C++ 操作数类型不兼容（“char”和“const char*”）
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Operand types are incompatible ("char" and "const char*")
提问by LeviTheDegu
I'm receiving the following error...
我收到以下错误...
Operand types are incompatible ("char" and "const char*")
操作数类型不兼容（“char”和“const char*”）
... when trying to perform an if statement. I'm assuming I'm not understanding how the input value is stored although I'm unsure if I can just cast it into the matching type?
... 尝试执行 if 语句时。我假设我不了解输入值是如何存储的，尽管我不确定是否可以将其转换为匹配类型？
Example code to reproduce is:
要重现的示例代码是：
char userInput_Text[3];
if (userInput_Text[1] == "y") {
// Do stuff.
}
I'm not sure what's causing this. It would appear that one type is a char and the other is a const char pointer although I'm unsure of what, for reference this error also occurs when I'm not using an array).
我不确定是什么原因造成的。看起来一种类型是 char 而另一种是 const char 指针，尽管我不确定是什么，作为参考，当我不使用数组时也会发生此错误）。
And tips / feedback would be much appreciated.
和提示/反馈将不胜感激。
回答by Karl Nicoll
Double quotes are the shortcut syntax for a c-stringin C++. If you want to compare a single character, you must use single quotes instead. You can simply change your code to this:
双引号是C++ 中c 字符串的快捷语法。如果要比较单个字符，则必须改用单引号。您可以简单地将代码更改为：
char userInput_Text[3];
if (userInput_Text[1] == 'y') { // <-- Single quotes here.
// Do stuff.
}
For reference:
以供参考：
"x"=const char *'x'=char
"x"=const char *'x'=char
