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If I have,
void foo(Bar c);
void foo(Bar&& c);
foo(Bar());
why is the call to 'foo' is ambiguous? Isn't Bar() in the foo argument clearly an rValue?
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There are two ways to resolve this ambiguity: Typecast char to float. Remove either one of the ambiguity generating functions float or double and add overloaded function with an int type parameter.
When the compiler is unable to decide which function it should invoke first among the overloaded functions, this situation is known as function overloading ambiguity. The compiler does not run the program if it shows ambiguity error.
You can pass an object to a function that takes an rvalue reference unless the object is marked as const . The following example shows the function f , which is overloaded to take an lvalue reference and an rvalue reference. The main function calls f with both lvalues and an rvalue.
Rvalue references allow programmers to avoid logically unnecessary copying and to provide perfect forwarding functions. They are primarily meant to aid in the design of higer performance and more robust libraries.
Binding to a reference is an "exact match", as is binding to a non-reference, so both overloads are equally good.
In Standardese, this is 13.3.3.1.4 ("Reference binding", [over.ics.ref]):
When a parameter of reference type binds directly (8.5.3) to an argument expression, the implicit conversion sequence is the identity conversion [...]
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