-fwrapv 有什么作用? [英] What does -fwrapv do?

问题描述

任何人都可以提供一些代码示例，它们在使用 fwrapv 与没有编译时表现不同.

Can anyone provide some code examples that act differently when compiled with fwrapv vs without.

它说 -fwrapv 应该假设加法、减法和乘法的有符号算术溢出，使用二进制补码表示进行换行."

It says -fwrapv is supposed to "assume that signed arithmetic overflow of addition, subtraction, and multiplication, wraps using two's-complement representation."

但每当我尝试溢出时，无论是否使用 fwrapv，结果都是一样的.

But whenever I try overflowing the result is the same with or without fwrapv.

推荐答案

想想这个函数:

int f(int i) {
    return i+1 > i;
}

从数学上讲，对于任何整数 i，i+1 应该总是大于 i.但是，对于 32 位 int，i 的一个值会使该语句为假，即 2147483647(即 0x7FFFFFFF，即INT_MAX).向该数字添加 1 将导致溢出，并且根据 2 的互补表示，新值将环绕并变为 -2147483648.因此，i+1>i 变为 -2147483648>2147483647 这是错误的.

Mathematically speaking, i+1 should always be greater than i for any integer i. However, for a 32-bit int, there is one value of i that makes that statement false, which is 2147483647 (i.e. 0x7FFFFFFF, i.e. INT_MAX). Adding one to that number will cause an overflow and the new value, according to the 2's compliment representation, will wrap-around and become -2147483648. Hence, i+1>i becomes -2147483648>2147483647 which is false.

当你编译不带-fwrapv时，编译器会假设溢出是'non-wrapping'并且它会优化那个函数总是返回1(忽略溢出情况).

When you compile without -fwrapv, the compiler will assume that the overflow is 'non-wrapping' and it will optimize that function to always return 1 (ignoring the overflow case).

用-fwrapv编译时，函数不会优化，会有加1比较的逻辑两个值，因为现在溢出是 'wrapping' (即溢出的数字将根据 2 的补码表示来换行).

When you compile with -fwrapv, the function will not be optimized, and it will have the logic of adding 1 and comparing the two values, because now the overflow is 'wrapping' (i.e. the overflown number will wrap according to the 2's compliment representation).

在生成的程序集中可以很容易地看到差异 - 在右侧窗格中，没有 -fwrapv，函数总是返回1 (true).

The difference can be easily seen in the generated assembly - in the right pane, without -fwrapv, function always returns 1 (true).

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