A program that prints even and odd characters from a string

this declaration is wrong:

char *str[41];

you're declaring 41 uninitialized strings. You want:

char str[41];

then, scanf("%40s" , str);, no & and limit the input size (safety)

then the loop (where your while (str[i]<41) is wrong, it probably ends at once since letters start at 65 (ascii code for "A"). You wanted to test i against 41 but test str[i] against \0 instead, else you get all the garbage after nul-termination char in one of odd or even strings if the string is not exactly 40 bytes long)

while (str[i]) {
    if (i % 2 == 0) {
        odd[j++] = str[i];
    } else {
        even[k++] = str[i];
    }
    i++;
}

if you want to use a pointer (assignement requirement), just define str as before:

char str[41];

scan the input value on it as indicated above, then point on it:

char *p = str;

And now that you defined a pointer on a buffer, if you're required to use deference instead of index access you can do:

while (*p) { // test end of string termination
    if (i % 2 == 0) {  // if ((p-str) % 2 == 0) { would allow to get rid of i
        odd[j++] = *p;
    } else {
        even[k++] = *p;
    }
    p++;
    i++;
}

(we have to increase i for the even/odd test, or we would have to test p-str evenness)

aaaand last classical mistake (thanks to last-minute comments), even & odd aren't null terminated so the risk of getting garbage at the end when printing them, you need:

even[k] = odd[j] = '\0';

(as another answer states, check the concept of even & odd, the expected result may be the other way round)

There are multiple problems in your code:

• You define an array of pointers char *str[41], not an array of char.
• You should pass the array to scanf instead of its address: When passed to a function, an array decays into a pointer to its first element.
• You should limit the number of characters read by scanf.
• You should iterate until the end of the string, not on all elements of the array, especially with (&str[i] < 41) that compares the address of the ith element with the value 41, which is meaningless. The end of the string is the null terminator which can be tested with (str[i] != '\0').
• You should read the characters from str with str[i].
• You should null terminate the even and odd arrays.

Here is a modified version:

#include <stdio.h>

int main(void) {
    char str[41];
    char odd[21];
    char even[21];
    int i = 0; 
    int j = 0;
    int k = 0;

    printf("Enter a string (40 characters maximum): ");
    if (scanf("%40s", str) != 1)
        return 1;

    while (str[i] != '\0') {
        if (i % 2 == 0) {
            odd[j++] = str[i];
        } else {
            even[k++] = str[i];
        }
        i++;
    }
    odd[j] = even[k] = '\0';

    printf("The even string is: %s\n", even);
    printf("The odd string is: %s\n", odd);

    return 0;
}

Note that your interpretation of even and odd characters assumes 1-based offsets, ie: the first character is an odd character. This is not consistent with the C approach where an even characters would be interpreted as having en even offset from the beginning of the string, starting at 0.

here is your solution :)

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

int main(void) 
{

    char str[41]; 
    char odd[21];
    char even[21];
    int i = 0; 
    int j = 0;
    int k = 0;

    printf("Enter a string (40 characters maximum): ");
    scanf("%s" , str); 

    while (i < strlen(str)) 
    {
        if (i % 2 == 0) {
                odd[j++] = str[i];
        } else {
                even[k++] = str[i];
        }
        i++;
    }
    odd[j] = '\0'; 
    even[k] = '\0';

    printf("The even string is:%s\n " , even);

    printf("The odd string is:%s\n " , odd);

    return 0;
}

solved the mistake in the declaration, the scanning string value, condition of the while loop and assignment of element of array. :)

Many answers all ready point out the original code`s problems.

Below are some ideas to reduce memory usage as the 2 arrays odd[], even[] are not needed.

As the "even" characters are seen, print them out.
As the "odd" characters are seen, move them to the first part of the array.

Alternative print: If code used "%.*s", the array does not need a null character termination.

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

int main(void) {
  char str[41];

  printf("Enter a string (40 characters maximum): ");
  fflush(stdout);
  if (scanf("%40s", str) == 1) {
    int i;
    printf("The even string is:");
    for (i = 0; str[i]; i++) {
      if (i % 2 == 0) {
        str[i / 2] = str[i];  // copy character to an earlier part of `str[]`
      } else {
        putchar(str[i]);
      }
    }
    printf("\n");
    printf("The odd string is:%.*s\n ", (i + 1) / 2, str);
  }
  return 0;
}

or simply

printf("The even string is:");
for (int i = 0; str[i]; i++) {
  if (i % 2 != 0) {
    putchar(str[i]);
  }
}
printf("\n");
printf("The odd string is:");
for (int i = 0; str[i]; i++) {
  if (i % 2 == 0) {
    putchar(str[i]);
  }
}
printf("\n");