g ++严格溢出，优化和警告 [英] g++ strict overflow, optimization, and warnings

问题描述

在使用严格溢出标志编译以下内容时，它在第二个测试中告诉我，r可能不是我认为的：

When compiling the following with the strict overflow flag, it tells me, on the 2nd test that r may not be what I think it could be:

    int32_t r(my_rand());
    if(r < 0) {
        r = -r;
        if(r < 0) {   // <-- error on this line
            r = 0;
        }
    }

错误是：

/build/buildd/libqtcassandra-0.5.5/tests/cassandra_value.cpp:
     In function 'int main(int, char**)':
/build/buildd/libqtcassandra-0.5.5/tests/cassandra_value.cpp:2341:13:
     error: assuming signed overflow does not occur when simplifying
     conditional to constant [-Werror=strict-overflow]
         if(r < 0) {
         ^

我不明白的是：为什么不会在之前的行上生成错误？因为真的溢出发生在我这样做，对吗？

What I do not understand is: why wouldn't the error be generated on the line before that? Because really the overflow happens when I do this, right?

    r = -r;

推荐答案

EDIT：我的第一个答案，因为它是无效的。这里是全新的版本。感谢@Neil Kirk指出我的错误。

I removed my first answer, because it was invalid. Here is completely new version. Thanks to @Neil Kirk for pointing out my errors.

回答问题在这里： http://stackoverflow.com/a/18521660/2468549

GCC总是假定，签名溢出从不会发生，假设，它总是优化内部 if（r <0）块。

GCC always assumes, that signed overflow does never occur, and, on that assumption, it (always) optimizes out the inner if (r < 0) block.

-Wstrict-overflow ，然后编译器发现， r = -r r < 0 可能仍然为真（如果 r == -2 ^ 31 最初），这导致错误（错误由基于假设溢出从不会发生，不是通过溢出可能性本身 - 这是 -Wstrict-overflow 工作原理）。

If you turn -Wstrict-overflow on, then compiler finds out, that after r = -r r < 0 may still be true (if r == -2^31 initially), which causes an error (error is caused by optimization based on assumption of overflow never occurring, not by overflow possibility itself - that's how -Wstrict-overflow works).

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