我试图准备最不可行的代码示例:
#include <stdio.h>
#include <inttypes.h>
#include <string.h>
typedef struct FECPUFlags {
uint16_t CF:1; // carry flag
uint16_t PF:1; // parity flag
uint16_t AF:1; // adjust flag
uint16_t ZF:1; // zero flag
uint16_t SF:1; // sign flag
uint16_t TF:1; // trap flag
uint16_t IF:1; // interrupt enable flag
uint16_t DF:1; // direction flag
uint16_t OF:1; // overflow flag
} FECPUFlags;
uint16_t fe_cpuflags_16(FECPUFlags cpuRegister) {
uint16_t result = 0;
result = result | cpuRegister.CF;
result = result | (cpuRegister.PF << 2);
result = result | (cpuRegister.AF << 4);
result = result | (cpuRegister.ZF << 6);
result = result | (cpuRegister.SF << 7);
result = result | (cpuRegister.DF << 10);
result = result | (cpuRegister.OF << 11);
return result;
}
int main() {
FECPUFlags flag;
#define ENUM(F) for(int i=0;i<=1;i++,flag.F=i)
ENUM(CF) ENUM(PF) ENUM(AF) ENUM(ZF) ENUM(SF) ENUM(DF) ENUM(OF) {
printf("0x%X\n",fe_cpuflags_16(flag));
}
return 0;
}
clang版
Apple LLVM version 7.0.2 (clang-700.1.81)
Target: x86_64-apple-darwin15.0.0
应用程序的输出因不同的优化模式而异。 clang -O0 vs clang -O3。
这是优化版本
的弊端_fe_cpuflags_16:
0000000100000d70 pushq %rbp
0000000100000d71 movq %rsp, %rbp
0000000100000d74 movl %edi, %eax
0000000100000d76 andl $0x1, %eax
0000000100000d79 leal (%rdi,%rdi), %ecx
0000000100000d7c andl $0x4, %ecx
0000000100000d7f orl %eax, %ecx
0000000100000d81 leal (,%rdi,4), %eax
0000000100000d88 andl $0x10, %eax
0000000100000d8b orl %ecx, %eax
0000000100000d8d shll $0x3, %edi
0000000100000d90 movl %edi, %ecx
0000000100000d92 andl $0x40, %ecx
0000000100000d95 orl %eax, %ecx
0000000100000d97 movl %edi, %eax
0000000100000d99 andl $0x80, %eax
0000000100000d9e orl %ecx, %eax
0000000100000da0 movl %edi, %ecx
0000000100000da2 andl $0x400, %ecx ## imm = 0x400
0000000100000da8 orl %eax, %ecx
0000000100000daa andl $0x800, %edi ## imm = 0x800
0000000100000db0 orl %ecx, %edi
0000000100000db2 movzwl %di, %eax
0000000100000db5 popq %rbp
0000000100000db6 retq
有趣的是,如果我在枚举之前将memset置于零 - 代码工作正常。
哪种优化可能会破坏此代码?或者这段代码可能已经以某种方式破坏了?
答案 0 :(得分:4)
您正在使用FECPUFlags flag;
而未初始化它。
这是undefined behavior。任何事情都可能发生。
答案 1 :(得分:3)
您有一个未初始化的变量,稍后会被访问,因此您有未定义的行为。
FECPUFlags flag;
(某些)标志的初始化仅在循环的第二次执行中发生:
#define ENUM(F) for(int i=0;i<=1;i++,flag.F=i)
^^^