets it right:
mov eax, 1
ret 0
]
Another thing is that, logically, since the actual type doesn''t matter,
it could be an incomplete type. However, if I just say
struct T;
before the foo''s body, compilation fails. Is it good and/or justified?
- Ark
愚蠢的机器是多么有趣。
你知道吗？你忘记了一个半冒号，他们搞砸了。
愚蠢不是吗？
1）如果x是双倍的话。如果x是NAN或INFinity，
结果不是一个而是NAN。
2）对于x == INTMAX，x + 1-x取决于操作的方式是由编译器订购的
。如果它使（x-1）+ x然后是1，
否则它会溢出并且存在未定义的行为。
如果INTMAX + 1回绕为负，结果是不是一个
3）如果x是无符号且等于UINT_MAX，则结果是-x。
4）这种优化的数量无限多
x + 1-x
x + 2-x
x + 3-x
x + 4-x + 2 * xxx
x + 5-x + 3 * xxxx + 5765 * x-5000 * x-765 * x
对于最后一个应该是5 Microsoft生成的表达式：
mov eax，DWORD PTR _x
It is amusing how stupid machines are.
You know? You forget a semi colon and they get all screwed up.
Stupid isn''t it?
1) If x is double. If x is a NAN or INFinity,
the result is not one but NAN.
2) For x == INTMAX, x+1-x depends on how the operations are
ordered by the compiler. If it makes (x-1)+x then is 1,
otherwise it overflows and there is undefined behavior.
If INTMAX+1 wraps around to negative the result is not one
either
3) If x is unsigned and equal to UINT_MAX, the the result is -x.
4) There is an infinite number of this kind of optimizations
x+1-x
x+2-x
x+3-x
x+4-x+2*x-x-x
x+5-x + 3*x-x-x-x + 5765*x-5000*x-765*x
For that last expression that should be 5 Microsoft generates:
mov eax, DWORD PTR _x
[ebp ]
添加eax，5
sub eax，DWORD PTR _x
[ebp]
add eax, 5
sub eax, DWORD PTR _x
[ebp]
mov ecx，DWORD PTR _x
[ebp]
mov ecx, DWORD PTR _x
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