pass struct array to a function by reference

Question

there have been a lot of resources on this but I could not really understand the concept of this.

Here is my attempt using pointers:

struct PEOPLE{
   int id;
   string name;
   float cash;
};

int main(){
   PEOPLE data[5];
   getdata(data);

   for(int i = 0; i < 5; i++){
      cout << data[i].id << " " << data[i].name << "  " << data[i].cash;
   }
   return 0;
}

void getdata(PEOPLE &*refdata[5]){
    refdata[0].id = 11;
    refdata[0].name = "John Smith";
    refdata[0].cash = 200.30;
    //and so on for index 1,2,3,4
}

Is this approach correct, I doubt it will work.

Shouldn't the structure definition end with a semicolon?

Aswin Murugesh
– Aswin Murugesh

2013-12-01 08:47:45 +00:00
Commented Dec 1, 2013 at 8:47 — Aswin Murugesh
– Aswin Murugesh, Commented Dec 1, 2013 at 8:47

juanchopanza · Accepted Answer · 2013-12-01 08:47:06Z

1

This would work for arrays of a size known at compile time:

template<size_t N >
void getdata(PEOPLE (&refdata)[N] )
{
   refdata[0] = ....;
}

To restrict to size 5,

void getdata(PEOPLE (&refdata)[5] )
{
   refdata[0] = ....;
}

answered Dec 1, 2013 at 8:47

juanchopanza

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16 Comments

juanchopanza Over a year ago

@edward That happens to be the required syntax for passing a (non-const) reference to a fixed size array. I must admit that doesn't make much sense to me, but the idea is to distinguish between a reference to an array and an array of references.

ETAN Over a year ago

thanks for your help, one more q, when i pass by reference, should I do like this getdata(data); or this getdata(data[5]);?

juanchopanza Over a year ago

@edward Use getdata(data);. getdata(data[5]) passes a single (invalid) element to the function. It is actually an out of bounds array access.

user529758 Over a year ago

But the question still remains: why would you ever do this? Arrays already decay into pointers; passing by reference doesn't bring much benefits, and it just causes confusion...

juanchopanza Over a year ago

@H2CO3 because it restricts it to a certain size, and you don't have to pass the size as a second parameter (second case), and it gives you a handle on the size in the function template case. It is making use of compile-time information lost when an array is allowed to decay into a pointer.

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yosim · Accepted Answer · 2013-12-01 09:22:44Z

0

You can simply define it as:

void getdata(PEOPLE refdata[])

and call:

 getdata(data);

answered Dec 1, 2013 at 9:22

yosim

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