Passing an array by reference

Question

How does passing a statically allocated array by reference work   void foo int   amp myArray  100        int main         int a 100       foo a       Does   amp myArray  100  have any meaning or its just a syntax to pass any array by reference  I don t understand separate parenthesis followed by big brackets here  Thanks

User · Answer

The following creates a generic function, taking an array of any size and of any type by reference:

template<typename T, std::size_t S>
void my_func(T (&arr)[S]) {
   // do stuff
}

play with the code.

User · Answer

It s a syntax for array references - you need to use   amp array  to clarify to the compiler that you want a reference to an array  rather than the  invalid  array of references int  amp  array 100      EDIT  Some clarification   void foo int   x   void foo int x 100    void foo int x       These three are different ways of declaring the same function  They re all treated as taking an int   parameter  you can pass any size array to them   void foo int   amp x  100      This only accepts arrays of 100 integers  You can safely use sizeof on x  void foo int  amp  x 100       error   This is parsed as an  array of references  - which isn t legal

User · Answer

It is a syntax  In the function arguments int   amp myArray  100  parenthesis that enclose the  amp myArray are necessary  if you don t use them  you will be passing an array of references and that is because the subscript operator    has higher precedence over the  amp  operator   E g  int  amp myArray 100     array of references  So  by using type construction    you tell the compiler that you want a reference to an array of 100 integers   E g int   amp myArray  100     reference of an array of 100 ints

User · Answer

Arrays are default passed by pointers  You can try modifying an array inside a function call for better understanding

User · Answer

It s just the required syntax   void Func int   amp myArray  100       Pass array of 100 int by reference the parameters name is myArray   void Func int  myArray      Pass an array  Array decays to a pointer  Thus you lose size information   void Func int   myFunc  double       Pass a function pointer  The function returns an int and takes a double  The parameter name is myFunc

[c++] Passing an array by reference

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