return statement

[swapnaoe]

Can anyone explain me the output of the following code

char *rev(int val);
void main()
{
extern char dec[];
printf("%c", *rev);
}
char *rev(int val)
{
char dec[]="abcde";
return dec;
}

[Banfa]

This code doesn't compile without warnings (assuming you are using a C file, in C++ is just doesn't compile). Unless you have a very good reason (like a legacy project with 100s of files) then you should always eliminate all warning messages by correcting the code.

However it does compile but since you have written main incorrectly you have invoked undefined behaviour, under undefined behaviour the program is not required to do(or not do) anything so it could output the entire works of Shakespear or it could output nothing. In my case it output a club (from cards) symbol.

Here is your code with my comments in bold

 /* No headers are included, stdio.h is required for printf 
   and stdlib.h is often a good idea too */
 
char *rev(int val);
void main()
 /* Main is declared incorrectly it should be
int main(int argc, char **argp)
 */
{
  extern char dec[];
 /* You have tried to declare a external reference to a 
   non-existent external variable.  dec in the function below 
   is local to the function and not available in this scope. */
  printf("%c", *rev);
 /* I assume this is meant to call the function rev but the 
   syntax is wrong, it should be something like
  printf("%c", *rev(0));
*/
}
 
char *rev(int val)
{ 
 /* variable val not used */
  char dec[]="abcde";
  return dec;
 /* You are returning a reference to a local variable, dec.  dec will no longer exist once the function has returned because it is an automatic variable (often placed on the stack) so the call function will be referencing a memory location that is not allocated resulting in undefined behaviour.  Declare dec outside the function or static.*/
}