C ++错误:赋值为'char *'到'char [2]的不兼容类型 [英] C++ Error: Incompatible types in assignment of ‘char*’ to ‘char [2]
问题描述
在我的头文件中我声明:
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>
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
赋值为'char *'到'char [2]'
我已经弄清楚 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。
p> C ++中的原始数组是一种恼人的,充满危险的。这就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
。
如其他人所说,首先, char [2]
与 char *
,或者至少不是通常的。 char [2]
是 char
和 char * $ c的大小2数组$ c>是指向
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。所以这样做:
char foo [2];
char * bar = foo;
但反之亦然:
const char * bar =hello;
const char foo [6] = bar; // ERROR
在声明函数参数时,添加到混淆, char [ ]
相当于 char *
。所以在你的构造函数中,参数 char short_name [2]
真的是 char * short_name
。
数组的另一个奇怪之处在于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)。所以例如我可以不这样做:
char foo [2] = {' a','b'};
char bar [2] = foo;
相反,我必须迭代 foo
并将它们复制到 bar
中,或者使用一些对我来说的功能,例如 std :: copy
:
char foo [2] = {'a','b'};
char bar [2];
// std :: begin和std :: end仅在C ++ 11中可用
std :: copy(std :: begin(foo),std :: end(foo),std: :开始(巴));
所以在你的构造函数中你必须手动复制 short_name
into short_name _
:
Territory :: Territory std :: string name,char * short_name,Player * owner,
char units):name_(name),owner_(owner),units_(units)
{
//注意std :: begin和std :: end可以*不*用于指针。
std :: copy(short_name,short_name + 2,std :: begin(short_name));
}
你可以看到这一切都很烦人,所以除非你有一个非常很好的理由,你应该使用 std :: vector
而不是原始数组(或在这种情况下可能 std :: string
)。
I have a bit of a problem with my constructor. In my header file I declare:
char short_name_[2];
- and other variables
In my constructor:
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_; }
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)
{ }
My error:
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]’
I already figured out that char[2] <=> char*
but I'm not sure how to handle this about my constructor and get/setters.
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
.
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;
But the reverse does not:
const char* bar = "hello";
const char foo[6] = bar; // ERROR
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
.
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;
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));
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));
}
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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