什么是＃运营商的宏吗？ [英] What does the # operator do in macros?

问题描述

#include <stdio.h>

#define foo(x, y) #x #y

int main()
{
    printf("%s\n", foo(k, l));
    return 0;
}

输出：结果
KL

我知道##做串联。从输出似乎＃也做串联。
我对么？

I know that ## does concatenation. From the output it seems that # also does concatenation. Am I correct?

如果我是正确的话是什么之间 ## 运营商和＃运营商？

If I am correct then what is the difference between ## operator and # operator?

推荐答案

＃打开参数为一个字符串。因此，美孚（K，L）变成K，L，这是一样的KL因为在C多重字符串是直接相邻的被视为一个单一的字符串。

# turns the argument into a string. So foo(k, l) becomes "k" "l", which is the same as "kl" because in C multiple string literals that are directly next to each other are treated as a single string literals.

如果＃并串联，你的printf调用将成为的printf（％S \\ n，KL）; 这将产生约误差 KL 没有被定义的。

If # did concatenation, your printf call would become printf("%s\n", kl); which would produce an error about kl not being defined.

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