在c中的循环中添加指向动态指针数组的指针

Question

I have a structure that represents a row in a table:

typedef struct {
    char *a;
    char *b;
} row;

and I have a function that initializes that row based on db data and returns a pointer to it

row* get_row(dbrow d) {
    row *r = malloc(sizeof(row));
    r->a = malloc(5);
    strcpy(r->a, d.a);
    r->b = malloc(5);
    strcpy(r->b, d.b);
    return r;
}

and finally, I have a function that has an row **rows as a parameter:

void get_rows(row **rows) {
    ...
    rows = malloc(rowNumber * sizeof(row*));
    int i;
    for (i = 0; i < rowNumber; i++) {
        rows[i] = get_row(dbrow);
    }
}

get_row works as expected and returns a pointer to valid row struct, but gdb shows that rows[0] (and all the others) never gets a new value, that is, it always points to the same address, almost as if the rows[i] = get_row(dbrow) line doesn't exist.

Answer 1

...gdb shows that rows[0] (and all the others) never gets a new value...

I'm assuming here that you are looking at the return value of your get_rows function, not at the value of its local variable rows . Here is the problem:

rows = malloc(rowNumber * sizeof(row*));

You are assigning a new value to a copy of the original pointer that the function received, not the original. This change will not be visible outside the function.

If you need to assign a new value to the argument then you will need to add another level of indirection. Remember; everything in C is passed by value. So, your function should take a row*** as its argument:

void get_rows(row ***rows) {
    if(!rows) {
        signal_some_error();
        return;
    }
    ...
    *rows = malloc(rowNumber * sizeof(row*));
    ...
}

---

Also, as user1700513 pointed out, you are assigning a row* to a row. That can't be your actual code as it would result in a compiler error.

Answer 2

change row r = malloc(sizeof(row)); to row* r = malloc(sizeof(row)); . I am wondering why you didn't get a compiler warning for this.