为什么不能烧焦**在C ++下面的函数的返回类型? [英] Why can't char** be the return type of the following function in C++?
问题描述
我在C ++中有以下功能:
I have the following function in C++ :
char** f()
{
char (*v)[10] = new char[5][10];
return v;
}
Visual Studio 2008中说以下内容:
Visual studio 2008 says the following:
error C2440: 'return' : cannot convert from 'char (*)[10]' to 'char **'
究竟应该怎样返回类型是,为了让此功能工作?
What exactly should the return type to be, in order for this function to work?
推荐答案
的char **
是不是同一类型为字符(*) 10]
。这两者都是不兼容的类型等字符(*)[10]
不能隐式转换为的char **
。因此,编译错误。
char**
is not the same type as char (*)[10]
. Both of these are incompatible types and so char (*)[10]
cannot be implicitly converted to char**
. Hence the compilation error.
函数的返回值类型看起来非常难看。你必须把它写成:
The return type of the function looks very ugly. You have to write it as:
char (*f())[10]
{
char (*v)[10] = new char[5][10];
return v;
}
现在它编译。
或者你可以使用的typedef
为:
typedef char carr[10];
carr* f()
{
char (*v)[10] = new char[5][10];
return v;
}
Ideone 。
基本上,字符(* V)[10]
定义一个指针指向大小为10的字符
阵列。这是相同的,如下:
Basically, char (*v)[10]
defines a pointer to a char
array of size 10. It's the same as the following:
typedef char carr[10]; //carr is a char array of size 10
carr *v; //v is a pointer to array of size 10
所以你的code变成等价于:
So your code becomes equivalent to this:
carr* f()
{
carr *v = new carr[5];
return v;
}
cdecl.org
帮助这里:
-
的char v [10]
内容申报V作为字符数组10
-
字符(* V)[10]
内容申报V作为字符指针数组10
char v[10]
reads asdeclare v as array 10 of char
char (*v)[10]
reads asdeclare v as pointer to array 10 of char
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