该警告是完全正确的。你想要的功能指针数组。你给它一个数组的数组。

The warning is exactly right. Your function wants an array of pointers. You're giving it an array of arrays.

预期：

 sep_foo:
 +------+       +-----+
 |char**|--> 0: |char*|-->"string1"
 +------+       +-----+
             1: |char*|-->"string2"
                +-----+
*sep_foo_qty-1: |...  |
                +-----+

您提供什么：

           sep_foo:
           +--------------------------------+
        0: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
        1: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
MAX_QTY-1: | ...                            |
           +--------------------------------+

与类型的元素 X 数组可以衰退为指针，以 - X 或 X * 。但 X值不允许在转换来改变。只有的有一个的衰变操作是允许的。你需要它来发生两次。在你的情况， X 是数组的 - MAX_STRING_LENGTH -chars。该函数要 X 是指针到字符。由于这些是不一样的，编译器会发出警告。我有点惊讶，这只是一个警告，因为没有什么好可以来自编译器允许发生的事情。

An array with elements of type X can "decay" into a pointer-to-X, or X*. But the value of X isn't allowed to change in that conversion. Only one decay operation is allowed. You'd need it to happen twice. In your case, X is array-of-MAX_STRING_LENGTH-chars. The function wants X to be pointer-to-char. Since those aren't the same, the compiler warns you. I'm a bit surprised it was just a warning since nothing good can come from what the compiler allowed to happen.

在你的功能，你可以写code：

In your function, you could write this code:

char* y = NULL;
*sep_foo = y;

这是合法的code，因为 sep_foo 是的char ** ，所以 * sep_foo 是的char * ，所以是是;你可以将它们分配。但你试图这样做， * sep_foo 不会的真正的是的char * ;将它指向字符的阵列。您code，实际上，将试图做到这一点：

That's legal code since sep_foo is a char**, so *sep_foo is a char*, and so is y; you can assign them. But with what you tried to do, *sep_foo wouldn't really be a char*; it would be pointing to an array of char. Your code, in effect, would be attempting to do this:

char destination[MAX_STRING_LENGTH];
char* y = NULL;
destination = y;

您可以不是指针分配到一个数组，所以编译器警告说，电话是没有好处的。

You can't assign a pointer into an array, and so the compiler warns that the call is no good.

有两种方法来解决这个问题：

There are two ways to solve this:

• 更改声明的方式，分配 sep_foo 呼叫方因此它匹配函数需要接受什么：

• Change the way you declare and allocate sep_foo on the calling side so it matches what the function expects to receive:

char** sep_foo = calloc(MAX_QTY, sizeof(char*));
for (int i = 0; i < MAX_QTY; ++i)
  sep_foo[i] = malloc(MAX_STRING_LENGTH);

或等价

char* sep_foo[MAX_QTY];
for (int i = 0; i < MAX_QTY; ++i)
  sep_foo[i] = malloc(MAX_STRING_LENGTH);

int parse(const char *foo, char sep_foo[MAX_QTY][MAX_STRING_LENGTH], int *sep_foo_qty);