Writing array directly to parameter gives error in c++

Question

#include <iostream>

using namespace std;

template <typename VAR, unsigned int N>
int size(VAR (&arr)[N])
{
return sizeof(arr) / sizeof(arr[0]);
}

int main()
{
cout << size("Test"); //This is working

int x[] = {7, 5, 43, 8};
cout << endl
     << size(x); //And this works too

cout << endl
     << size({7, 9, 7, 9}); //When i try this its give me "no matching function for call to 'size'" error

return 0;
}

Parameter takes strings and arrays that modified outside the parameter. but i need to write the array directly inside the function like the code above. size({some integers});

This function already is in the standard library: en.cppreference.com/w/cpp/iterator/size — MadScientist
– MadScientist, Commented Jan 28, 2022 at 11:57
A non-const lvalue reference can not bind to an rvalue. Here size({7, 9, 7, 9}), the argument {7, 9, 7, 9} is an rvalue. — digito_evo
– digito_evo, Commented Jan 28, 2022 at 12:08
Unrelated: You could return directly N instead of sizeof(arr) / sizeof(arr[0]) — Ch3steR
– Ch3steR, Commented Jan 28, 2022 at 12:23

Vlad from Moscow · Accepted Answer · 2022-01-28 12:06:04Z

6

Declare the function parameter like

int size( const VAR (&arr)[N])

That is you may not bind a temporary object with a non-constant lvalue reference.

edited Jan 28, 2022 at 12:06

answered Jan 28, 2022 at 11:52

Vlad from Moscow

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Comments

Jeff Schaller · Accepted Answer · 2022-01-30 15:48:36Z

5

{1,2,3,4} is not an array, but a list initialization.

Anyhow, don't reinvent the wheel. We already have std::size; for example:

#include <iostream>

int main()
{
    std::cout << std::size("Test") << '\n'; //This is working

    int x[] = {7, 5, 43, 8};
    std::cout << std::size(x) << '\n'; //And this works too

    std::cout << std::size({7, 9, 7, 9}) << '\n'; // this too

    return 0;
}

edited Jan 30, 2022 at 15:48

Jeff Schaller

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answered Jan 28, 2022 at 11:56

JHBonarius

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Comments

Neeraj Jain · Accepted Answer · 2022-01-28 12:21:03Z

1

Error is because you parameter take address of array(that is rvalue) and you give whole array(that is lvalue ) that's why code give error .

For solve this you make parameter const

`

 #include <iostream>


using namespace std;


 template <typename VAR, unsigned int N>


int size(const VAR (&arr)[N])
   {
 return sizeof(arr) / sizeof(arr[0]);
}

int main()
  {
cout << size("Test"); //This is working

 int x[] = {7, 5, 43, 8};
 cout << endl
 << size(x); //And this works too

 cout << endl
 << size({7, 9, 7, 9}); //When i try this its give me "no matching function for 
 call to 'size'" error

  return 0;
   }`

I hope you get Your answer .

answered Jan 28, 2022 at 12:21

Neeraj Jain

111 bronze badge

1 Comment

JHBonarius Over a year ago

Welcome to Stack Overflow. Good first answer! But can you clean up the layout a bit?

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