Why do I need to add const to the argument?

Question

I have this function :

void
SpookyBoy( bool Troublemaker, const Glib::ustring& name, HorrorPunkBand& band );

if I remove the word const I got this error :

 no known conversion for argument 2 from ‘const char [5]’ to ‘Glib::ustring&’

I just wonder what does the compiler, can someone explains to me why with the word const the compiler can cast ?

even if I have to admit the two questions are similar for someone who already knows the answer (at fortiori), but certainly not for someone who does not know yet the answwer (at priori).

Does your function declaration really look like that? According to the error message the `ustring` parameter is apparently declared as `const Glib::ustring& name`, i.e. as a reference. — AnT stands with Russia, May 27 '15 at 02:48
possible duplicate of [Why const for implicit conversion?](http://stackoverflow.com/questions/3895647/why-const-for-implicit-conversion) — AnT stands with Russia, May 27 '15 at 02:52

score 3 · Accepted Answer · edited May 23 '17 at 12:06

You are apparently calling your function with a string literal as an argument

SpookyBoy(..., "1234", ...);

The compiler can implicitly convert the string literal (which in this case has type const char [5]) to type Glib::ustring. But the result of that implicit conversion is a temporary object. In C++ language only const lvalue references can be bound to temporary objects. Non-const references cannot be bound to temporary objects.

Without that const in function declaration you'd have to call your function as

Glib::ustring name("1234");
SpookyBoy(..., name, ...);

i.e. by explicitly introducing a named object of Glib::ustring type and passing it as an argument.

P.S. See also Why const for implicit conversion?

Why do I need to add const to the argument?

1 Answers1