malloc和gcc优化2 [英] malloc and gcc optimization 2

问题描述

while(count < 30000000){
    malloc(24);
    count++;
}

以上代码在我用gcc -O0编译的计算机上运行约170毫秒.但是，使用x> 0的-Ox进行编译时，优化器巧妙地指出所请求的内存将永远不会被使用，因此将其从优化的可执行文件中排除.它是如何做到的?

the above code runs in about 170 ms on my computer compiled with gcc -O0. However, compiling with -Ox where x > 0, the optimizer cleverly figures out that the memory being requested will never be used and so it is excluded from the optimized executable. How does it do this?

推荐答案

编译器发现从未使用malloc返回值，因此对其进行了优化.如果您想即使在-O3中也无法优化malloc调用，则可以使用volatile限定符:

Well the compiler sees malloc return value is never used so it optimizes it out. If you want to prevent malloc call to be optimzed out even in -O3 you can use the volatile qualifier:

while(count < 30000000){
    void * volatile p = malloc(24);
    count++;
}

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