addition of two large integers of different size

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C++

Mint Mint::operator+=(const Mint &rhs) {
	int sum;
	int carry = 0;
	unsigned int i;
	unsigned int len = num.size();
	if (len < rhs.num.size())
	{
		len = rhs.num.size();
	}
	for (i=0;i<len;i++)
	{
		sum = num[i] + rhs.num[i] + carry;
		carry = sum / 10;
		sum = sum % 10;
		if (i < len)
		{
			num[i] = sum;
		}
		else
		{
			num.push_back(sum);
		}

	}
	if(carry != 0 )
	{
		num.push_back(carry);
	}
	return *this;
}

hello everyone
can you please check out this implementation of the += operator
i'm trying to write a class that deals with large integer numbers so i wrote this for the += operator and it works just fine for numbers that have the same size but when i try to add 2 numbers of different size i get some weird result

any advice

Posted 31-Dec-15 3:14am

Member 12223678

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Richard MacCutchan · Answer 1 · 2015-12-31T03:49:00

Solution 1

A simple method of 'normalisation' would be to take the shorter of the two numbers and make it the same as the longer one. Just expand it on the left with zeroes until they are both the same size. The addition should then work.

Posted 31-Dec-15 3:49am

Richard MacCutchan

Comments

Member 12223678 31-Dec-15 9:56am

than you for the answer but i'm not sure how would that work since i'm passing the rhs as const (const Mint &rhs)

Richard MacCutchan 31-Dec-15 10:06am

You would need to copy it to a temporary location. You also need to take into account the situation when the LHS is the shorter number.

Member 12223678 · Answer 2 · 2015-12-31T03:59:00

C++

Mint Mint::operator+=(const Mint &rhs) {
	int sum;
	int carry = 0;
	unsigned int i;

	if (num.size() < rhs.num.size())
	{
		while(num.size() < rhs.num.size())
		{
			num.push_back(0);
		}
	}
	unsigned int len = num.size() ;

	for (i=0;i<len;i++)>
	{
		sum = num[i] + rhs.num[i] + carry;
		carry = sum / 10;
		sum = sum % 10;
		if (i < len)
		{
			num[i] = sum;
		}
		else
		{
			num.push_back(sum);
		}

	}
	if(carry != 0 )
	{
		num.push_back(carry);
	}
	return *this;
}

would something like this work.?