Overloading operator += => String and char

**GaPe** · 05-07-2005

Hi!

I have another overloading operator problem.

I want to use my String class with type char.

Example:
String temp = "";
char z = 'a';
temp += z;

.hpp file

Code:

String &operator+=(const char &character);

.cpp file

Code:

String &String::operator+=(const char &character) {
String newString = *this + character;
setString(newString);
return *this;
}

How can I make it work?

**7stud** · 05-07-2005

...and for some reason you deem this line:

setString(newString);

totally irrelevant?

**CornedBee** · 05-07-2005

You should implement + in terms of +=, not the other way round.

**GaPe** · 05-07-2005

How do I do that?

**Thantos** · 05-07-2005

Lets assume that your String class has the following data members:
a : a pointer that to the memory used to store the data
size : the amount of data being stored in a
alloc : the amount of memory allocated to a

Code:

String &String::operator+=(const char &character) {
  // Note INDENTATION
  if ( size == alloc ) // Ok we need more memory
    getMoreMemory(); // Call to private member function that gets more memory for us
  a[size++] = character;
  return *this;
}

Then you simply have operator + like so

Code:

String operator+ (const String& s, const char &c) {
  return String(s) += c;
}

Now before you ask how to write getMoreMemory() I'd suggest you search the boards this time as I won't be so nice again.

**CornedBee** · 05-07-2005

Typo in the second code example, Thantos. Ought to be "operator +", not "operator +=".

**MrWizard** · 05-07-2005

edit: Beaten to the typo..

But note, that you should prefer to return a const object for operator + to prohibit behavior such as

(String + String) = String

So,

Code:

const String operator+(const String& s, const char& c)

**Thantos** · 05-07-2005

Originally Posted by CornedBee

Typo in the second code example, Thantos. Ought to be "operator +", not "operator +=".

Thanks, I just copied his header and forgot to remove the =

Originally Posted by MrWizard

edit: Beaten to the typo..

But note, that you should prefer to return a const object for operator + to prohibit behavior such as

(String + String) = String

So,

Code:

const String operator+(const String& s, const char& c)

What? You example of

Code:

(String + String) = String

Makes no sense since it wouldn't be legal.

Thanks like saying:

Code:

int x = 5, y = 7, z = 10;
x + y = z;

Both would be illegal syntax. Adding the const on it doesn't change anything since its still by value.

**MrWizard** · 05-07-2005

Originally Posted by Thantos

Thanks, I just copied his header and forgot to remove the =

What? You example of

Code:

(String + String) = String

Makes no sense since it wouldn't be legal.

Thanks like saying:

Code:

int x = 5, y = 7, z = 10;
x + y = z;

Both would be illegal syntax. Adding the const on it doesn't change anything since its still by value.

Actually because in your example you don't return a const object you do let people write code like my example and it WILL compile. Of course it doesn't make sense , that's why you return a const object, to disallow that behavior! Here is a simple example if you still don't believe me:

Code:

struct MyInt
{
  int m_val;

  MyInt(int val) : m_val(val) { }
  MyInt& operator+=(int val)
  {
	m_val += val;
	return *this;
  }
};

MyInt operator +(const MyInt lhs, int rhs)
{
  return MyInt(lhs) += rhs;
}

int main(void)
{
  MyInt v1(100), v2(200), v3(-1);

  (v1 + 10) = v3;
}

Play around with that and try switching to return a const object and note that it will fix the problem.

**GaPe** · 05-09-2005

Thanks! It's working now.

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