C++ 错误:将‘char*’赋值给‘char [2] 时的类型不兼容

2021-12-26 00:00:00 error-handling char c++

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‘Territory::Territory(std::string,char*, Player*, char)': Territory.cpp:15:33: 错误: 不兼容的类型将‘char*’赋值给‘char [2]’

    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]’

    我已经发现 char[2] <=>char* 但我不知道如何处理这个关于我的构造函数和 get/setter 的问题.

    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很好的理由,否则你应该使用 std::vectorstd::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]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为指向第一个元素的指针.所以这是有效的:

    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;
    

    但反过来不行:

    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*.所以在你的构造函数中,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).例如,我可以不能做这样的事情:

    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,或者使用一些为我执行此操作的函数,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,所以除非你有充分的理由,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::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).

相关文章