Add element to array member of a struct

Question

I have an union:

typedef union { int arr1[5]; char arr2[5]; } type

and a struct:

typedef struct { int size; type elt; } foo

and finally a variable:

foo bar = { 3, {1,2,3} }.

I want to add an element to bar. Therefore I defined:

void myfun(foo x, int new) 
{
      x.elt.arr1[x.size] = new;
      x.size += 1;
      return;
}

If I call myfun(bar,4) it should change bar appropriately. But if I view the elements of bar before and after the call of myfun it will print bar without the additional element.

Edit: My view function is:

void myprint(foo x)
{
     for ( int i = 0; i < x.size; i++) {
           printf("%d ", x.elt.arr1[i]);
      }
      return;
}

What's my fault?

How do you "view the elements of bar"? Can you provide a stackoverflow.com/help/mcve ? — aschepler
– aschepler, Commented Sep 6, 2016 at 17:49
Function arguments are passed by value, so you aren't modifying the original structure but a copy. — interjay
– interjay, Commented Sep 6, 2016 at 17:51
change your call by reference (&) and modifiy a bit your function to use a pointer , and it should be ok — Gar
– Gar, Commented Sep 6, 2016 at 18:28

L. Zeda · Accepted Answer · 2016-09-06 18:13:03Z

1

As interjay wrote, you are not modifying bar, but just a coppy (that is created separately in memory everytime when function myfun is called). Look up difference between functions called by value and functions called by reference. What you need is:

void myfun(foo *x, int new) 
{
    x->elt.arr1[x->size] = new;
    x->size += 1;
    return;
}

and then:

myfun(&bar,4)

That way, variable bar will be edited.

edited Sep 6, 2016 at 18:13

answered Sep 6, 2016 at 18:06

L. Zeda

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1 Comment

L. Zeda Over a year ago

Yeah, my bad. Edit that post if you find something wrong, I can't check it on this PC and I am horrible with syntax :D.

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