How to pass array by reference in my actual parameters

Question

I came across an error when I tried to pass array by reference using int(& a)[5]. But when I wanna implement it ,we should pass a reference which I initialize as int & refer=*a; Following is my code ,and two ways to pass reference and pointer to the calling function . How to explain that I can't simply pass a reference into it?

 #include <iostream>

using namespace std;

void add(int (&)[5],int );
int main()
 {


int a[5]={5,4,3,2,1};
int len=sizeof(a)/sizeof(a[0]);
cout<<"len="<<len<<endl;
cout<<"a="<<a<<endl;// the name of array is actually the pointer

int& refer=*a;

add(a,len);    //which is correct,we pass a pointer to the calling function;
add(refer,len);   //which is wrong,we pass a reference to the calling function;
for(int i=0;i<len;i++){
    cout<<a[i]<<" ";
}

return 0;
}



void add(int (&a)[5],int len){
  for(int i=0;i<len;i++){
   a[i]=a[i]+10;
  }
}

If add only takes an array of size 5, why do you pass the length? You only need int a[] to pass an array for modification in a function. In C++ you can use std::vector<int>& and take advantage of the fact a vector knows its size. — Neil Kirk
– Neil Kirk, Commented Oct 2, 2014 at 1:53
Because I wanna know how many times I should loop. When I pass a[] into a calling function, a[] decays to the pointer to the first element.In that sense, I would lose my information about the size of this array,which means I should add the length of my array together. — user3570984
– user3570984, Commented Oct 2, 2014 at 2:01
@user3570984: With this signature, a doesn't decay to pointer, so you know the size (and may remove int len). — Jarod42
– Jarod42, Commented Oct 2, 2014 at 7:14

Cheers and hth. - Alf · Accepted Answer · 2014-10-02 01:57:18Z

2

int& refer=*a;

This is not a reference to the array, but to its first element. Try:

int (&refer)[5] = a;
add(refer,len);

edited Oct 2, 2014 at 1:57

Cheers and hth. - Alf

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answered Oct 2, 2014 at 1:53

AlexD

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3 Comments

user3570984 Over a year ago

As for int (&refer)[5] = a; I don't quite understand why you can assign pointer,like a,to a reference.

Cheers and hth. - Alf Over a year ago

@user3570984: Because a is not a pointer. It's an array. The expression a decays to pointer in certain contexts, essentially where a pointer is expected. Also note that this is not assignment. It's initialization, although expressed with an = sign.

AlexD Over a year ago

@user3570984 You can have a reference to an array (or a pointer) pretty much the same way as a reference to an integer. The line int (&refer)[5] = a; says " refer is a reference to a".

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