c - 如何在函数中传递参数数组指针？

Question

我有这个代码:

// *** foo.c ***

#include <stdlib.h> // malloc

typedef struct Node {
    char (*f)[20];
    //...
} Node;

int function(char f[30][20]){
    // Do some stuff with f...
}

int main(){
    Node * temp = (Node*)malloc(sizeof(Node));
    function(temp->f[20]);
}

我想将 temp->f 数组指针传递给函数，但在编译时出现以下错误:

$ gcc foo.c -o foo
foo.c: In function ‘main’:
foo.c:16:5: warning: passing argument 1 of ‘function’ from incompatible pointer type [enabled by default]
     function(temp->f[20]);
     ^
foo.c:10:5: note: expected ‘char (*)[20]’ but argument is of type ‘char *’
    int function(char f[30][20]){
        ^

我该如何解决？

Answer 1

expected 'char (*)[20]' but argument is of type 'char *

将调用函数的方式更改为:

function(temp->f);

还有:

• Why you should not cast the result of malloc()
• Why you should always check the result of malloc()