什么是强力清除？ [英] What is strlen elision?

问题描述

我可以看到它列出的内容作为现代C ++习惯用法之一，但是究竟是什么呢？

这只是复制省略？

解决方案

您可能知道 std :: strlen 遍历整个字符串。在某些情况下，这可能效率不高，因为这意味着CPU必须开始对字符进行计数，从而减少了其他内容的缓存位置。

但是在大​​多数情况下，编译器能够优化 std :: strlen 并计算字符数字符串本身而不是使结果程序执行该操作。

A std :: strlen 的简单示例完全可以完全优化。

  #include< cstring> 
 
 int main（）{
 return std :: strlen（ hi）; 
} 
  

-O3以下得到的程序集是

  main：＃@main 
 mov eax，2 
 ret 
  

即使 -O0 不生成循环！

现在这是一个非常简单的示例，但是它可以工作甚至对于 std :: strlen 的用法也更复杂。

  #include< cstring> 
 
 int main（int argc，char **）{
 const char * string; 
 if（argc == 1）
 string = hello; 
 else if（argc == 2）
 string = world; 
 else if（argc == 3）
 string =！; 
 
 return std :: strlen（string）; 
} 
  

这里的字符串完全是已优化，以及 std :: strlen 调用。

I can see it listed as one of the modern C++ idioms, but what it is exactly?

Is it just a type of copy elision?

解决方案

You might know that std::strlen loops over the whole string. This can be inefficient in some cases, as it means that the CPU has to start counting characters, which reduces cache locality for other stuff.

But in most cases, the compiler is able to optimize std::strlen and count the number of characters of the string itself instead of making the resulting program do it. This is sometimes called strlen elision (because it elides the call to strlen).

A simple example of std::strlen would simply get optimized completely.

#include <cstring>

int main() {
    return std::strlen("hi");
}

Under -O3 the resulting assembly is

main: # @main
  mov eax, 2
  ret

Even under -O0 no loop is generated!

Now this is a rather simple example, but it works even for a bit more complicated usages of std::strlen.

#include <cstring>

int main(int argc, char**) {
    const char *string;
    if (argc == 1)
        string = "hello";
    else if (argc == 2)
        string = "world";
    else if (argc == 3)
        string = "!";

    return std::strlen(string); 
}

Here the strings are completely optimized away, as well as the std::strlen call.

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