C++ 初始化字符**
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Initialize char**
提问by Boyang
I'm very new to C++. I'm trying to call a function that takes in char**:
我对 C++ 很陌生。我正在尝试调用一个接受 char** 的函数：
bool func(char** a) {
//blablabla
}
So it takes in an array of c-strings. I need to create a char**, but nothing works.
所以它需要一个 c 字符串数组。我需要创建一个字符**，但没有任何效果。
char** a = char[255][255]; // error: type name is not allowed
char** a = new char[255][255]; // error: a value of type "char (*)[255]" cannot be used to initialize an entity of type "char **"
char a[][] = {"banana", "apple"};
char** b = &a; // error: a value of type "<error-type> (*)[2]" cannot be used to initialize an entity of type "char **"
At the end I need to do:
最后我需要做的是：
char* a[] = {"banana", "apple"};
Why the first few didn't work and why the last one worked?
为什么前几个不起作用，为什么最后一个起作用？
Thanks in advance.
提前致谢。
采纳答案by Shoe
There's a lot wrong in your code.
你的代码有很多错误。
char** a = char[255][255]; // error: type name is not allowed
First of all this is not even valid C++ (or C for that matter). Maybe you meant:
首先，这甚至不是有效的 C++（或 C）。也许你的意思是：
char a[255][255];
In any case always remember that the type of a bi-dimensional dynamically allocated array is not **but (*)[N]which is very different.
在任何情况下都要记住，二维动态分配数组的类型不是**但这(*)[N]是非常不同的。
char** a = new char[255][255]; // error: a value of type "char (*)[255]" cannot be used to initialize an entity of type "char **"
The error message you provide in the comment explains exactly what I said earlier.
您在评论中提供的错误消息完全解释了我之前所说的内容。
char a[][] = {"banana", "apple"};
In the above code the correct type of the variable ashould be char* a[]. Again, arrays and pointer (for what the type is concerned) are very different things. A chararray may decay to pointer (if NULL?terminated), but for the rest, except with explicit casts, you can't use pointers and arrays like you are doing.
在上面的代码中，变量的正确类型a应该是char* a[]. 同样，数组和指针（对于类型而言）是非常不同的东西。一个char阵列可以衰减到指针（如果NULL？终止），但其余，除具有显式转换，则不能使用指针和数组一样，你在做什么。
The last one worked because, like I said earlier, char* []is the correct type for an array of C-strings.
最后一个有效，因为正如我之前所说，它char* []是 C 字符串数组的正确类型。
Anyway, if you just doing homework, it is ok to learn this things. But in future development using C++: try not to use "features" that start with C-, like C-strings, C-arrays, etc. C++'s standard library gives you std::string, std::array, std::vectorand such for free.
不管怎样，如果你只是做功课，学习这些东西是可以的。但在使用C ++未来的发展：尽量不要使用与启动“特色” C-，像C字符串，C-阵列等C ++标准库给你std::string，std::array，std::vector和这样的免费。
If you reallyneed to allocate dynamic memory (with newand delete, or new[]and delete[]) please use smart pointers, like std::shared_ptror std::unique_ptr.
如果您确实需要分配动态内存（使用newanddelete或new[]and delete[]），请使用智能指针，例如std::shared_ptror std::unique_ptr。
回答by Hyman
You say you are working in C++. Then you can easily ignore const char*and char**and focus about what you can use:
你说你正在使用 C++。然后，您可以轻松忽略const char*并char**专注于您可以使用的内容：
#include <string>
#include <vector>
std::vector<std::string> arrayOfStrings;
arrayOfStrings.push_back("foo");
bool func(const std::vector<std::string>>& a) {
..
}
If you know the size at compile time you can even use std::array:
如果您知道编译时的大小，您甚至可以使用std::array：
std::array<255, std::string> fixedArrayOfStrings
EDIT: since you need to build an array of C strings in any case you can easily do it starting from the vector:
编辑：因为您需要在任何情况下构建一个 C 字符串数组，您可以从向量开始轻松地完成它：
const char **arrayOfCstrings = new const char*[vector.size()];
for (int i = 0; i < vector.size(); ++i)
arrayOfCstrings[i] = vector[i].c_str();
func(arrayOfCstrings);
delete [] arrayOfCstrings;
回答by Zaffy
char**
is ambiguous - it can mean:
不明确 - 它可能意味着：
- pointer to pointer
- array of c-strings - experienced programmer would write
char* arr[]instead
- 指向指针的指针
- C-字符串数组-有经验的程序员会写
char* arr[]，而不是
In the first case it is quite simple:
在第一种情况下，它非常简单：
char* niceString = GetNiceString();
func(&niceString);
however in the second case it is slightly more complex. The function will not know the length of the array so you need to end it explicitly with a NULL, just like for example environis:
然而，在第二种情况下，它稍微复杂一些。该函数不知道数组的长度，因此您需要以 a 显式结束它NULL，例如environ：
char* a[3] = { "One", "Two", NULL }; /* note that this is possibly dangerous
because you assign const char* (READ-ONLY) to char* (WRITABLE) */
func(a); // char*[] gets downgraded to char** implicitly
回答by dare
so char** a = char[255][255];it's weird to C and C++
and if you want a static 2d array just
所以char** a = char[255][255];这对 C 和 C++ 来说很奇怪，如果你想要一个静态的二维数组
char a[255][255];
