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decltype error C2440 cannot convert from 'int *' to 'int *&'

Tags:

c++

decltype

The following is a contrived example of the actual code:

int** Ptr = 0;
decltype(Ptr[0]) Test = (int*)0;

I get the error:

error C2440: 'initializing': cannot convert from 'int *' to 'int *&'

I'm not sure why I'm getting that since from my understand of decltype (correct me if I'm wrong) it just takes whatever expression you give it and resolve it to its actual type. In this case Ptr[0] is an int* so I'm expecting: int* Test = (int*)0;

What am I missing? Why am I getting that error?

like image 977
vexe Avatar asked Nov 09 '15 01:11

vexe


2 Answers

If we go to the draft C++ standard section 7.1.6.2 Simple type specifiers [dcl.type.simple] and see what the cases are, for decltype it starts out saying:

For an expression e, the type denoted by decltype(e) is defined as follows:

We see that in this case the expression is not an id-expression nor a class member access, which would give the result you expected(emphasis mine):

  • if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;

but the result is an lvalue:

  • otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

which results in a reference.

As M.M point out std::remove_reference could be used to obtain the result you want:

std::remove_reference<decltype(Ptr[0])>::type Test = (int*)0;

As T.C. points out std::decay is also an option and is shorter:

std::decay<decltype(Ptr[0])>::type Test = (int*)0;
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Shafik Yaghmour Avatar answered Sep 21 '22 16:09

Shafik Yaghmour


In addition to all the other answers, you may also just use

int ** ptr = 0;
decltype(+ptr[0]) test = (int*)0;
// (+*p) is now an r-value expression of type int, rather than int&

Some of the rules used here are:

  • if the value category of expression is lvalue, then decltype yields T&;
  • if the value category of expression is prvalue, then decltype yields T.

Also note that if the name of an object is parenthesized, it is treated as an ordinary lvalue expression, thus decltype(x) and decltype((x)) are often different types. [an excerpt from https://en.cppreference.com/w/cpp/language/decltype]

seen at http://www.cplusplus.com/forum/beginner/149331/
more on decltype here : https://en.cppreference.com/w/cpp/language/decltype

like image 31
Alex Vask Avatar answered Sep 22 '22 16:09

Alex Vask