HELP! Placing an extra copy of the 4th digit.

[kaye0822]

Hi, my goal is to write a function called doubleFour that places an extra copy of the 4th digit right after that digit in an integer parameter. The problem is after I placed an extra copy of the 4th digit, I'm stuck of adding back the remaining number. I know there has to be some thing to do with x%10 and x/10, but I'm just stuck of putting it in. PLEASE help !

  #include <iostream>
using namespace std;

int doubleFour (int x);

int main() {
int x = doubleFour(19683);
cout << x << endl; // prints 196883
cout << doubleFour(271828)<< endl; // prints 2718828
cout << doubleFour(314159)<<endl; // prints 3141159
return 0;
}


int doubleFour(int x){
	int y,z,c, temp=x%100;
	if (x<=9999){
		y=((x/10)*10)+(x%10);
		z=(y*10)+(y%10);
		return (z*10);
	}
	else return doubleFour(x/10);

}

[snowright]

I am not quite sure I understand what you mean. Test this, it might be just what you want.

int doubleFour(int x) {
	return 10 * ( x - x % 1000 ) + x % 10000;
}

Otherwise could you explain in a different way "places an extra copy of the 4th digit right after that digit"?

[JockX]

@snowright - it looks like your code works almost as expected, but there is a catch - you are doubling 4th digit from the right, and the examples show that it is fourth digit from the left that needs doubling.
I have put together a formula that will do just that - double n-th digit from the left, so it is a bit more advanced than requested. But basically, all the calculations are done in single line ;P
Hope this helps <trollface>

#include <cmath>
#include <iostream>
using namespace std;
/***
 * This will take two parametes: 
 * x - is the number that is to be worked on 
 *     (please mind the int capacity - it will 
 *     not  work for 10+ digit integers!)
 * y - it is which digit (from the left!) is 
 *     going to be doubled. If y is more than
 *     total number of digits, nothing will be
 *     changed.
 */
unsigned int doubleDigit(unsigned int x, int y)
{
	// Function will need the number of digits
	int nDigits = 1;
	while (pow(10.0, nDigits) <= x)
		nDigits++;

	if(nDigits < y)
		return x;

	// Based on number of digits, x and y result is returned:
	return 
		(int)(x/(pow(10.0,nDigits-y)))*(int)pow(10.0,nDigits-y+1)
		+(int)(x/(pow(10.0,nDigits-y)))*(int)pow(10.0,nDigits-y)-
		(int)(x/(pow(10.0,nDigits-y+1)))*(int)pow(10.0,nDigits-(y-1))
		+(x-(int)(x/(pow(10.0,nDigits-y)))*(pow(10.0,nDigits-y)));
}

int main()
{
	unsigned int yourNumber;
	cout << "Enter your number:";
	cin >> yourNumber;

	// Will double each of first 9 digits:
	for(int i = 1; i < 10; i++)
	{
		cout << "Doubling digit #" << i << ": ";
		cout << doubleDigit(yourNumber, i) << endl;
	}
	system("pause");
}

[kaye0822]

Hi. what i mean is, to repeat whatever digit is in the 4th place of the number.

123456 become 1234456
356768 become 3567768

and so.

Thanks for advance.

[ngbeslhang]

Maybe try using the arrays instead?
I'm still some kind of new to C++ thought.

[snowright]

kaye0822, thanks for the clarification, solution will follow.
JockX, I have a better idea, how do you like it?

int doubleFour( int x ) {
	int e = floor( log10( x ) );
	return 10 * ( x - x % ( int ) pow( 10, e - 3 ) ) + x % ( int ) pow( 10, e - 2 );
}

[ShodanHo]

Using a collection class:

#include <iostream>
#include <vector>

#define STR(x) #x << '=' << x

int doubleFour(unsigned x)
{
  std::vector<unsigned> digits;
  while (x) {
    digits.push_back(x % 10);
    x /= 10;
  }
  std::size_t where = digits.size() + 1 - 4;
  while (digits.size()) {
    x *= 10;
    x += digits.back();
    if (digits.size() == where) {
      x *= 10;
      x += digits.back();
    }
    digits.pop_back();
  }
  return x;
}

int main()
{
  std::cout << STR(doubleFour(0)) << std::endl;
  std::cout << STR(doubleFour(1)) << std::endl;
  std::cout << STR(doubleFour(12)) << std::endl;
  std::cout << STR(doubleFour(123)) << std::endl;
  std::cout << STR(doubleFour(1234)) << std::endl;
  std::cout << STR(doubleFour(12345)) << std::endl;
  std::cout << STR(doubleFour(123456)) << std::endl;
  std::cout << STR(doubleFour(1234567)) << std::endl;
  std::cout << STR(doubleFour(356768)) << std::endl;
  return 0;
}

gives:

doubleFour(0)=0
doubleFour(1)=1
doubleFour(12)=12
doubleFour(123)=123
doubleFour(1234)=12344
doubleFour(12345)=123445
doubleFour(123456)=1234456
doubleFour(1234567)=12344567
doubleFour(356768)=3567768

[dukhi x]

int dnum(int x)
{
	int n=0,t=x;int arr[10];int ret=0;
	for(n;t>0;n++)
	{
		arr[n]=t%10;
		t/=10;
	}
	n--;//index of array is 1 less than size of array
	for(int i=0;i<=n;i++)
	{
		ret=ret*10+arr[n-i];
		if(i==3)
		{
			ret=ret*10+arr[n-i];
		}
	}
	cout<<"ret : "<<ret;
	return 0;
}

[JockX]

snowright wrote:
I have a better idea, how do you like it?

int doubleFour( int x ) {
    int e = floor( log10( x ) );
    return 10 * ( x - x % ( int ) pow( 10, e - 3 ) ) + x % ( int ) pow( 10, e - 2 );
}

I don' t like it, because it is better, clearer and shorter than mine :P. It's ideal for the task. My only comfort is that it crashes for numbers lower than 1000 and only covers digit at position 4 :)

[snowright]

Well, you know, I tend not to provide working solutions for people often since they are most likely here learning to do things themselves. I only provide ideas. But yes, you would probably want to check e >= 3 given you do not know the number is not large enough already before using something similar I provided straight. Also replacing e - 3 and e - 2 with e - n + 1 and e - n + 2 correspondingly will provide a general solution for n-place digit ( combined with the check e >= n - 1 if that is needed ).

kaye0822, needed you explanation or assist in understanding any solution, it will be provided after asked.

[ngbeslhang]

Why don't use std::string?
i means input, l means length and f means four.

#include <iostream>

int main() {
    std::string i;

    std::cin >> i;
    int l = i.length();

    while (l < 4) {
        std::cin >> i;
    }

    std::string f = i.substr(3, 1);
    i.insert(3, f);

    std::cout << i;
    return 0;
}

I've learnt how to use std::string and found that one of it's method, insert() can solve this question easily.
Anyway I'm still learning C++ yet. (And also I'm using mobile, which the screen is to small and makes me hard to type the code... :( )
Here's the online example: http://ideone.com/7topbn