CODEWITHSUNDEEP
c++
user310712
user310712

Reputation: 51

Function overloading

I found this code , and i m not sure that whether overloading should happen or not.

void print( int (*arr)[6], int size );

void print( int (*arr)[5], int size );

what happens if I pass pointer to an array of 4 elements , to it should come...

any thread will be helpful.

Upvotes: 5

Views: 255

Answers (3)

sbi
sbi

Reputation: 224159

KennyTM's answer is the correct one. Here's an additional thought, though, based on the fact that your question comes with a C++ tag. In C++, you can use templates with non-type arguments to find out array dimensions: 

#include <iostream>

template< std::size_t N >
void print(int (&arr)[N]) {std::cout << N << '\n';}

int main()
{
    int arr[6];
    print(arr);
    return 0;
}

Upvotes: 6

Akanksh
Akanksh

Reputation: 1310

The call would be ambiguous as none of the two overloads would be able to convert to int (*arr)[4]. You need to pass in an element of 5 or 6 elements explicitly.

VS2008 gives:

error C2665: 'print' : none of the 2 overloads could convert all the argument types
(2088): could be 'void print(int (*)[5],int)'
(2093): or       'void print(int (*)[6],int)'
 while trying to match the argument list '(int (*)[4], int)'

Hope that helps.

Upvotes: 1

kennytm
kennytm

Reputation: 523654

Overloading will happen, and passing the pointer to the array of 4 int's will not match either function. It's clearer if you write them as the equivalent form:

void print( int arr[][6], int size );
void print( int arr[][5], int size );

An N×4 array can be decayed to a pointer to array of 4 int's. And it's well known that 2D arrays having different 2nd dimensions are incompatible.

Upvotes: 10

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