gcc何时在编译时评估函数? [英] When is gcc evaluating functions at compile time?

问题描述

当我有以下代码时

int main(void) {
        printf("%zd\n", strlen("Hello World!"));

        return 0;
}

并用-O3进行编译，strings将显示字符串"Hello World！".二进制文件缺少它，因为它是在编译时求值的.

and compile it with -O3, strings will show that the string "Hello World!" is missing from the binary since it got evaluated at compile time.

如果我改用自己的功能

static inline size_t my_strlen(const char *s) {
        const char *tmp = s;

        while (*++tmp);

        return tmp - s;
}

int main(void) {
        printf("%zd\n", my_strlen("Hello World!"));

        return 0;
}

使用相同的选项，仍可以在二进制文件中找到该字符串.

with the same options, the string can still be found in the binary.

这是为什么?

推荐答案

因为它是优化符合标准.

在某些系统上，strlen最终被扩展为GCC编译器已知的_builtin_strlen.在我的计算机上，/usr/include/x86_64-linux-gnu/bits/string.h(由<string.h>包含的间接)具有

On some systems, strlen is finally expanded to _builtin_strlen which is known by the GCC compiler. On my machine /usr/include/x86_64-linux-gnu/bits/string.h (which is indirectly included by <string.h>) has

# define strlen(str) \
  (__extension__ (__builtin_constant_p (str)         \
                 ? __builtin_strlen (str)            \
                 : __strlen_g (str)))

实际上是通过GNU glibc和gcc

So actually is it done by the mix of GNU glibc and gcc

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