Converting a [char] variable to [int] then back to [char] - c++

Question

I am using Visual C 6
I am trying to convert a character array (single-quotation) into an integer, then incrementing the value by 1, then storing the result back into a different character array..
But I keep getting an unexpected value when converting back to character..
Here is my code
char char_array[4];
char_array[0] = '1';
char_array[1] = '2';
char_array[2] = '3';
char_array[3] = '\0'; //Terminating character
int my_number = atoi(char_array);
printf("my_number = %d" , my_number); // output is 123
my_number++; // works and my_number is incremented =124
printf("now: my_number = %d" , my_number); // output is 124
char result[4]; //declared to store the result
result = itoa(my_number); // Output is unexpected.
printf("%c", result[0]); // Output is 2 instead of 1
printf("%c", result[1]); // Output is 2
printf("%c", result[2]); // Output as 3 instead of 4
It seems that the function itoa() somehow knows the original value 123 and in some weird way knows that I have incremented that value.. but the addition is done to the wrong digit. Instead of adding 1 to the least significant digit, the addition is done to the most significant digit.

Answer 1

I find it really difficult to believe that your compiler is letting this code through:
char result[4]; //declared to store the result
result = itoa(my_number); // Output is unexpected.
For one reason, you're attempting to reseat an array. Which shouldn't be allowed. For another, itoa() normally takes three arguments. It's prototype should look like:
char *itoa(int value, char * str, int base);
So you should be calling it as:
char result[4];
itoa(my_number, result, 10);
Or, if you'd like to use portable functions that don't have possible buffer overflows:
char result[4];
snprintf(result, 4, "%d", my_number);

Answer 2

itoa is not a standard C library function.
You can use
char result[sizeof(int) * CHAR_BIT / 10 * 3 + 4]; // '-1', '\0', max sizeof int on my 4 byte machine
// 10 bits are roughly equal to 3 digits at decimal base, extra 4 for '-', '\0', extra digit and safe character
sprintf(result, "%d", my_number);
If you still want to use itoa, consult the documentation of this function (in library/compiler documentation)

Answer 3

my_number is incremented and hence if you are using itoa() then it will know the new value of my_number which is 124.
Check the code below:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
char char_array[4];
char_array[0] = '1';
char_array[1] = '2';
char_array[2] = '3';
char_array[3] = '\0'; //Terminating character
int my_number = atoi(char_array);
printf("my_number = %d" , my_number); // output is 123
my_number++; // works and my_number is incremented =124
printf("now: my_number = %d" , my_number); // output is 124
char result[4]; //declared to store the result
snprintf(result,4,"%d",my_number);
printf("%c", result[0]);
printf("%c", result[1]);
printf("%c", result[2]);
return 0;
}

Answer 4

First the itoa(my_number) maybe wrong, I only know the followed function:
char * itoa ( int value, char * str, int base );
str should be an array long enough to contain any possible value:
(sizeof(int)*8+1) for radix=2 i.e. 17 bytes in 16-bits platforms and
33 in 32-bits platforms.