Is it possible in C++ passing function inout parameter with default value

Question

Can we write function in c++, something like this, may be not exactly like this, syntax may be different

void FunctionTest(int param1, _Inout_ int& param2 = 10)
{
    param2 = 20;
}

int main() {
    int a = 100, int b = 200;
    FunctionTest(10);
    FunctionTest(a, b);
}

Not necessary it should be pass by reference, may other alternative will also work pointer or any other way.

Is it possible in C++ passing function inout parameter with default value? No. — Eljay
– Eljay, Commented Jul 2, 2024 at 18:20
Compiler quite reasonable says "error: non-const lvalue reference to type 'int' cannot bind to a temporary of type 'int'" to your snippet. OTOH, int foo = 10; void FunctionTest(int param1, int& param2 = foo) . . . is just fine. ` — Avi Berger
– Avi Berger, Commented Jul 2, 2024 at 18:23
Inout is Microsoft annotation, it would be better if you just forget about their existence — Gene
– Gene, Commented Jul 2, 2024 at 18:24

Quentin · Accepted Answer · 2024-07-03 10:41:48Z

5

Yes, you can actually do it quite concisely with a helper function:

template <class T>
T &as_lvalue(T &&o) { return o; }

void FunctionTest(int param1, int& param2 = as_lvalue(10))
{
    param2 = 20;
}

Do keep in mind that the referenced object will only live until the end of the full-expression containing the call to FunctionTest. This is what makes non-const-reference parameters with default values quite an odd sight: why provide an output parameter if you won't be able to examine the results before it vanishes?

If you want to make a function that may provide additional output besides its return value, maybe a cleaner pattern would be a (natively nullable) pointer parameter and a separate overload for the reference:

void FunctionTest(int param1, int *param2 = nullptr)
{
    if(param2)
        *param2 = 20;
}

void FunctionTest(int param1, int &param2)
{
    return FunctionTest(param1, &param2);
}

edited Jul 3, 2024 at 10:41

answered Jul 2, 2024 at 18:32

Quentin

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Comments

463035818_is_not_an_ai · Accepted Answer · 2024-07-02 20:10:29Z

It is unclear why you think this would be needed. What would be the "out" when the default is used?

You can overload functions:

void FunctionTest(int param1,int& param2) {
    param2 = 20;
}

void FunctionTest(int param1) {}

int main() {
    int a = 100;
    int b = 200;
    FunctionTest(10);
    FunctionTest(a, b);
}

However, I cannot imagine this to be a good use case for overloading, because the two calls do something very different.

Only for the sake of completeness. A default arguemnt for reference can be a global:

   int foo;
   void function(int& b = foo);

Or a function local static or static class member:

   struct moo { static int x; }
   void function(int& b = moo::x);

   int& bar() { 
         static int x = 0;
         return x;
   }
   void function(int& b = bar::x);

Or more generally a function that returns a (valid) reference to any of the above.

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