Shift all non-null elements to the left - c

Question

The elements must be shifted to the left if they are 0. I tried a loop that starts from the end of the 2D Matrix Array, but I can't seem to get my head around why it doesn't work.
The input must be in range -999 to 999.
I assign the element that is != 0 to the first index and afterwards if have to move that temporary integer one step forward. That is my main issue, since I can't get my head around it. Perhaps a different approach is wanted?
(I'm trying to solve it without Pointers)
int main()
{
int M[3][3];
int j = 0;
int i = 0;
for(i = 0; i < 3; ++i){
for(j = 0; j < 3; ++j){
printf("Give me the number for [%d][%d]: ", i, j);
scanf("%d", &M[i][j]);
if(-999 > M[i][j] || M[i][j] > 999) j--;
// a)
if((i + j) % 2 == 0) {
M[i][j] = 0;
}
}
}
// Print the result of a)
for(i = 0; i < 3; ++i){
for(j = 0; j < 3; ++j){
printf("%d ", M[i][j]);
}
printf("\n");
}
// Shifting the elements to the left b)
int temp = M[0][0];
for(i = 2; i >= 0; i--)
{
for(j = 2; j >= 0; j--){
if(M[i][j] != 0)
{
M[i][j] = temp;
}
// change the value of temp
}
}
// Print the array after a) and b)
for(i = 0; i < 3; ++i){
for(j = 0; j < 3; ++j){
printf("%d ", M[i][j]);
}
}
}
The result of the a) is just the matrix after the first step where if (i=j) converts the value at that place 0.
For example if you input: 1 2 3 4 5 6 7 8 9
The result would be: 0 2 0 4 0 6 0 8 0
The final result of b) is yet to be done, but should look like:
2 4 6 8 0 0 0 0 0.

Answer 1

Perhaps a different approach is wanted?
As a [3][3] array is in memory like a [1][3*3] one, simple shift M[0] as if it had 9 elements.
unsigned nonzero_index = 0;
for(i = 0; i < 3*3; ++i){
if (M[0][i]) {
M[0][nonzero_index++] = M[0][i];
}
}
while (nonzero_index < 3*3) { // zero fill the rest
M[0][nonzero_index++] = 0;
}