CODEWITHSUNDEEP
c++pass-by-reference
nogeek001
nogeek001

Reputation: 723

pass by reference : no known conversion for argument 6 from 'int' to 'int&'

I have a function where I pass an argument by reference since I expect the function to edit it. This function is called at several places and I only care about the ref value when called at a particular instance . Pseudocode: 

test_fn(int a, int b, inc , int d, int e, int& ref)
{
//bunch of other functionalities 
//.
//.
ref = (a*b+c)*(d+e);
}

test_fn(1,2,3,4,5,0)//everywhere that I do not care about ref 
int value = 0;
test_fn(1,2,3,4,5, value)//I care about value here and would use it in the remainder of the code .

Why can I not pass a 0 directly ? I tried passing a NULL as well and that has a long int to an int conversion error.

Why is this wrong ? And what is the best way to achieve the expected outcome here?

Upvotes: 2

Views: 2706

Answers (5)

Aaron McDaid
Aaron McDaid

Reputation: 27133

Consider this much simpler example:

test_fn(int& ref)
{
    ref = 3;
}
int main() {
    test_fn(0);
}

This is effectively trying to set 0 to 3. i.e:

int main() {
    0 = 3;
}

But that's nonsense. An int & (as opposed to a const int&) can only accept something that is modifiable.

(As @nogeek001 points out, const int& wouldn't allow us to modify ref anyway.)

Upvotes: 2

PaulMcKenzie
PaulMcKenzie

Reputation: 35440

As others have stated, you can't pass a literal as a reference.

What you can do is pass an address, and then check if it is NULL within the function:

test_fn(int a, int b, inc , int d, int e, int* ref)
{
    int someValue = (a*b+c)*(d+e);
    if ( ref )
       *ref = someValue;
}
//...
test_fn(1,2,3,4,5,0);
int value = 0;
test_fn(1,2,3,4,5, &value)

Upvotes: 2

user_4685247
user_4685247

Reputation: 2985

in order to pass variable by reference it has to exist, passing 0 or NULL means you're sending in a constant. You cannot edit the value of a constant, as it is actualy not a variable.

As for solving your problem, you probaby should use pointers to achieve that, then check if the pointer is set to 0, NULL or if you use C++11, nullptr

Upvotes: 2

edmz
edmz

Reputation: 8494

ref is a reference not a pointer. 0 may be passed if it were, meaning a pointer to null; reference can't point to nothing and must be bound to an lvalue.

Upvotes: 0

CinchBlue
CinchBlue

Reputation: 6190

A regular int& means that it needs to be assigned to a variable already; it needs to be an lvalue.

0 is not assigned to a variable; it's a "free variable," which means that it's unattached to a label. This means that it's an rvalue, a temporary variable that is not bound to a variable. It's denoted by int&&.

rvalues can be converted to lvalues if you make it const int&. It makes sense that a constant can be converted to a reference to int constant (reading right to left).

However, that would be pointless, as you want to modify the variable; therefore, the answer is to follow your own convention and don't pass in things that are not already in "existence" and bound to a label/name, like constants or moved variables.

Upvotes: 2

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