问题
I know it's possible to do something like :
int foo(int a = 0, int b = 1) { return a + b; }
and then use it without the default parameters eg.:
foo(); // a = 0, b = 1 -> 1
or with the last one as default eg.:
foo(2); // a = 2 and b = 1 default -> 3
But my question is: Is it possible to use the default value for the first argument (a) and give the value of the second (b)
My first thought was doing it like (which doesn't work!):
foo(,2); // a = 0 default and b = 2
Does a syntax for this exist or is this just not possible ?
回答1:
No, it is not possible in current syntax.
回答2:
Alternatively from specifying default parameter values you can use multiple function overloads like:
int foo(int a, int b){return a+b; }
int foo(int b){return foo(0,b); }
int foo(){return foo(0,1); }
回答3:
I know that it is not really an answer to your question, but you could try to use boost named parameters. It might be useful for your use case.
回答4:
It is not possible directly. However, it is possible somehow if you declare a struct for the input parameters that encapsulates the default values and has individual setters:
#include <iostream>
struct Foo{
int a,b;
Foo() : a(0),b(1) {};
Foo setA(int x) { a=x; return *this;}
Foo setB(int x) { b=x; return *this;}
};
void foo(Foo f = Foo()){ std::cout << f.a << " " << f.b << std::endl; }
int main() {
foo(); // uses default values for a and b
foo(Foo().setA(3)); // uses default value for b
foo(Foo().setB(5)); // uses default value for a
}
This is of course quite some overhead. On the other hand, imho it anyhow makes sense quite often to encapsulate parameters into a struct and provide default values there instead of on the function declaration.
回答5:
Once we provide a default value for a parameter, all subsequent parameters must also have default values. For example,
// Invalid
void add(int a, int b = 3, int c, int d);
// Invalid
void add(int a, int b = 3, int c, int d = 4);
// Valid
void add(int a, int c, int b = 3, int d = 4);
来源:https://stackoverflow.com/questions/39838204/default-parameters-can-only-the-last-arguments-be-left