C ++比较char和字符串文字 [英] C++ comparing a char to a string literal

问题描述

这里是初学者...

我正在为我的计算机科学课编写一个非常简单的程序，遇到一个我想知道的问题更多关于。这是我的代码：

I'm writing a very simply program for my computer science class and I ran into an issue that I'd like to know more about. Here is my code:

#include <iostream>

using namespace std;

int main(int argc, const char * argv[])
{
    char courseLevel;

    cout << "Will you be taking graduate or undergraduate level courses (enter 'U'"
            " for undergraduate,'G' for graduate.";
    cin >> courseLevel;

    if (courseLevel == "U")
    {
        cout << "You selected undergraduate level courses.";
    }

    return 0;
}

我的if语句收到两条错误消息：
1 ）未指定与字符串文字的比较结果（请改用strncmp）。
2）指针和整数（'int'和'const char *'）之间的比较。

I'm getting two error messages for my if statement: 1) Result of comparison against a string literal is unspecified (use strncmp instead). 2) Comparison between pointer and integer ('int' and 'const char*').

我似乎已经通过将U括起来解决了这个问题。单引号，或者该程序至少仍然可以运行。但是，正如我所说的，我只是想了解为什么会收到错误消息，以便可以更好地了解自己的工作。

I seem to have resolved the issue by enclosing my U in single quotes, or the program at least works anyway. But, as I stated, I'd simply like to understand why I was getting the error so I can get a better understanding of what I'm doing.

推荐答案

您需要使用单引号代替。

You need to use single quotes instead.

在C语言中，（以及许多其他语言）a 字符常量 是一个单个字符 ¹ 包含在单引号中

In C, (and many other languages) a character constant is a single character¹ contained in single quotes:

'U'

而 字符串文字 > 是双引号中包含的任意数量的字符：

While a string literal is any number of characters contained in double quotes:

"U"

您将 courseLevel 声明为单个字符： char courseLevel ; 因此，您只能将其与另一个 char 进行比较。

You declared courseLevel as a single character: char courseLevel; So you can only compare that to another single char.

if（courseLevel == U ），左侧是 char ，而右侧是 const char * -指向该字符串文字中第一个 char 的指针。您的编译器告诉您：

When you do if (courseLevel == "U"), the left side is a char, while the right side is a const char* -- a pointer to the first char in that string literal. Your compiler is telling you this:

指针与整数（' int '和' const char * '）

---

因此，您的选择是：

---

So your options are:

if (courseLevel == 'U')       // compare char to char

或者，例如：

if (courseLevel == "U"[0])    // compare char to first char in string

---

---

1. 完整性提示：您可以具有多字符常量：

int a ='abcd'; //在GCC中为0x61626364

int a = 'abcd'; // 0x61626364 in GCC

但这绝对不是您想要的。

But this is certainly not what you're looking for.

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