Why is there a `-Wint-conversion` warning?

November 10, 2022

I’m not new to C, but this doesn’t make sense in my mind. in my char encrypt function, I get this warning:

crypt.h: In function ‘encrypt’:
crypt.h:32:9: warning: returning ‘char *’ from a function with return type ‘char’ makes integer from pointer without a cast [-Wint-conversion]
   32 |   return(encrypted_string);
      |         ^

Please note: This is supposed to return char not char*

I have fixed this seemingly by changing it to char *encrypt. But that doesn’t make sense.
Can somebody explain how and why this works this way? The code seems to work, but clarity would be nice.

Here’s my code:

char encrypt(char string[])
{
  // allocating new buffer for encrypted string
  // note: normal string
  char encrypted_string[strlen(string)];

  // comparing string to cipher
  for(int i = 0; i < strlen(string); i++)
  {
    for(int j = 0; j < strlen(CHARSET); j++)
    {
      if(string[i] == CHARSET[j])
      {
        encrypted_string[i] = CIPHER[j];
        break;
      }
    }
  }
  return(encrypted_string);// returns pointer?
}

>Solution :

char encrypted_string[strlen(string)]; is an array, that decays into a char* whenever used in an expression.
Your function returns char, an integer type.
Thus: "returning ‘char *’ from a function with return type ‘char’"

Please note: This is supposed to return char not char*

That really doesn’t make any sense.

I have fixed this seemingly by changing it to char *encrypt. But that doesn’t make sense.

Well true, but maybe not for the reason you thought…You can’t return a pointer to a local array inside a function, as explained by any decent beginner-level learning material (and thousands of other SO posts).

You can fix your program by using caller allocation instead and return the result through a parameter. Or return a char* to a dynamically allocated string, but in that case mind memory leaks.

Also possible bugs here: for(int j = 0; i < strlen(CHARSET); i++). Should probably be j not i.