C++ 错误:将‘char*’分配给‘char [2]时的类型不兼容

问题描述

我的构造函数有点问题.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:

In my cpp file:

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:

Territory.cpp: 在构造函数中'Territory::Territory(std::string,char*, Player*, char)’: Territory.cpp:15:33: 错误: 不兼容的类型将‘char*’分配给‘char [2]’

我已经知道 char[2] <=>char* 但我不确定如何处理我的构造函数和 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

C++ 中的原始数组有点烦人，而且充满危险.这就是为什么除非你有充分的理由应该使用 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，正如其他人所说，char[2] 与 char* 不同，或者至少通常不同.char[2] 是一个大小为 2 的 char 数组，char* 是一个指向 char 的指针.他们经常感到困惑，因为数组会在需要时衰减到指向第一个元素的指针.所以这行得通:

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

更令人困惑的是，在声明函数参数时，char[] 等价于 char*.因此，在您的构造函数中，参数 char short_name[2] 实际上是 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.

数组的另一个怪癖是它们不能像其他类型一样被复制(这是为什么函数参数中的数组被视为指针的一种解释).所以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;

相反，我必须遍历 foo 的元素并将它们复制到 bar 中，或者使用一些对我有用的函数，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));

因此，在您的构造函数中，您必须手动将 short_name 的元素复制到 short_name_ 中:

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));
}

正如你所看到的，这一切都很烦人，所以除非你有很好的理由，否则你应该使用 std::vector 而不是原始数组(或者在这种情况下可能是 std::字符串).

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).

这篇关于C++ 错误:将‘char*’分配给‘char [2]时的类型不兼容的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！