Why does this code (unassigned temporary variable constructed from a const char* variable):
class A
{
public:
A(const char*) {}
};
int main()
{
const char* constCharPointerVariable = "StringLiteral";
A(constCharPointerVariable);
return 0;
}
Give these errors?
error C2512: 'A' : no appropriate default constructor available
error C2040: 'constCharPointerVariable' : 'A' differs in levels of indirection from 'const char *'
Whereas this code (assigned temporary variable constructed from a const char* variable):
class A
{
public:
A(const char*) {}
};
int main()
{
const char* constCharPointerVariable = "StringLiteral";
A a(constCharPointerVariable);
return 0;
}
Gives no errors.
And this code (unassigned temporary variable constructed from a const char* variable static_cast to a const char*):
class A
{
public:
A(const char*) {}
};
int main()
{
const char* constCharPointerVariable = "StringLiteral";
A(static_cast<const char*>(constCharPointerVariable));
return 0;
}
Gives no errors.
Bonus points if you can provide the section number in the C++ specification that specifies the first code sample to be not allowed.
A(constCharPointerVariable);
This is actually a declaration of a variable of type A
named constCharPointerVariable
. It does not create a temporary object.
If you used clang, you'd get the more helpful error message:
error: redefinition of 'constCharPointerVariable' with a different type
A(constCharPointerVariable);
^
As a simpler example, the following is invalid because it declares two int
objects in the same scope, both named x
:
int x(0);
int (x);
As for why the code is parsed this way, you can find the syntax rules for Declarators in §A.7 of C++11. Basically, when you declare a variable, you can enclose its name in any number of parentheses.
Relevant productions include:
(
ptr-declarator )
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