Want to Convert Integer to String without itoa function

Question

i want to convert int to char* in C without using itoa() function.

Because on my Linux Systems i itoa function is not there. i am using this code which i found from here

I want to run this function on Embedded devices also which are using Linux.

So i am looking for without use of itoa.

I dnt want to use sprintf also because its uses for just prints.

So any body please help me to figured out this problem.

Thanks

Answer 1

Here is another solution (I used the function that user1089679 posted above and addressed the comments that most other users have made). Any feedback on the i_to_a function (and ideally on the whole program) would be appreciated. Please feel free to be as critical and detailed as you want (if you are nice while doing it, would be great too :)):

#include <stdlib.h>
#include <stdio.h>

int no_of_digits( int num, int radix );
char *i_to_a( char *str, int digit_count, int num, int radix );

int main()
{
  int num, radix;

  printf( "Enter a decimal number: " );
  scanf( "%d", &num );
  printf( "Enter radix: " );
  scanf( "%d", &radix );

  int digit_count = no_of_digits( num, radix );
  char *str = malloc( sizeof( char ) * ( digit_count + 1 ) );
  if( str == NULL )
  {
    printf( "Malloc failed in main, exiting.\n" ); //do additional error handling as needed
    return 0;
  }
  str = i_to_a( str, digit_count, num, radix );
  printf( "\nThe converted number as a string is: %s\n", str );
  free( str );
  return 0;
}

int no_of_digits( int num, int radix )
{
  int digit_count = 0;

  if( num < 0 )
    digit_count++;

  while( num != 0 )
  {
    digit_count++;
    num /= radix;
  }
  return digit_count;
}

char *i_to_a( char *str, int digit_count, int num, int radix )
{
  if( num == 0 )
  {
    *str = '0';
    return str;
  }
  if( num == -2147483648 )
  {
    str[0] = '-';
    str[1] = '1';
    for( size_t i = 2; i < digit_count; ++i )
        str[i] = '0';
    return str;
  }
  
  
  if( num < 0 )
  {
    num = -1 * num;
    str[0] = '-';
  }
  
  str[digit_count] = '\0';
  while( num > 0 )
  {
    str[digit_count-1] = num % radix + '0';
    num /= radix;
    digit_count--;
  }

  return str;
}

Answer 2

#include <stdio.h>
#include <string.h>

#if 0
char *strrev(char *str){
    char c, *front, *back;

    if(!str || !*str)
        return str;
    for(front=str,back=str+strlen(str)-1;front < back;front++,back--){
        c=*front;*front=*back;*back=c;
    }
    return str;
}
#endif

char *itoa(int v, char *buff, int radix_base){
    static char table[] = "0123456789abcdefghijklmnopqrstuvwxyz";
    char *p=buff;
    unsigned int n = (v < 0 && radix_base == 10)? -v : (unsigned int) v;
    while(n>=radix_base){
        *p++=table[n%radix_base];
        n/=radix_base;
    }
    *p++=table[n];
    if(v < 0 && radix_base == 10) *p++='-';
    *p='\0';
    return strrev(buff);
}

int main ()
{
  int i;
  char str[33];

  printf ("Enter a number: ");
  scanf ("%d",&i);

  itoa (i,str,10);
  printf ("decimal: %s\n", str);

  itoa (i, str, 16);
  printf ("hexadecimal: %s\n", str);

  itoa (i, str, 2);
  printf ("binary: %s\n", str);

  return 0;
}

Answer 3

I found solution regarding this..

I am Happy to and i want which i expected.

#include <string.h>
#include <stdlib.h>

char *i_to_a(int num);

int main()
{

    char *str = i_to_a(4567);
    printf("%s",str);
    free(str);
    str = NULL;
return 0;

}
int no_of_digits(int num)
{
    int digit_count = 0;

    while(num > 0)
    {
        digit_count++;
        num /= 10;
    }

    return digit_count;
}


char *i_to_a(int num)
{
    char *str;
    int digit_count = 0;

    if(num < 0)
    {
        num = -1*num;
        digit_count++;
    }

    digit_count += no_of_digits(num);   
    str = malloc(sizeof(char)*(digit_count+1));

    str[digit_count] = '\0';

    while(num > 0)
    {
        str[digit_count-1] = num%10 + '0';
        num = num/10;
        digit_count--;
    }

    if(digit_count == 1)
        str[0] = '-';

    return str;
}

Answer 4

Here is a simple snippet you can use. There are more elegant and advanced ways, but this gets the job done.

In embedded projects in the past, I have measured this to be approximately 1000 times more efficient than sprintf(). This code is also MISRA-C compliant.

void getDecStr (uint8_t* str, uint8_t len, uint32_t val)
{
  uint8_t i;

  for(i=1; i<=len; i++)
  {
    str[len-i] = (uint8_t) ((val % 10UL) + '0');
    val/=10;
  }

  str[i-1] = '\0';
}

Answer 5

Thing is snprintf is the perfect function for this:

char str[LEN];
snprintf(str, sizeof(str), "%d", num);