Specifying Default Parameter When Calling C++ Function

Specifying default parameter when calling C++ function

As workaround, you may (ab)use boost::optional (until std::optional from c++17):

void f(boost::optional<int> oa = boost::none,
       boost::optional<int> ob = boost::none,
       boost::optional<int> oc = boost::none)
{
    int a = oa.value_or(0); // Real default value go here
    int b = ob.value_or(0); // Real default value go here
    int c = oc.value_or(0); // Real default value go here

    //...Some Code...
}

and then call it

f(a, boost::none, c);

C default arguments

Not really. The only way would be to write a varargs function and manually fill in default values for arguments which the caller doesn't pass.

Is there a simple way to call a function with default arguments?

You can't do this as part of the natural language. C++ only allows you to default any remaining arguments, and it doesn't support named arguments at the calling site (cf. Pascal and VBA).

An alternative is to provide a suite of overloaded functions.

Else you could engineer something yourself using variadic templates.

Change only a specific Default Parameter on a function

When you pass a value for a particular parameter that has a default argument, you have to pass values for all the default parameters before it. Otherwise, the value you have passed will be taken as the value for the first default parameter.

So you have to do this:

newAddress = QInputDialog::getText(
             0,
             "Enter an Address to Validate",
             "Adress: comma separated (e.g Address line 1, Address line 2, City, Postal Code)", 
             QLineEdit::Normal, 
             QString(), 
             &ok);

You can leave out passing values for the parameters after bool * parameter.

The C++ standard states in [dcl.fct.default]/1

Default arguments will be used in calls where trailing arguments are missing.

Where to put default parameter value in C++?

Default parameter values must appear on the declaration, since that is the only thing that the caller sees.

EDIT: As others point out, you can have the argument on the definition, but I would advise writing all code as if that wasn't true.

How to Call function with default parameter?

The syntax doesn't allow to omit parameter values at arbitrary positions, only those at the very end of the list.

You have to overload 2 versions:

int myfunction(VpTR*& viewporttable,  OpenMode f=fR) {
    return myfunction(viewporttable,L"*Active",L"*Active",f);
}

int myfunction(VpTR*& viewporttable,  wchar* vpname=L"*Active", OpenMode f=fR);

Then you can call

myfunction(myviewporttable, fR);

Default values on arguments in C functions and function overloading in C

No, Standard C does not support either. Why do you feel you need to convert your C++ code to C? That could get quite tricky - I'd have thought writing wrappers would be the way to go, if your C++ must be callable from C.

How to call a function with default parameter through a pointer to function that is the return of another function?

Defaulted arguments are a bit of C++ syntactic sugar; when calling the function directly with insufficient arguments, the compiler inserts the default as if the caller had passed it explicitly, so the function is still called with the full complement of arguments (Mult(4) is compiled into the same code as Mult(4, 2) in this case).

The default isn't actually part of the function type though, so you can't use the default for an indirect call; the syntactic sugar breaks down there, since as soon as you are calling through a pointer, the information about the defaults is lost.

Default parameter and function pointer as function arguments C++

The fundamental principle behind default arguments is that they apply to the right-most arguments in the argument list.

void f(int, double = 0, void* = 0); // okay
void g(int, double = 0, void*); // error: third argument needs a default value

Similarly, when calling a function with default arguments, only the rightmost arguments get their default values:

f(3); // okay; second and third arguments get default values
f(3, 1.0); // okay; third argument gets default value
int x;
f(3, 1.0, &x); // okay; no default arguments used

You can't skip arguments, even if it's possible to determine which one you meant:

f(3, &x); // illegal: second argument has type double, and &x can't be converted to double

This can get confusing if one of the arguments is convertible to the appropriate type, and it looks like that might be what's happening in the code in the question that compiles. For example,

void h(int, bool = false, int* = 0);
int x;
h(3, &x);   // okay; &x gets converted to bool
h(3, true); // same as the previous one
h(3, false, &x); // okay, all three arguments provided

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