C将char与"\ n"进行比较警告:指针与整数之间的比较 [英] C comparing char to "\n" warning: comparison between pointer and integer

问题描述

我有C代码的以下部分:

I have the following part of C code:

char c;
int n = 0;
while ( (c = getchar()) != EOF ){
    if (c == "\n"){
        n++;
    }
}

在编译过程中，编译器告诉我

during compilation, compiler tells me

warning: comparison between pointer and integer [enabled by default]

问题是，如果将"\n"替换为'\n'，则完全没有警告. 谁能解释我的原因?另一个奇怪的事情是我根本没有使用指针.

The thing is that if to substitute "\n" with '\n' there are no warnings at all. Can anyone explain me the reason? Another strange thing is that I am not using pointers at all.

我知道以下问题

• 警告:指针与整数之间的比较[默认情况下启用]在c
中
• 警告:C语言中指针与整数之间的比较

• warning: comparison between pointer and integer [enabled by default] in c
• warning: comparison between pointer and integer in C

但我认为它们与我的问题无关.

but in my opinion they are unrelated to my question.

PS.如果不是char c而是int c仍会警告.

PS. If instead of char c there will be int c there will be still warning.

推荐答案

• '\n'被称为字符文字，并且是标量整数类型.

  • '\n' is called a character literal and is a scalar integer type.

    "\n"被称为 string 文字，并且是数组类型.请注意，数组会衰减到指针，所以这就是为什么会出现该错误的原因.

    "\n" is called a string literal and is an array type. Note that arrays decay to pointers and so that's why you're getting that error.

    这可以帮助您了解:

    // analogous to using '\n'
    char c;
    int n = 0;
    while ( (c = getchar()) != EOF ){
        int comparison_value = 10;      // 10 is \n in ascii encoding
        if (c == comparison_value){
            n++;
        }
    }
    
    // analogous to using "\n"
    char c;
    int n = 0;
    while ( (c = getchar()) != EOF ){
        int comparison_value[1] = {10}; // 10 is \n in ascii encoding
        if (c == comparison_value){     // error
            n++;
        }
    }
    

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