ENV31-C. Do not rely on an environment pointer following an operation that may invalidate it

If an environment variable might have changed during program execution, get a fresh copy using getenv(). If you rely on your old variable, you could be left with incorrect data or a dangling pointer.

Non-Compliant Coding Example

After a call to setenv(), environment pointers to the old value and copies of the old value will be incorrect.

char *temp;
char *copy;

if ((temp = getenv("TEST_ENV")) != NULL) {
  copy = malloc(strlen(temp) + 1);
  if (copy != NULL) {
    strcpy(copy, temp);
  }
  else {
    /* handle error condition */
  }
}
/* ...program code... */
setenv("TEST_ENV", var, 1);
/* ...program code... */


printf("TEST_ENV: %s\n", temp);
printf("TEST_ENV: %s\n", copy);

Neither of the print statements will be correct.

Compliant Solution

You should always fetch fresh copies of environment variables, especially if you know that a value has changed.

char *temp;
char *copy;

if ((temp = getenv("TEST_ENV")) != NULL)

else {
/* handle error condition */
}
}
/* ...program code... */
setenv("TEST_ENV", var, 1);
/* ...program code... */

if ((temp = getenv("TEST_ENV")) != NULL) {
copy = malloc(strlen(temp) + 1);
if (copy != NULL)

}
else

}
printf("TEST_ENV: %s\n", temp);
printf("TEST_ENV: %s\n", copy);
This will provide us with the current value of the environment variable.

Risk Assessment

The program would not be using current environment values, causing unexpected results.

References

[[ISO/IEC 9899-1999:TC2]] Section 7.20.4, "Communication with the environment"