Passing argument by reference

Question

Having an array:

int a = { 1 ,2 , 3}

I can pass a pointer to a function in order to modify it.

int modify( int *a ){
   a[0] = 10;
}

But why can't I pass a reference to int to modify it? e.g

int modify( int &a ){
   a[0] = 10;
}

How does compiler manipulate with a reference? By using pointers, we pass the memory (in this case, the first element of the array ). But what happens with a reference? Why, for example, does this work?

vector<int > a 

void mod( vector<int> & a ){
      a[0] = 10;
     //a.push_back(10)
}

`int modify(int &a ){ a = 10; }` – nwp Nov 07 '16 at 22:14 — nwp, Nov 07 '16 at 22:14
you don't have an array. it should be `int a[] = {1,2,3};` – Bryan Chen Nov 07 '16 at 22:15 — Bryan Chen, Nov 07 '16 at 22:15

R Sahu · Answer 1 · 2016-11-07T22:20:40.120

0

but why cant i pass reference to int to modify it?

You sure can.

The syntax needs to be a bit different for that.

int modify( int (&a)[3] ){
   a[0] = 10;
}

The difference between using

int modify( int *a ){ ... }

and

int modify( int (&a)[3] ){ ... }

is that you can call the first one using arrays of different sizes while the second one can be called only with arrays of size 3.

edited Nov 07 '16 at 22:20

answered Nov 07 '16 at 22:13

R Sahu

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Passing argument by reference

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