C ++错误：在'char *'到'char [2]的赋值中不兼容的类型 [英] C++ Error: Incompatible types in assignment of ‘char*’ to ‘char [2]

问题描述

我有一个问题，我的构造函数。
在我的头文件中我声明：

I have a bit of a problem with my constructor. In my header file I declare:

char short_name_[2]; 

• 和其他变量



在我的构造函数中：

Territory(std::string name, char short_name[2], Player* owner, char units);
void setShortName(char* short_name);
inline const char (&getShortName() const)[2] { return short_name_; }

在我的cpp文件中：

Territory::Territory(std::string name, char short_name[2], Player* owner, 
                     char units) : name_(name), short_name_(short_name), 
                    owner_(owner), units_(units)
{ }

我的错误： / p>



My error:




Territory.cpp：在构造函数'Territory :: Territory（std :: string，
char *，Player *，char） '：Territory.cpp：15：33：error：incompatible types
在'char *'到'char [2]'的赋值

Territory.cpp: In constructor ‘Territory::Territory(std::string, char*, Player*, char)’: Territory.cpp:15:33: error: incompatible types in assignment of ‘char*’ to ‘char [2]’

我已经知道 char [2]< => char * 但我不知道如何处理这个关于我的构造函数和get / setters。

I already figured out that char[2] <=> char* but I'm not sure how to handle this about my constructor and get/setters.

推荐答案

p> C ++中的原始数组有点恼人，充满了危险。这是为什么除非你有一个很好的理由，你应该使用 std :: vector 或 std :: array 。

Raw arrays in C++ are kind of annoying and fraught with peril. This is why unless you have a very good reason to you should use std::vector or std::array.

首先，如其他人所说， char [2] 不同于 char * ，或至少不是通常。 char [2] 是 char 和 char * 是指向 char 的指针。他们经常会感到困惑，因为数组会衰减到指向第一个元素的指针，只要他们需要。所以这个工作：

First off, as others have said, char[2] is not the same as char*, or at least not usually. char[2] is a size 2 array of char and char* is a pointer to a char. They often get confused because arrays will decay to a pointer to the first element whenever they need to. So this works:

char foo[2];
char* bar = foo;

但反之则不然：

const char* bar = "hello";
const char foo[6] = bar; // ERROR

添加到混淆，当声明函数参数时， char [ ] 等效于 char * 。所以在你的构造函数中，参数 char short_name [2] 真的是 char * short_name 。

Adding to the confusion, when declaring function parameters, char[] is equivalent to char*. So in your constructor the parameter char short_name[2] is really char* short_name.

另一个奇怪的数组是，它们不能像其他类型一样复制（这是为什么函数参数中的数组被视为指针的一个解释）。例如，我可以不做这样的事情：

Another quirk of arrays is that they cannot be copied like other types (this is one explanation for why arrays in function parameters are treated as pointers). So for example I can not do something like this:

char foo[2] = {'a', 'b'};
char bar[2] = foo;


$ b c>并将它们复制到 bar 中，或者使用一些函数，例如 std :: copy ：

Instead I have to iterate over the elements of foo and copy them into bar, or use some function which does that for me such as std::copy:

char foo[2] = {'a', 'b'};
char bar[2];
// std::begin and std::end are only available in C++11
std::copy(std::begin(foo), std::end(foo), std::begin(bar));

所以在你的构造函数中，你必须手动复制 short_name into short_name _ ：

So in your constructor you have to manually copy the elements of short_name into short_name_:

Territory::Territory(std::string name, char* short_name, Player* owner, 
                     char units) : name_(name), owner_(owner), units_(units)
{ 
    // Note that std::begin and std::end can *not* be used on pointers.
    std::copy(short_name, short_name + 2, std::begin(short_name));
}

正如你可以看到，这是非常讨厌，所以除非你有一个非常好的原因你只是应该使用 std :: vector 而不是原始数组（或在这种情况下可能 std :: string ）。

As you can see this is all very annoying, so unless you have a very good reason you just should use std::vector instead of raw arrays (or in this case probably std::string).

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