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c++c++11standardstype-traits

Why is a variable not an lvalue in C++?

#include <type_traits>

int main()
{    
    int n;
    n = 0;

    // failure!
    static_assert(std::is_lvalue_reference_v<decltype(n)>); 
}

n can be put on the left side, so it should be obviously an lvalue.

Why does the static_assert fail?

Solution

  • decltype has special rules for id-expressions, deducing their type without regard to value category. If you want it to deduce a type based on the value category an id-expression normally has, you can surround the id-expression in parenthesis:

    static_assert(std::is_lvalue_reference_v<decltype((n))>);

    (n) has the same type and value category as n in the type system, but isn't treated specially by decltype. Since the expression is an lvalue, the deduced type will be of an lvalue reference type.