c ++：function arg char **不同于char * [] [英] c++: function arg char** is not the same as char*[]

问题描述

我使用g ++。我使用的代码，有一个 main（int，char **），重命名，所以我可以调用它。我查看了 http://stackoverflow.com/questions / 779910 / should-i-use-char-argv-or-char-argv-in-c ，其中 char ** char * [] 。这在c ++函数调用中似乎不是真的。例如：

I am using g++. I am using code that had a main(int,char**), renamed so I can call it. I looked at http://stackoverflow.com/questions/779910/should-i-use-char-argv-or-char-argv-in-c, where char** is said to be equivalent to char* []. This does not appear to be true in c++ function calls. For example:

void f1(char** p){;}

void f2(char* p[]){

   f1(p);

 //...`

}

失败与编译器抱怨不能转换 char（*）[] 到 char ** ...引用我看看数组被转换为指针的调用，但是似乎不是这样的情况：

fails with the compiler complaining "cannot convert char (*)[] to char**..." The references I look to say that arrays are converted to pointers for the call, but this does not seem to be the case as:

void f3(char* [] p);

char caa[16][16];  
f3(caa);

也会失败。我假设只要间接级别相同（例如 char *** ptr 和 char [] [] [] carray ）类型是可以互换的。

also fails. I had assumed that as long as the levels of indirection were the same (e.g. char*** ptr and char[][][] carray ) the types were interchangeable.

有人可以提供我可以审查的参考，澄清这些问题吗？

Can someone provide a reference I can review that clarifies these issues?

谢谢。

推荐答案

这在C ++中仍然成立。如果你的编译器抱怨你描述的第一种情况，它是不一致的。

This still holds true in C++. If your compiler complains as you describe for your first case, it is non-conformant.

为了解释你的第二种情况，重要的是要了解实际发生了什么。数组类型的表达式可隐式转换为相应的指针类型，即： T [n] - > T * 。但是，如果 T 本身是一个数组，则不会特别处理这种情况，并且数组到指针衰减不会传播 。所以 T * [n] 衰减到 T ** ，但 T [x] y] 只会衰减到 T [y] * ，并且不会进一步。

To explain your second case, it is important to understand what actually happens. An expression of array type is implicitly convertible to a corresponding pointer type, i.e.: T[n] -> T*. However, if T itself is an array, this case isn't treated specially, and array-to-pointer decay does not propagate. So T*[n] decays to T**, but T[x][y] will only decay to T[y]*, and no further.

从实现的角度来看，这是有意义的，因为进一步衰减，如果允许，将给予 T ** ，这是指针的指针;而2D C阵列不实现为锯齿状阵列（即，指向阵列的指针阵列） - 它们形成单个连续的存储器块。所以，没有 T * 里面数组取地址给你一个 T ** 。对于允许的情况，典型的实现只是将数组的地址作为一个整体，并将其转换为指向单个元素的指针类型（当底层指针表示对于所有类型都是相同的，通常情况下，这种转换是一个

From implementation perspective this makes sense, because decaying further, if allowed, would give T**, which is pointer to pointer; whereas 2D C arrays aren't implemented as jagged arrays (i.e. array of pointers to arrays) - they form a single contiguous memory block. So, there's no T* "inside" the array to take an address of to give you a T**. For the allowed cases, a typical implementation simply takes the address of the array as a whole and converts it to type of pointer to single element (when underlying pointer representation is the same for all types, as is usually the case, this convertion is a no-op at run time).

这里的规范性引用是ISO C ++ 03，4.2 [conv.array] / 1：

The normative reference here is ISO C++03, 4.2[conv.array]/1:

类型NT数组或T的未知边界数组的左值或右值可以转换为类型指向T的指针的右值。 result是指向数组的第一个元素的指针。

An lvalue or rvalue of type "array of N T" or "array of unknown bound of T" can be converted to an rvalue of type "pointer to T." The result is a pointer to the first element of the array.

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