C++ passing array using reference operator

Question

I have a question about passing array with reference operator. I want to write code that pass array using reference operator. Then I tried

void swap(int &testarray[3]){
 // code
}

It gives me error. It says,

/main.cpp:5: error: declaration of 'testarray' as array of references

However when changed my code with

void swap(int (&testarray)[3]){
  // code
  }

It runs properly. Only difference is having bracket.

Why it needs bracket and what is the difference between int (&testarray)[3] and int &testarray[3]

Thanks for helping.

@Cornstalks Why wouldn't you want to pass an array by reference? @burakim: Since you've tagged this C++11, you could switch to std::array<int, 3> and avoid such vagaries of C declarator syntax. — Praetorian
– Praetorian, Commented Jan 6, 2014 at 18:31
@Praetorian: because arrays can't be assigned to, and already decay to pointers. — Cornstalks
– Cornstalks, Commented Jan 6, 2014 at 18:57
@Cornstalks I don't see how not being assignable is relevant, you can still assign to individual elements. And the decaying to pointer loses size information, which makes it inconvenient for most use cases. — Praetorian
– Praetorian, Commented Jan 6, 2014 at 19:01

Jarod42 · Accepted Answer · 2014-01-06 18:29:10Z

7

void foo(int &testarray[3]) is interpreted as void foo((int &)testarray[3]) due to priority. And array of references are illegal.

Whereas void foo(int (&testarray)[3]) is interpreted as you want. (reference of an array of 3 int).

void foo(int testarray[3]) is equivalent to void foo(int testarray[]) which decays to void foo(int *testarray). (int pointer).

answered Jan 6, 2014 at 18:21

Jarod42

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Comments

Vlad from Moscow · Accepted Answer · 2014-01-06 18:21:17Z

0

In fact this construction

int & testarray[3]

defines an array of references to integral objects. The C++ Standard does not allow to define arrays of references to objects.

answered Jan 6, 2014 at 18:21

Vlad from Moscow

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