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

Question

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?

If you are looking to pass NULL as an option, have you looked into using pointers over references? http://stackoverflow.com/a/57492/1658810 — Lilith Daemon, Jul 21 '15 at 20:38

score 3 · Answer 1 · answered Jul 21 '15 at 20:35

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.

Aaron McDaid · Answer 2 · 2015-07-21T20:41:22.677

3

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.)

edited Jul 21 '15 at 20:41

answered Jul 21 '15 at 20:35

Aaron McDaid

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(as opposed to a const int&) ? Using a const int& however will not let the function modify the ref value - correct? – nogeek001 Jul 21 '15 at 20:39

user_4685247 · Accepted Answer · 2015-07-21T20:40:24.927

2

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

edited Jul 21 '15 at 20:40

answered Jul 21 '15 at 20:35

user_4685247

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score 2 · Answer 4 · answered Jul 21 '15 at 20:40

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)

score 0 · Answer 5 · answered Jul 21 '15 at 20:37

0

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.

answered Jul 21 '15 at 20:37

edmz

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

5 Answers5