编译器优化方案导致运行速度慢

本文介绍了编译器优化方案导致运行速度慢的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有下面这段code的，我在C.其写下相当简单，因为它正好位移位 X 为每一个循环。

I have the following piece of code that I wrote in C. Its fairly simple as it just right bit-shifts x for every loop of for.

int main() { int x = 1; for (int i = 0; i > -2; i++) { x >> 2; } }

现在正在发生奇怪的是，当我刚刚编译它没有任何的优化或第一级优化（ -O ），它运行得很好（我定时可执行文件及其有关 1.4S 与 -O 和 5.4S 无任何优化。

Now the strange thing that is happening is that when I just compile it without any optimizations or with first level optimization (-O), it runs just fine (I am timing the executable and its about 1.4s with -O and 5.4s without any optimizations.

现在，当我加入 -O2 或 -O3 开关编译和时间生成的可执行文件，这不是' ŧ停止（我已经测试高达 60 ）。

Now when I add -O2 or -O3 switch for compilation and time the resulting executable, it doesn't stop (I have tested for up to 60s).

这是什么可能导致此任何想法？

Any ideas on what might be causing this?

推荐答案

优化循环产生无限循环是根据签署整数溢出你的结果。有符号整数溢出是ç未定义行为，不应该依赖。不仅可以将它混淆开发它也可以由编译器优化出

The optimized loop is producing an infinite loop which is a result of you depending on signed integer overflow. Signed integer overflow is undefined behavior in C and should not be depended on. Not only can it confuse developers it may also be optimized out by the compiler.

大会（不优化）：的gcc -std = C99 -S -O0的main.c

Assembly (no optimizations): gcc -std=c99 -S -O0 main.c

_main: LFB2: pushq %rbp LCFI0: movq %rsp, %rbp LCFI1: movl $1, -4(%rbp) movl $0, -8(%rbp) jmp L2 L3: incl -8(%rbp) L2: cmpl $-2, -8(%rbp) jg L3 movl $0, %eax leave ret

大会（优化的3级）：GCC -std = C99 -S -O3的main.c

_main: LFB2: pushq %rbp LCFI0: movq %rsp, %rbp LCFI1: L2: jmp L2 #<- infinite loop