如何使用指向指针的指针在 main 中返回？

Question

void    ft_destroy(char ***factory);

int     main()
{
    char name[] = "sebastian";
    char *pt1 = &name;
    char **pt2 = &pt1;
    char ***pt3 = &pt2;
    printf("%s", ft_destroy(pt3));
    return (0);
}

error: incompatible pointer types initializing 'char *' with an expression of type 'char (*)[10]'
      [-Werror,-Wincompatible-pointer-types]
        char *pt1 = &name;

Answer 1

name is an array of 10 char , which is a char [10] . So &name is a pointer to an array of char , which is a char (*)[10] . Since pt1 is a char * , you should assign to it a pointer to a char .

Since name is an array of char , name[0] is a char , and &name[0] is a pointer to a char . So you can do char *pt1 = &name[0]; .

If you just use name , C will automatically convert it to a pointer to its first element, so you can also do char *ptr1 = name; .

Answer 2

char *pt1 = &name;

&name is address of type char[10] you need char (*pt1)[10] to store it.

Perhaps you can do as below.

char *pt1 = &name[0]; //or char *pt1 = name;

in this case &name[0] is of type char * .