Int + Unsigned Int

Let’s see some example code first.

Example 1

int main()
{
	unsigned int X = 10;
	int Y = 0;
	
	if( X > Y )	cout << "X >  Y" << endl;
	else		cout << "X <= Y" << endl;

	return 0;
}

Output

X >  Y

Example 2

int main()
{
	unsigned int X = 10;
	int Y = -1;
	
	if( X > Y )	cout << "X >  Y" << endl;
	else		cout << "X <= Y" << endl;

	return 0;
}

Output

X <= Y

Isn’t X always larger than Y in these two cases?

1. Analyse

Two different type variable addition, will lead to implicit type conversion.

unsigned int + int –> unsigned int

Look back to the examples:

X(10) > Y(0) 
==> (unsigned int)X(10) > (unsigned int)Y(0) 
==> true

X(10) > Y(-1) 
==> (unsigned int)X(10) > (unsinged int)Y(-1) 
==> (unsigned int)X(10) > (unsinged int)Y(4294967295) 
==> false

2. Trap

Why I write this artical? Because we will fell into this trap if we are not carefully enough. Look at the following code:

int main()
{
	cout << "Start Program" << endl;

	string str("hello world");
	for( int i=-1; i<str.size()-1; i++ )	
		cout << str[i+1];

	cout << "End Program" << endl;
	return 0;
}

Output

Start Program
End Program

Because string::size() will return size_t which is unsigned int. It is equal to example 1 in the previous.

We should pay more attension to the different data type operation.