Convert an int array to a single int

Question

I'm trying to convert array[]={1,2,3} into a int number=123; how can I do that ?

my code is this:

int main()
{

    int array[]={1,2,3};
    int number;

    for (int i =0; 3<i ; i++){
        int val=1;
        for(int j=0; j<i; j++ ){
                val*=10;
        }
        number += array[i] *val;

    }
    cout<<number;

    while(1);
    return 0;
}

Answer 1

You're adding numbers in the wrong "direction". To keep the digits in order, you need to multiply your number variable by 10 and then add array[i] instead of multiplying array[i] and adding it to number.

You also need to initialize number to zero before you use it, because a variable has a random value until it's explicitly given one.

You also need to do i < 3 ("loop while i is less than three") instead of 3 < i ("loop while 3 is less than i", which is never true).

int array[] = {1,2,3};
int number = 0;

for (int i = 0; i < 3; i++) {
    number *= 10;
    number += array[i];
}

cout << number;

Let's walk through what happens.

• In the beginning, number is equal to zero.
• We get to the loop. First, i equals 0.
  • We multiply number by 10. For the first iteration, number is still zero after that.
  • We add array[0] (1) to number. Number is now 1.
• i now increments and is equal to 1. 1 is less than 3, so we go in the loop body again.
  • We multiply number again by 10 to make room for the next digit. number is now equal to 10.
  • We add array[1] (2) to number. Number is now 12.
• i increments and is equal to 2. 2 is less than 3, so we repeat.
  • We multiply number by 10, again to make room for the next digit. It's now 120.
  • We add array[2] (3) to number, making it 123, the desired result.
• i increments and becomes 3. 3 is obviously not less than 3, so we exit the loop.
• We print number (123) to the console.

Answer 2

I would do this using streams, as your case is really about lexical interpretation.

int number;
std::array<int, 3> arr { 1,2,3 };
std::stringstream ss;
for(int i : arr) ss << i;
ss >> number;
std::cout << number;

or course if you don't need the number itself you can just use std::cout.

std::array<int, 3> arr { 1,2,3 };
for(int i : arr) std::cout << i;

Answer 3

#include <iostream>
#include <cmath>

int main()
{
    int numarray[] = {1,2,3};
    int num = 0;

    for(int i = 2; i>=0; i--)
    {
        num += numarray[2-i]*pow(10,i);
    }

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

Replace 2 with the maximum size (m) of a number array minus 1 for different sized arrays.

Answer 4

Looks like your first for loop is messed up. Try this:

for (int i =0; i<3 ; i++){
    int val=1;
    for(int j=0; j<(3-i); j++ ){
            val*=10;
    }
    number += array[i] *val;

}