How to add items to array in a function using pointers? - c

Question

I am trying to write this function that takes an array input and assigns random numbers between 1-52 to the array. Then prints back the numbers. I get an error message on the 4th line that says expression is not assignable. Is my use of pointers incorrect or is it not possible to add content to array with pointers?
void shuffleDeck(int *deck[]) {
int i;
for(i=0;i<=52;i++)
rand()%53 = deck[i]; // I get an error message on this line
}
int main() {
srand(time(NULL));
int deck[2000];
shuffleDeck(deck);
int i;
for(i=0;i<=52;i++){
printf("%d", deck[i]);
}
return 0;
}

Answer 1

Assigment direction is from right to left
So in this expression
deck[i] = rand()%52 + 1;
First is called function rand()
Then the result of function %52
Then 1 is added
And finnally its assigned to deck[i]
Array (what you have in main) is
0 1 2 3 4
+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 |
+---+---+---+---+---+
^-------------------------- arr[0]
While Array of pointers (what your function expects) is
0 1 2 3 4
+---+---+---+---+---+
| 0 | 0 | 0 | 0 | 0 | 0
+---+---+---+---+---+
------^
| +---+---+---+---+---+
| | 0 | 0 | 0 | 0 | 0 | 1
| +---+---+---+---+---+
| +---+---+---+---+---+
| | 0 | 0 | 0 | 0 | 0 | 2
| +---+---+---+---+---+
|
----------------------------- arr[0][0]
When array is passed to function, then arr[] and *arr are the same. But only as function arguments!
%53 returns 0-52,so if you need 1-52, use %52 + 1 (0-51 + 1 = 1-52)
If you cant solve it, here is code with fixed bugs.
void shuffleDeck(int deck[]) {
int i;
for(i=0;i<=52;i++)
deck[i] = rand()%52 + 1;
}
int main() {
srand(time(NULL));
int deck[2000];
shuffleDeck(deck);
int i;
for(i=0;i<=52;i++){
printf("%d", deck[i]);
}
return 0;
}