C++ 错误：只读变量的赋值
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C++ error: assignment of read-only variable
提问by mxmolins
I have some C++ code that returns this error:
我有一些返回此错误的 C++ 代码：
error: assignment of read-only variable ‘parking'
The code:
编码：
char const * const parking= "false";
if (phidgets.value(PHIDGET3V_1) > 1000) {
parking = "true";
}
else{
parking = "false";
}
What does this error mean and how do I fix it?
这个错误是什么意思，我该如何解决？
回答by Gabriel
parkingis set to be const(char const * const parking = "false") so it cannot be modified.
parking设置为const( char const * const parking = "false") 所以它不能被修改。
When you do parking = "true"it raises compile time error.
当你这样做parking = "true"时会引发编译时错误。
How to reproduce the problem very simply to illustrate:
如何复现问题很简单来说明：
#include <iostream>
int main(){
const int j = 5;
j = 7;
}
constmeans constant, meaning you are not allowed to change it:
const表示常量，意思是你不能改变它：
error: assignment of read-only variable ‘j'
回答by Vlad from Moscow
You declared parking as constant pointer.
您将停车声明为常量指针。
char const * const parking= "false";
So it will point only to string literal "false"and may not be changed.
所以它只会指向字符串文字"false"并且可能不会改变。
Also this statement
还有这个说法
char const * const message = "Value: "+ parking +" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxy";
is invalid. There is no addition operator for pointers.
是无效的。指针没有加法运算符。
回答by shuttle87
In your code you set up the parking variable with a const, this is telling the compiler that it will not be modified later. You then modify parking later by setting it to true or false.
在您的代码中，您使用 a 设置了停车变量const，这告诉编译器以后不会修改它。然后您稍后通过将其设置为 true 或 false 来修改停车。
Using std::stringis far more idiomatic c++ though. So I would do this instead:
不过，使用std::stringC++ 更为惯用。所以我会这样做：
#include<string>
std::string parking = "false";
if (phidgets.value(PHIDGET3V_1) > 1000) {
parking = "true";
//leds_on(LEDS_RED);
} else {
parking = "false";
//leds_off(LEDS_RED);
}
std::string message = "Value: "+ parking +" ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxy";
std::stringoverloads +to do concatenation so it does what you think it does in the last line. Previously you were adding some pointers and that probably doesn't do what you think it does.
std::string重载+以进行连接，因此它执行您认为在最后一行中执行的操作。以前您添加了一些指针，但这可能与您认为的不一样。
