Wsrtjtyk

I have a 3 digit integer myInt:

int myInt = 809;

I need an output string *myStr which consists of the 2 last digits. Therefore

char *mystr = "09";

What would be the easiest solution to do that?

Here is a simple solution:

#include <stdio.h>



int main(void) {

    char buf[3];

    char *myStr;

    unsigned int myInt = 809;



    buf[0] = myInt / 10 % 10 + '0';  /* tens digit */

    buf[1] = myInt % 10 + '0';       /* unit digit */

    buf[2] = '';                   /* null terminator */

    myStr = buf;



    puts(mystr);

    return 0;

}

You can do like this:

#include <stdio.h>

#include <string.h>



int main(int argc, char **argv)

{

    char buf[32];

    int myInt = 809;

    char* mystr;

    snprintf(buf, sizeof buf, "%d", myInt);

    if (strlen(buf) > 2)

        mystr = buf + strlen(buf) - 2;

    else

        mystr = buf;

    fputs(mystr, stdout);

    return 0;

}

For the specific case of turning "magic numbers" into strings, you can as well do that at compile-time. That is, when you have an integer constant rather than a run-time value:

#include <stdio.h>



#define VAL 809



#define STRINGIFY(x) #x

#define STR(i) STRINGIFY(i)

#define SUBSTR(i, n) &(STRINGIFY(i))[n]



int main (void)

{

  int whatever = VAL;



  puts(STR(VAL));

  puts(SUBSTR(VAL, 1));



  return 0;

}

Output:

809

09

No pointer magic necessary here. Just use plain math (the remainder operator: %), along with some suitable formatting:

#include <stdio.h>



int main(void)

{

  char a[3]; /* 2 characters for any value >= 0 and < 100. So we rely on the 100 below.

                +1 character for the `0`-terminator to make this a C-string*/

  int i = 809;



  assert(i >= 0); 



  sprintf(a, "%02d", i % 100);



  puts(a);

}

or avoiding magic numbers:

#include <stdio.h>



long long power_integer(int base, int x)

{

  if (0 == x)

  {

    return 1;

  }



  return base * power_integer(x - 1);

}



#define CUTOFF_BASE (10)

#define CUTOFF_DIGITS (2)



int main(void)

{

  char a[3]; 

  char f[8]; /* To hold "%01lld" to "%019lld". 

                (A 64 bit integer in decimal uses max 19 digits) */



  int i = 809;



  assert(i >= 0); 



  sprintf(f, "%%0%dlld*, CUTOFF_DIGITS);

  sprintf(a, f, i % power_integer(CUTOFF_BASE, CUTOFF_DIGITS));



  puts(a);

}

Output would be:

09

You could try something like this:

#include <stdio.h>

#include <string.h>

int main(int argc, char **argv){

    int myInt = 809;

    char buf[16];  // buffer big enough to hold the string representation

    sprintf(buf, "%d", myInt); // write the string to the buffer

    char* myStr = &buf[strlen(buf) - 2]; // set a pointer to the second last position

    printf("myStr is %sn", myStr);

    return 0;

}