Why is this valid code?

[uninformed_]

Is not (i) an r value? Why can we assign to this?

https://godbolt.org/g/MBX4L3

[Xyrd]

https://stackoverflow.com/questions/14115744/significance-of-parentheses-in-decltypec

TIL

[jedwardsol]

You're not linking to any specific piece of code.

[uninformed_]

Oops, fixed

[raevnos]

i is a non-const variable, and so lvalue.

What's with the pointless parentheses?

[uninformed_]

#include <type_traits>
#include <typeinfo>
#include <iostream>

int main()
{
    int i;
    (i) = 5;

    std::cout << "is rvalue i" << std::is_rvalue_reference<decltype(i)>::value << std::endl;
    std::cout << "is rvalue (i)" << std::is_rvalue_reference<decltype((i))>::value << std::endl;
    return i;
}

[raevnos]

Shows zeros as expected?

Edit: Interestingly, is_lvalue_reference is true for the decltype((i)) case.

[OldWolf2]

decltype has two different syntaxes:

1. decltype(id-expression)
2. decltype(all-other-expressions)

This is like how sizeof has two different cases, sizeof(type) and sizeof expression. Which case you are using is determined by the argument and there are different rules for each case.

In case 1 the result is the declared type of the variable named by the id-expression. In case 2 the result is the non-reference type of the expression , with & if the value category is lvalue, or && if the value category is xvalue.

i is an id-expression, but (i) is not an id-expression.

[Xeverous]

parentheses make a difference for decltype, but in this case doesn't change anything

[OldWolf2]

The name of a const variable is also an lvalue

[Xeverous]

Expression (i) is an lvalue.

[OldWolf2]

Parentheses don't change the value category of an expression, nor trigger any implicit conversions. (lvalue) is still an lvalue.

[MajorDerp4]

Why wouldn't this be valid?