Why is a hard-coded string constant an lvalue?

Asked Dec 06 '12 at 01:03

Active Dec 06 '12 at 01:22

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Possible Duplicate:
Why are string literals l-value while all other literals are r-value?

In What are move semantics? appears the following code snippet and comment:

void some_function(std::string&& r);

some_function("hello world");

with the comment:

In the above example, "hello world" is an lvalue of type const char[12].

Why is the hard-coded constant "hello world" an lvalue of type const char[12], rather than an rvalue of type const char[12]?

edited May 23 '17 at 12:12

Community

asked Dec 06 '12 at 01:03

Dan Nissenbaum

1

It is a duplicate. Thanks, the linked question (and its answers) does answer my question. – Dan Nissenbaum Dec 06 '12 at 01:09
2

There are no array rvalues in C++. String literals are no exception. (Unless you say `throw "string";`.) – Kerrek SB Dec 06 '12 at 01:09
@KerrekSB In the example `throw "string"`, `"string"` is an rvalue? – Dan Nissenbaum Dec 06 '12 at 01:15
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@DanNissenbaum: No, but it's an exception... *ba-dish dum*. – Kerrek SB Dec 06 '12 at 01:15
@KerrekSB Indeed. My head is too much in the box. :) – Dan Nissenbaum Dec 06 '12 at 01:17
@KerrekSB: Yes there are array rvalues in C++. `Alias{1,2,3}` will create an `int[3]` rvalue. It's just that *string literals* are always lvalues. – Xeo Dec 06 '12 at 01:55

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