How do I Create and Pass string to function by reference

Question

I am having a constructor like this:

class class_foo{
    std::string s_;
    class_foo(std::string& s) : s_(s){};
}

I know I can do:

std::string s = "test";
cf = class_foo(s);

Is there a way to do:

cf = class_foo("test");

But is says: note: candidate constructor not viable: expects an l-value for 3rd argument

@eerorika was right. I can simply use a const like here: link

Possible duplicate : [Why does const allow implicit conversion of references in arguments?](https://stackoverflow.com/questions/48576011/why-does-const-allow-implicit-conversion-of-references-in-arguments) — François Andrieux, Jan 24 '20 at 19:53

score 1 · Accepted Answer · answered Jan 24 '20 at 19:55

I am having a constructor like this:
void class_foo(std::string& s) : s_(s){};

A constructor cannot have a return type (even void).

Is there a way to do:
cf = class_foo("test");

Not if you want to keep the argument as non-const reference. It is however unclear why you want it to be non-const reference. Perhaps it doesn't need to be a non-const reference? If you don't intend to modify the argument, then you should use a reference to const. Then your suggested construction would work.

Good catch. I updated that. Oh yeah you are right. I can just declare it const — Julian, Jan 24 '20 at 23:24

How do I Create and Pass string to function by reference

1 Answers1