Point of declaration for auto keyword
c++, c++11
Solution
auto x = x; // inner x
is ill-formed.
To quote from the C++11 standard (emphasis mine):
7.1.6.4 auto specifier
...
3 Otherwise, the type of the variable is deduced from its initializer. The name of the variable being declared shall not appear in the initializer expression. ...
And so because `x` after `=` resolves to the `x` in `auto x` (as explained in the question you linked), that above piece of code is ill-formed.
Problem
I had a Q&A before: Point of declaration in C++. The rule point-of-declaration nicely is applicable on many situations. Now, I confused on usage of `auto` in combination of this rule. Consider these two codes: i. Declaring `x` by itself (we don't expect it to work): ``` { auto x = x; } ``` ii. Declaring the inner `x` by the outer `x` (It makes error in gcc 4.8.x): ``` { int x = 101; // the outer x { auto x = x; // the inner x } } ``` According to the rule of point-of-declaration, it should work but it doesn't. It seems there is another rule in the standard that I missed it. The question is, Where is the point-of-declaration when using `auto`? There are two possibilities: i. If the point of declaration is after `=`, at the end of statement: ``` auto object = expression; ^ Is it here? If it is, why gcc complains? ``` So the second declaration is valid and must work, because there is no `x` but that outer one (which is declared before). Therefore `auto x=x` is valid and the inner `x` should be assigned to `101`. ii. If the point of declaration is before `=` : ``` auto object = expression; ^ ``` Well, it doesn't make any sense because `auto` has to wait until see the following expression. For example `auto x;` is invalid. Update: I need an answer which explains it by the rule point of declaration.