c - Dynamically allocating array explain

Question

This is sample code my teacher showed us about "How to dynamically allocate an array in C?". But I don't fully understand this. Here is the code:

int k;
int** test;
printf("Enter a value for k: ");
scanf("%d", &k);
test = (int **)malloc(k * sizeof(int*));
for (i = 0; i < k; i++) {
    test[i] = (int*)malloc(k * sizeof(int)); //Initialize all the values
}

I thought in C, to define an array you had to put the [] after the name, so what exactly is int** test; isn't it just a pointer to a pointer? And the malloc() line is also really confusing me.....

Answer 1

According to declaration int** test; , test is pointer to pointer, and the code pice allocating memory for a matrix of int values dynamically using malloc function.

Statement:

test = (int **)malloc(k * sizeof(int*));
    //                ^^------^^-------
    //  allocate for  k  int*  values    

Allocate continue memory for k pointers to int (int*). So suppose if k = 4 then you gets something like:

 temp      343  347  351  355
+----+    +----+----+----+----+
|343 |---?| ?  | ?  | ?  |  ? |
+----+    +----+----+----+----+

I am assuming addresses are of four bytes and ? means garbage values.

temp variable assigned returned address by malloc, malloc allocates continues memory blocks of size = k * sizeof(int**) that is in my example = 16 bytes.

In the for loop you allocate memory for k int and assign returned address to temp[i] (location of previously allocated array).

test[i] = (int*)malloc(k * sizeof(int)); //Initialize all the values
//                     ^^-----^^----------
//       allocate for  k   int  values    

Note: the expression temp[i] == *(temp + i). So in for loop in each iterations you allocate memory for an array of k int values that looks something like below:

   First malloc                     For loop   
  ---------------                  ------------------
       temp
      +-----+
      | 343 |--+
      +-----+  |
               ▼                    201   205   209    213  
        +--------+                +-----+-----+-----+-----+
 343    |        |= *(temp + 0)   |  ?  |  ?  |  ?  | ?   |  //for i = 0
        |temp[0] |-------|        +-----+-----+-----+-----+
        | 201    |       +-----------▲
        +--------+                  502   506  510    514
        |        |                +-----+-----+-----+-----+
 347    |temp[1] |= *(temp + 1)   |  ?  |  ?  |  ?  | ?   |  //for i = 1
        | 502    |-------|        +-----+-----+-----+-----+
        +--------+       +-----------▲
        |        |                  43    48    52    56
 351    | 43     |                +-----+-----+-----+-----+
        |temp[2] |= *(temp + 2)   |  ?  |  ?  |  ?  | ?   |  //for i = 2
        |        |-------|        +-----+-----+-----+-----+
        +--------+       +-----------▲
 355    |        |
        | 9002   |                 9002  9006   9010 9014
        |temp[3] |                +-----+-----+-----+-----+
        |        |= *(temp + 3)   |  ?  |  ?  |  ?  | ?   |  //for i = 3
        +--------+       |        +-----+-----+-----+-----+
                         +-----------▲

Again ? means garbage values.

Additional points:

1) You are casting returned address by malloc but in C you should avoid it. Read Do I cast the result of malloc? just do as follows:

test = malloc(k* sizeof(int*));
for (i = 0; i < k; i++){
    test[i] = malloc(k * sizeof(int));
}

2) If you are allocating memory dynamically, you need to free memory explicitly when your work done with that (after freeing dynamically allocated memory you can't access that memory). Steps to free memory for test will be as follows:

for (i = 0; i < k; i++){
    free(test[i]);
}
free(test);

3) This is one way to allocate memory for 2D matrix as array of arrays if you wants to allocate completely continues memory for all arrays check this answer: Allocate memory 2d array in function C

4) If the description helps and you want to learn for 3D allocation Check this answer: Matrix of String or/ 3D char array