为什么功能（的char * []数组）有效的函数定义，但不（CHAR（*数组）的C []？ [英] Why is function(char * array[]) a valid function definition but not (char (*array)[] in C?
问题描述
我的认为的，这是因为前者是一个指针数组为char，后者是一个指向字符数组，我们需要正确指定对象的大小被指向为我们的函数定义。在前者;
I think that it is because the former is an array of pointers to char and the latter is a pointer to an array of chars, and we need to properly specify the size of the object being pointed to for our function definition. In the former;
function(char * p_array[])
的
the size of the object being pointed to is already included (its a pointer to char), but the latter
function(char (*p_array)[])
需要阵列p_array点的大小为p_array定义的一部分？
我在哪里，我一直在思考这个太久，刚刚混淆自己的舞台，有人请让我知道如果我的推论是正确的。
needs the size of the array p_array points to as part of p_array's definition? I'm at the stage where I've been thinking about this for too long and have just confused myself, someone please let me know if my reasoning is correct.
推荐答案
两者都是有效的用C ，但不是C ++ 。你的一般的是正确的：
Both are valid in C but not C++. You would ordinarily be correct:
char *x[]; // array of pointers to char
char (*y)[]; // pointer to array of char
然而，阵列衰变的指针，如果它们显示为函数参数。因此，他们成为：
However, the arrays decay to pointers if they appear as function parameters. So they become:
char **x; // Changes to pointer to array of pointer to char
char (*y)[]; // No decay, since it's NOT an array, it's a pointer to an array
在C中的数组类型，尺寸之一是允许不指定。这必须是最左边的一（哎呦，我最右边的说在第一次）。因此，
In an array type in C, one of the sizes is permitted to be unspecified. This must be the leftmost one (whoops, I said rightmost at first). So,
int valid_array[][5]; // Ok
int invalid_array[5][]; // Wrong
（你可以连他们......但我们很少有这样做的原因......）
(You can chain them... but we seldom have reason to do so...)
int (*convoluted_array[][5])[][10];
有一个陷阱，并美中不足的是，与 [] 在一个数组类型是的不完全型。的可以围绕一个指针传递给一个不完整的类型，但某些操作将无法正常工作，因为他们需要一个完整的类型。例如，这是不行的：
There is a catch, and the catch is that an array type with [] in it is an incomplete type. You can pass around a pointer to an incomplete type but certain operations will not work, as they need a complete type. For example, this will not work:
void func(int (*x)[])
{
x[2][5] = 900; // Error
}
这是一个错误，因为为了找到的地址x [2] ，编译器需要知道有多大 X [0] 和 X [1] 是。但 X [0] 和 X [1] 具有类型 INT [] - 不完全类型有没有关于它有多大的信息。这将成为你想象的类型未腐烂的版本会是这样，这是更清晰的 INT X [] [] - 显然无效C.如果你想各地传递一个二维数组中的C，你有几种选择：
This is an error because in order to find the address of x[2], the compiler needs to know how big x[0] and x[1] are. But x[0] and x[1] have type int [] -- an incomplete type with no information about how big it is. This becomes clearer if you imagine what the "un-decayed" version of the type would be, which is int x[][] -- obviously invalid C. If you want to pass a two-dimensional array around in C, you have a few options:
-
传递的尺寸参数的一维数组。
Pass a one-dimensional array with a size parameter.
void func(int n, int x[])
{
x[2*n + 5] = 900;
}
使用指针数组的行。这有点笨拙，如果你有真正的2D数据。
Use an array of pointers to rows. This is somewhat clunky if you have genuine 2D data.
void func(int *x[])
{
x[2][5] = 900;
}
使用一个固定大小的。
Use a fixed size.
void func(int x[][5])
{
x[2][5] = 900;
}
使用可变长度阵列（C99而已，所以它可能不会与微软的编译器工作）。
Use a variable length array (C99 only, so it probably doesn't work with Microsoft compilers).
// There's some funny syntax if you want 'x' before 'width'
void func(int n, int x[][n])
{
x[2][5] = 900;
}
这是一个常见的问题，甚至区域对C退伍军人。许多语言缺乏内在的乱用的真正的，可变大小的支持，多维数组（C ++，Java和Python）的虽然有少数语言确实有它（Common Lisp的，哈斯克尔，Fortran语言）。你会看到很多使用数组的数组或手动计算数组偏移code的。
This is a frequent problem area even for C veterans. Many languages lack intrinsic "out-of-the-box" support for real, variable size, multidimensional arrays (C++, Java, Python) although a few languages do have it (Common Lisp, Haskell, Fortran). You'll see a lot of code that uses arrays of arrays or that calculates array offsets manually.
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