Types in template functions.

[Vladimir Grigoriev]

Let consider the function

template <typename T>

void check_const( T i )

{

++i;

}

If then it will be called such a way



check_const<const int>( 0 );

the compiler ( VC++ 2005 EE) issue the error

"error C3892: 'i' : you cannot assign to a variable that is const"

However if it will be called the following way

const int value = 0;

check_const( value );

the compiler does not issue the error.

Why does not the compiler issue the error in the second caase?



Vladimir Grigoriev

[Igor Tandetnik]

Post by Vladimir Grigoriev
Let consider the function
template <typename T>
void check_const( T i )
{
++i;
}
If then it will be called such a way
check_const<const int>( 0 );
the compiler ( VC++ 2005 EE) issue the error
"error C3892: 'i' : you cannot assign to a variable that is const"
However if it will be called the following way
const int value = 0;
check_const( value );
the compiler does not issue the error.

14.8.2.1 Deducing template arguments from a function call
1 Template argument deduction is done by comparing each function template parameter type (call it P) with the type of the corresponding argument of the call (call it A) as described below.
2 If P is not a reference type:
...
- If A is a cv-qualified type, the top level cv-qualifiers of A’s type are ignored for type deduction.

So, T is deduced as int, not as const int.

[Vladimir Grigoriev]

Thanks, Igor.
Only the question: cv in cv-qualified means const or volatile?

Vladimir Grigoriev

Post by Vladimir Grigoriev
Let consider the function
template <typename T>
void check_const( T i )
{
++i;
}
If then it will be called such a way
check_const<const int>( 0 );
the compiler ( VC++ 2005 EE) issue the error
"error C3892: 'i' : you cannot assign to a variable that is const"
However if it will be called the following way
const int value = 0;
check_const( value );
the compiler does not issue the error.

14.8.2.1 Deducing template arguments from a function call
1 Template argument deduction is done by comparing each function template
parameter type (call it P) with the type of the corresponding argument of
the call (call it A) as described below.
2 If P is not a reference type:
...
- If A is a cv-qualified type, the top level cv-qualifiers of A’s type are
ignored for type deduction.

So, T is deduced as int, not as const int.

[Igor Tandetnik]

Post by Vladimir Grigoriev
Only the question: cv in cv-qualified means const or volatile?

Correct.