Non-aggregate type

[MarkoKlacar]

Hi,

I have two problems.

I have a class a called Edit that contains the following:

Header

void setCharAt(int index,char ch);
void rubout(int index) const;
char text[];
 

Problem 1:
Now, the rubout function takes an index and it's supposed to do something with this index in the char text[]. To be more specific, it's supposed to remove the char at the index position and pack the array.

The thing is when I try to get the length of the char tex[] I get this:

request for member `length' in `this->Edit::text', which is of on-aggregate type `char[0]'

Problem 2:
When I call the setCharAt(int index, char ch) it looks like this:

Edit::setCharAt(1,'v');

The error I get for this is:

passing `const Edit' as `this' argument of `void Edit::setCharAt(int, char)' discards qualifiers

What am I doing wrong?

Cheers.

[RRick]

Its hard to tell what is going on without the actual code that is causing the problem. Please, we don't want to see all of your code, but a bit more is needed.

#1 looks like you are treating char [] like a string. I.e. ...myEdit.text.length(). That won't work. My sugestion is to make text a std::string.

#2 Somehow your edit object is a const and since since setCharAt is not a const method, the compiler will complain.

Finally, rubout is const, but it sounds like it will modify the text. That won't work either.

[MarkoKlacar]

Hi,

Thanks for your response. As you mention, the methods rubout should not be const.

This is probably a noob issue. Could the fact that I'm calling a non-const method from a const method have anything to do with it?

I con provide you with more code, but I think you've gotten the problem.

Thanks in advance....

/M

[weaknessforcats]

text is a char array. It is not an object. There is no length member. Hence the error.

Maybe you meant you use a string object? If so, use the size() method for the length rather than the length() method. The length() method is deprecated.

[MarkoKlacar]

Hey guys,

Thanks for the replies. But how do I tackle the other problem?

From a const method call a non-const method?
That's causing me problems...

Cheers
/M

[weaknessforcats]

From a const method call a non-const method?

A non-const method (or any function for that matter) is one that can change variables outside the function. Like functions with pointer arguments.

You make a function const by making the arguments const:

void fx(const int* arg)
{
    *arg = 15;          //ERROR. the int is const.
}
 

That is, the const arguments reassures the compiler that the function cannot change anything outside the scope of the function:

void fx(const int* arg)
{
    cout << *arg << end;
}
 

This sort of const function can be called by a const member function.

Now for a const member funciton:

void MyClass::AMethod() const
{
      fx(this->APointerMember);     //OK. fx cannot change what APoionterMember points at
}
 

Remember, that member functions have an invisible first argument that is the this pointer. When you call a member function that has no arguments, there is actually one supplied by the compiler: The this pointer.

A member function can use the this pointer to chnage other class members unless the member function is const. A const member function is not allowed to change a class member.

Therefore, from within a const member function, you can call any function that has const pointer arguments, or other const member functions. And that's all.