Array of pointers to arrays

Question

I am new to C programming and this is my problem:

I want to store the first value of each array in a new array, then the second value of each array in a new array an

Answer 1

You have stored the data as you intended, you just need to access it properly

for (i=0; i<4;i++)
{
  for (j = 0; j < 4; j++) {
    int* temp = tab[i];    
    printf("%d\t", temp[j]);      // or try the next line...
    printf("%d\t", *(temp + j));  // prints same value as above line
    printf("%d\t", tab[i][j];     // the same value printed again
  }
}   

All of the above print the same value, it is just different ways of accessing that value using pointer arithmetic. Each element of tab is a int* the value of each is the address of your other defined int[] arrays at the start

Edit: In response to the comment of Jerome, you can achieve that by declaring 4 arrays

int tab1[4]={*t1,*t2,*t3,*t4};
int tab2[4]={*(t1+1),*(t2+1),*(t3+1),*(t4+1)};
int tab3[4]={*(t1+2),*(t2+2),*(t3+2),*(t4+2)};
int tab4[4]={*(t1+3),*(t2+3),*(t3+3),*(t4+3)};

Now tab1 contains the first elements of each array, tab2 the second elements, and so on. Then you can use

 int *tttt[4]={tab1,tab2,tab3,tab4};
 for (i=0; i<4;i++) {
   for (j = 0; j < 4; j++) {
     printf("%d\t", tttt[i][j]);   
   }
 } 

to print what you wanted. If you declared another pointer array like you did at the start

  int* tab[4] = {t1,t2,t3,t4};

then essentially in matrix terms, tttt is the transpose of tab