confused on output. returning two variables and still prints

Question

int addmult(int ii, int jj){
  int kk, ll;
  kk = ii + jj;
  ll = ii * jj;
  return (kk, ll);
}
void main(void){
  int i=3, j=4, k, l;
  k = addmult(i, j);
  l = addmult(i, j);
  printf("%d, %d\n", k, l);
}

I thought that it wasn't possible to return two variables from a function. How does the compiler know to print ll instead of kk? I know that in the function ii=3 and jj=4, k=7 and l=12, but then it goes on to return two variables. Could someone please elaborate on why it ends up printing:

12, 12

unDeadHerbs · Accepted Answer · 2017-11-01T20:56:27.877

1

The function isn't returning two values, you have stumbled upon the Comma operator.

/**
 *  Assigns value of b into i.
 *  Results: a=1, b=2, c=3, i=2
 */
int a=1, b=2, c=3;              
int i = (a, b);

The function is just returning the value of ll, you want to use reference parameters to return the two values.

void addmult(int ii, int jj, int* kk,int* ll){
  *kk = ii + jj;
  *ll = ii * jj;
}
void main(void){
  int i=3, j=4, k, l;
  addmult(i, j, &k, &l);
  printf("%d, %d\n", k, l);
}

edited Nov 01 '17 at 20:56

answered Nov 01 '17 at 20:53

unDeadHerbs

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this isn't C, is it? – Stephan Lechner Nov 01 '17 at 20:54
woops sorry, needs to pass by pointers, fixed – unDeadHerbs Nov 01 '17 at 20:55
@undeadherbs Thank you! i didnt know that you could use the reference parameter to return both values. I would verify your answer but the site redirected me to a question on the comma operate... Thank you for answering tho! – Alex Nov 01 '17 at 20:58
@undeadherbs never mind it says i can verify it in 3 minutes XD – Alex Nov 01 '17 at 20:59

confused on output. returning two variables and still prints

1 Answers1