convert integer to array, c++

Question

I would like to convert an integer into an array, so that it looks like the following:

int number = 123456 ;
int array[7] ;

with the result:

array[0] = 1 
array[1] = 2
...
array[6] = 6

Answer 1

The easiest way I can imagine now is:

char array[40];
int number = 123456;

memset(array, 0x00, sizeof(array));

sprintf(array, "%d", number);

Additionally you can convert each digit to int just subtracting the char value by 0x30.

EDIT: If this is a homework, your teacher you probably ask you to write the program using % operator though (example 12 % 10 = 2). If this is the case, good homework ;-)

Answer 2

You can extract the last digit of the number this way:

int digit = number % 10;
number /= 10;

Note that you should also check whether number is positive. Other values require additional handling.

Answer 3

You can use modulus to determine the last digit.

And you can use division to move another digit to the last digit's place.

Answer 4

just use modular arithmetic:

int array[6];
int number = 123456;
for (int i = 5; i >= 0; i--) {
    array[i] = number % 10;
    number /= 10;
}

Answer 5

Perhaps a better solution is to work backwards:

123456 % 10 = 6

123456 / 10 = 12345

12345 % 10 = 5

12345 / 10 = 1234

Answer 6

Take the log10 of the number to get the number of digits. Put that in, say pos, then, in a loop, take the modulo of 10 (n % 10), put the result in the array at position pos. Decrement pos and divide the number by 10. Repeat until pos == 0

What did you want to do with the sign if it's negative?

Answer 7

You can't simply "convert" it. The integer is not represented in software in decimal notation. So the individual digits you want don't exist. They have to be computed.

So, given an arbitrary number, how can you determine the number of ones?

We could divide by ten, and then take the remainder: For 123, the division would give 12, and then there's a remainder of 3. So we have 3 ones. The 12 tells us what we have past the ones, so it can be our input for the next iteration. We take that, divide by 10, and get 1, and a remainder of 2. So we have 2 in the tens place, and 1 left to work with for the hundreds. Divide that by 10, which gives us zero, and a remainder of 1. So we get 1 in the hundreds place, 2 in the tens place, and 3 in the ones place. And we're done, as the last division returned zero.

Answer 8

thank you all for your help. first of all: this is not my homework, it's part of a braintester i'm doing just for fun. the methos, proposed by ehuhtala works good with my programm! thank you all for your help :)

Answer 9

See SO question Language showdown: Convert string of digits to array of integers? for a C/C++ version (as well as other languages).

Answer 10

if this is really homework then show it your teacher - just for fun ;-)

CAUTION! very poor performance, clumsy way to reach the effect you expect and generally don't do this at home(work) ;-)

#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>

typedef std::vector< int > ints_t;

struct digit2int
{
    int operator()( const char chr ) const
    {
        const int result = chr - '0';
        return result;
    }
};

void foo( const int number, ints_t* result )
{
    std::ostringstream os;
    os << number;
    const std::string& numberStr = os.str();
    std::transform(
        numberStr.begin(),
        numberStr.end(),
        std::back_inserter( *result ),
        digit2int() );
}

int main()
{
    ints_t array;
    foo( 123456, &array );
    std::copy(
        array.begin(),
        array.end(),
        std::ostream_iterator< int >( std::cout, "\n" ) );
}

Answer 11

but what if i need a 30 digit long !!! i cant put it in a variable!