如何防止gcc优化器产生不正确的位操作？ [英] How can I prevent the gcc optimizer from producing incorrect bit operations?

问题描述

请考虑以下程序。

#include <stdio.h>

int negative(int A) {
    return (A & 0x80000000) != 0;
}
int divide(int A, int B) {
    printf("A = %d\n", A);
    printf("negative(A) = %d\n", negative(A));
    if (negative(A)) {
        A = ~A + 1;
        printf("A = %d\n", A);
        printf("negative(A) = %d\n", negative(A));
    }
    if (A < B) return 0;
    return 1;
}
int main(){
    divide(-2147483648, -1);
}

如果在没有编译器优化的情况下进行编译，则会产生预期的结果。

When it is compiled without compiler optimizations, it produces expected results.

gcc  -Wall -Werror -g -o TestNegative TestNegative.c
./TestNegative
A = -2147483648
negative(A) = 1
A = -2147483648
negative(A) = 1

使用编译器优化进行编译时，会产生以下错误输出。

When it is compiled with compiler optimizations, it produces the following incorrect output.

gcc -O3 -Wall -Werror -g -o TestNegative TestNegative.c
./TestNegative 
A = -2147483648
negative(A) = 1
A = -2147483648
negative(A) = 0

我正在运行 gcc版本5.4.0 。

我是否可以在源代码中进行更改，以防止编译器在 -O3下产生此行为？ ？

Is there a change I can make in the source code to prevent the compiler from producing this behavior under -O3?

推荐答案

1. -2147483648 并没有执行您认为的操作。 C没有负常数。包含 limits.h 并使用 INT_MIN 代替（几乎每个 INT_MIN 在二进制补码机上的定义将其定义为（-INT_MAX-1）是有原因的。）

1. -2147483648 does not do what you think it does. C doesn't have negative constants. Include limits.h and use INT_MIN instead (pretty much every INT_MIN definition on two's complement machines defines it as (-INT_MAX - 1) for a good reason).

A =〜A + 1; 调用未定义的行为，因为〜A + 1 导致整数溢出。

A = ~A + 1; invokes undefined behavior because ~A + 1 causes integer overflow.

不是编译器，而是您的代码。

It's not the compiler, it's your code.

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