Combine two numbers into one. Example: 123 and 456 become 123456

Question

In C++, how do I combine (note: not add) two integers into one big integer?

For example:

int1 = 123;
int2 = 456;

Is there a function to take the two numbers and turn intCombined into 123456?

EDIT:

My bad for not explaining clearly. If int2 is 0, then the answer should be 123, not 1230. In actuality though, int1 (the number on the left side) would only have a value if int2 goes over the 32 bit limit. So when int2 is 0, then int1 is 0 (or garbage, i'm not sure).

Answer 1

You could convert them into strings, combine them and then convert them back to an int?

Answer 2

You can turn them into strings and concat them.

itoa(num, buf, 10)

http://www.cplusplus.com/reference/clibrary/cstdlib/itoa/

Answer 3

You could use stringstream:

string Append(int _1, int _2){
    stringstream converter;

    converter << _1 << _2;

    return converter.str();
}

then call atoi on the returned string.

Answer 4

For each digit in int2, you can multiple int1 by 10 and then add int2:

// merge(123, 0) => 1230
int merge(int int1, int int2)
{
    int int2_copy = int2;
    do
    {
        int1 *= 10;
        int2_copy /= 10;
    } while (int2_copy);

    return int1 + int2;
}

You could get rid of the loop using log10 and ceil.

Answer 5

Assuming both ints are non-negative, and int1 goes on the left and int2 goes on the right, you need to figure out how many digits long int2 is, multiply int1 by 10 a bunch of times, and then add them.

unsigned int int1 = blah;
unsigned int int2 = blah;

unsigned int temp = int2;

do
{
    temp /= 10;
    int1 *= 10;
} while (temp >0)

unsigned int newInt = int1 + int2;

Answer 6

The power of ten, that you need to multiply the first number with, is the smallest one, that is bigger than the second number:

int combine(int a, int b) {
   int times = 1;
   while (times <= b)
      times *= 10;
   return a*times + b;
} 

Answer 7

Another option that works for C too:

#include <stdio.h>

int CombineInt(int int1, int int2)
{
  char cResult[32];

  sprintf(cResult, "%d%d", int1, int2);
  return atoi(cResult);
}

Answer 8

The following is essentially sth's accepted solution but with the b==0 fix, and the loop replaced with an expression to calculate the scale directly:

#include <math.h>

int combine(int a, int b) 
{
    int times = 1;
    if( b != 0 )
    {
        times = (int)pow(10.0, (double)((int)log10((double)b)) + 1.0);
    }
    return a * times + b ;
}

In some circumstances (such as a target with an FPU and a good maths library) the expression might be faster than the loop, but I have not tested that hypothesis.