对三重指针的作用极为困惑

Question

int main()
{
    char** subject_array;
    char** courses_array;    

    initialize(subject_array, courses_array);
}


void initialize(char*** subject_array, char*** courses_array)
{
    int i;

    *subject_array = (char**) malloc(100 * sizeof(char*)); // 100 char pointers
    *courses_array = (char**) malloc(100 * sizeof(char*));

    for(i = 0; i < 100; i++) //malloc for subject_array
    {
        (*subject_array)[i] = (char*) malloc(4 * sizeof(char)); // 4 chars for each
        (*courses_array)[i] = (char*) malloc(6 * sizeof(char)); // char pointer
    }

} //void initialize

My question is: Why is there a discrepancy between the declared double pointers in main and the triple pointers in the initialize function? I think the triple pointer allows me to modify where the 2d array points.

My assignment gives these requirements:

Since you will be changing to where the pointers point, this will involve triple pointers for subjects and courses!

The above code takes care of bolded requirement but I have absolutely no idea what this even means. Can someone break it down to me? Does that mean I can modify contents of the 2d arrays of subject_array and courses_array? Such as swapping? But I can do that already with just double pointers making a 2d array! WHY triple pointers?

Answer 1

First of all, the code is incorret, because initialize asks for char*** 's and main gives it char** 's. To add another "pointer level", you take the address of whatever you want to give it to. So, the last line of main shall read initialize(&subject_array, &courses_array) (or shall it be return 0 ?)

After all, if you don't need to modify the variable subject_array and courses_array themselves, not what they point to , you may remove those pointers. And you actually do modify those variables themselves by means of *var = malloc(...) . And, as I already noted, the only problem is that the arguments to initialize are char***, char*** , not char**, char** as main is doing.