class foo {
public:
static int func(const char*& p) const;
};
This is illegal, why?
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Answer / rafal dzbek
the 2nd const used in the example is invalid because it can
be used only with member functions which have a hidden
argument called this. The 2nd const would be applied to
this.
The funct is static member function so it hasn't any this
pointer.
Actually it is not obvious what is exactly inccorect.
We can remove static keyword, then we get syntactically
correct class definition, or we can remove 2nd const and
again the class can be considered valid.
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / jaroosh
This code is obviously wrong, and here is why :
declaring a method to be const, means :
this method cannot CHANGE values of any member variables,
but while the method already is static, it has no means of
changing values of member variables, because simply - it
cannot see them (its belongs to a CLASS, not any specific
OBJECT).
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / abdur rab
A member function can be declard as Const by considering
the fact that they contain a hidden "this pointer" to be a
pointer to a const object. However Static methods do not
have the "this pointer", and hence can't be const or
virtual.
Hence it is illegal to declare a static function as const.
The C++ language standard stipulates at section 9.4.1
that "[...] A static
member function shall not be declared const
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / revathy
static functions can access static data only
Thus this is illegal
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / john gummadi
"const" cannot be used along with reference. And the function itself is const which means it cannot change any parameters. So no place for the reference here. If you really need reference, then get id of consts.
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / john gummadi
There is nothing to do with class members here, we don't
see any. The function takes a constant pointer as a
parameter, when it is constant it cannot be changed
(although technically we can change by casting), then you
cannot use reference (&).
But who knows, compilers may accept, I haven't tested.
| Is This Answer Correct ? | 1 Yes | 2 No |
Answer / rohit
the 2nd const used in the example is invalid because it can
be used only with member functions which have a hidden
argument called this. The 2nd const would be applied to
this.
The funct is static member function so it hasn't any this
pointer.
Actually it is not obvious what is exactly inccorect.
We can remove static keyword, then we get syntactically
correct class definition, or we can remove 2nd const and
again the class can be considered valid.
A member function can be declard as Const by considering
the fact that they contain a hidden "this pointer" to be a
pointer to a const object. However Static methods do not
have the "this pointer", and hence can't be const or
virtual.
Hence it is illegal to declare a static function as const.
The C++ language standard stipulates at section 9.4.1
that "[...] A static
member function shall not be declared const
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / kalpana.y
this is illegal because
->here the class name is foo
->but,func is declared
->const is declared at outside
| Is This Answer Correct ? | 0 Yes | 4 No |
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