我需要帮助来了解这一行代码来动态创建数组吗？

Question

I am starting to get more practice with malloc and although there is nothing wrong with the code it executes properly.

int *arr = (int *)malloc(x * y * sizeof(int));
int i, j, r;

for(i=0; i<x; i++){
    for(j=0; j<y; j++){
        r = 0 + rand() % 7;
        *(arr + i*y + j) = r; 
   //I dont understand the left hand portion of the above line.

//x and y are both 5000

I found it online and before I found it, I had tried doing the exact same thing but I guess my syntax was wrong. Anyways, I need help understanding the line that has the comment next to it

Answer 1

*(arr + i*y + j) is trying to store value in the array element denoted by array[i][j] .

What value you receive from r in the previous line, is being fed to the array[i][j] ^th element of the array.

arr locates the base element of the array, adding i*y locates to the (i+1)th row of the array (index count starts from 0 in C) , and adding j to that produces the exact column(j+1 th column) in which array[i][j] is located.

Answer 2

The block of memory allocated in the first line is being used as a two-dimensional matrix. For more information, read this: row/column-major order .

Here is a diagram:

That diagram probably should be transposed based on the use of y in the expression, but the concept is the same. The numbers in each box is the linear offset. As you can see, the offset into the linear block of memory is equal to the row index multiplied by the width plus the column index, or i*y + j .

Answer 3

Normally what you do is declare a 2D array int arr[5000][5000]; or the equivalent malloc code.

What is done in this case, is to declare a single dimensional array, but to use it as a 2-D array. This way is also possible, but it involves the extra overhead of constantly checking the location.