Spliting string to an array in C - c

Question

int main()
{
int longNum = 12345, tempNum[5], i;
for (i = 0; i <= 5; i++)
{
tempNum[i] = longNum[i] ; // not valid, how do i make this work?
}
printf("%d\n", tempNum);
return 0;
}
Im trying to go through all the digits of longNum and push them into tempNum[].

Answer 1

You may try the modulo operator:
int main()
{
int longNum = 12345, tempNum[5], i;
for (i = 4; i >= 0; i--)
{
tempNum[i] = longNum % 10;
longNum /= 10;
}
for( i = 0; i < 5; i++ )
{
printf("%d\n", tempNum[i]);
}
return 0;
}

Answer 2

For a dynamic length:
#include <stdio.h>
#include <stdlib.h>
int main()
{
long long longNum = 1234512345; // can be any size within `long long` limit.
int tempNum[5], i;
int num_digits = 0;
long temp = longNum;
while( temp > 0 )
{
temp /= 10;
num_digits++;
}
printf( "num_digits = %d\n", num_digits );
// Allocate dynamic array.
int *pos = malloc( num_digits * sizeof( int ) );
for( i = num_digits - 1; i >= 0; i-- )
{
pos[i] = longNum % 10;
longNum /= 10;
}
for( i = 0; i < num_digits; i++ )
{
printf("%d\n", pos[i]);
}
return 0;
}

Answer 3

You could try this:
int main( )
{
int longNum = 12345, tempNum[5], i;
char numstr[99];
itoa( longNum, numstr, 10 );
for ( i = 0; i < 5; i++ )
{
tempNum[i]=numstr[i] - '0';
printf( "\n%d", tempNum[i] );
}
return 0;
}

Answer 4

#include <stdio.h>
int main()
{
int longNum = 12345, tempNum[5], i;
for (i = 1; i <= 5; i++)
{
tempNum[5-i] = longNum % 10;
longNum = longNum/10;
}
printf("%d%d%d%d%d\n", tempNum[0], tempNum[1], tempNum[2], tempNum[3], tempNum[4]);
return 0;
}