CODEWITHSUNDEEP
c++istream
AlesSvetina
AlesSvetina

Reputation: 463

Running function with std::istream

I am trying to input a value in my function, which looks like this:

int funkcija( std::istream & in ) {
    int value(0);
    in >> value;
    if(not in) throw std::exception();
    if( value%2 == 0 ) {
        return (value/2);
    }
    else return (value*3)+1;
}

When I try to run it:

int i(0);
std::cout << "Input a number: ";
std::cin >> i;
funkcija(i);

I get an error: ..\working.cpp:17:14: error: invalid initialization of reference of type 'std::istream& {aka std::basic_istream&}' from expression of type 'int' ..\working.cpp:7:5: error: in passing argument 1 of 'int funkcija(std::istream&)'

What does it mean and how to solve it? Thank you!

Upvotes: 0

Views: 117

Answers (2)

Daniel Frey
Daniel Frey

Reputation: 56903

You are trying to pass the integer you've already read, try:

std::cout << "Input a number: ";
int i = funkcija(std::cin);
std::cout << i << " ";

While this would work it seems strange. Consider separating the input- and output-handling from the calculation to improve your design. Change the function to:

int funkcija( int value ) {
   if( value%2 == 0 ) {
       return (value/2);
   }
   else return (value*3)+1;
}

and maybe call it like this:

std::cout << "Input a number: ";
int i;
if( !( std::cin >> i ) ) throw std::exception();
do {
    i = funkcija( i );
    std::cout << i << " ";
} while( i != 1 );

Upvotes: 3

cristobalito
cristobalito

Reputation: 4282

i is of type int not istreeam - you're passing i to the function, therefore its complaining that it's an int and not an istream. You can probably pass in std::cin directly to the function.

Upvotes: 0

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