Pass array from one function to another

Question

This is what I want to do:

int randomtal(void){
  int loop;
  int tal[10];
  srand(time(0));

  for(loop = 0; loop < 10; loop++){
      tal[loop] = rand() % 10+1;
  }
  return tal;
}


int upp1(int argc, const char * argv[]) {
   int Tal[10];
   Tal[] = randomtal();
   return 0;
}

Simply the randomtal() function generates 10 numbers and put these in an array. I then want to pass this array to the upp1() function and put these in the Tal array. Basically make a copy of tal[] in function randomtal() and pass this to Tal in function upp1().

Tal[] is the array I want to store all numbers in function upp1 and randomtal(); is me calling that function to generate numbers — J. Doe
– J. Doe, Commented Jan 25, 2018 at 14:40

dbush · Accepted Answer · 2018-01-25 14:41:35Z

2

You can change your function to accept an array as a parameter. And because an array as a function argument is actually a pointer to the first element of the array, changes to the array in the function are reflected in the caller.

void randomtal(int tal[10]){
  int loop;
  srand(time(0));

  for(loop = 0; loop < 10; loop++){
      tal[loop] = rand() % 10+1;
  }
}


int main(int argc, const char * argv[]) {
   int Tal[10];

   randomtal(Tal);

   return 0;
}

answered Jan 25, 2018 at 14:41

dbush

233k27 gold badges261 silver badges334 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

J. Doe Over a year ago

Worked like a charm. Thank you!

Collectives™ on Stack Overflow

Pass array from one function to another

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest