Thread: Return value string

I was just reading some code (after I realized that a string -- array -- is actually a pointer to the first element in the string) and if had a function:

Code:

char* new_function(int x, int y)
{
    /* Code */
}

And I'm not really sure what the 'char*' means. Does it mean that new_function returns a string? If not, what is the return value from new_function that is signified by 'char*'? Thanks.

Code:

char* rtnString(char string[])
{
    printf("%s", string);
    string[0] = 'A';
    return (string);
}

int main()
{
    char string[] = "New string";
    string = rtnString(string);
    printf("%s", string);
    return (0);
}

This code would first print out (from rtnString) "New string" to the screen. Then after going through the function and returning to main, would it print out "Aew string"? I imported the string and then returned it with the return value of 'char *'. Would this work? Thanks.

I don't understand why you had to create pointer to the string in main. After all, a string is a pointer to the first element in the array of characters, right? So, wouldn't the levels of indirection be off? When I'm passing string into rtnString:

Code:

string = rtnString(string);

string is already a pointer, right? So, string would be *(string), which is a pointer. So, wouldn't I be assigning a char * to a char *? Your variable "s" would be a pointer to a pointer, right? Is that level of indirection off? Thanks.

I think I might understand what you are saying.

Code:

string = rtnString(string);

Do you mean that when I say string is equal the return value, that I'm just derefercing string and taking the value? But, aren't I working on the object? So, this is assigning the char * (a string) to a value? I guess that is what I don't understand. How (in my example) is string just a char? Or, is string just the first char (string[0])? Is that what you are saying? So, in my example, string is really just the first character? Not the char *, but just char? I think I'm partially understanding. Thanks...

Garfield

Like I said, pointers are fun, no?

Okay, here's the long and short of it. Your function is supposed to return a char* or pointer to a char, and that's fine because if you return an entire array then that's what you're passing, so far so good.

However, when you assign the returned array to itself back in the main, you are actually trying to assign an array to a single element of the same array.

string = rtnString(string);

This is pretty much saying, take the array that is given to me by rtnString and assign it to the first element of string. It's like this because when you put in just simply 'string' in an assignment you are really saying, here's the first element, put my value there as opposed to passing it or printing 'string' where it means, this is a pointer to the address of this array. So, to make a long story short:

string = rtnString(string); //assigning char* to char
is the same as:
string[0] = rtnString(string); //also assigning char* to char

so you get an indirection error, or an LValue error because an array can't be placed into a char. So what I did was create a pointer to your array in main and then assign the char* that you returned from rtnString and placed it there, which was doing the same thing you wanted to do, just without the errors.

char *s = string;
s = rtnString(string); //char* to string being assigned the edited
//string, also of char* == good.

If you're confused, don't worry. We all went through the whole 'WTF!' phase about pointers

-Prelude

OOOOHHHH!!!! I understand. string is equal to string[0], which is the first element in the array. The function returns and assigns a pointer (a string) to one element. Thanks! You couldn't have explained it better!

Garfield

phew, and here I thought I was being impossibly cryptic. It's nice to know someone understands my endless rants ;D

-Prelude

> No, that wouldn't work because you're trying to assign a char*
> to a char. Remember that just the name of the array with
> nothing else is a pointer to the first element, essentially it's
> saying array[0]. It's already dereferenced, so you're looking at
> the value, not the address

No. You are wrong. Well actually, you're right and wrong at the
same time.

It is legal to do the following:

Code:

char* myfun( char *s ) //duplicate a string
{
   //this code is irrelevent.
   //pay attention to the actual use of the function
   char *d = NULL;

   if ( s == NULL ) return NULL;
   d = malloc ( strlen( s ) + 1 );
   strncpy( d, s, strlen(s) );
   return d;
}

int main ( void )
{
   char array[SIZE];
   char *s;

   //valid
   s = myfun( "hello" );
   puts( s );
   free( s );

   //valid
   strncpy( array, "hi", 2 );
   s = myfun( array );
   puts( s );
   free( s );

   //valid, assuming 's' is pointint to usable space
   //note, this would actually cause a memory leak
   //because 's' isn't being freed inside the function,
   //so it'd end up pointing to the new string, "losing"
   //the old string. however, it is still legal
   s = myfun( s );

   //invalid.
   //actually, it _is_ valid. You could do this. It however,
   //is wrong. since 'array' is an array, even though it
   //is treated as a pointer, it is wrong in this case because
   //we're actually telling it to point some place else. this
   //is the reason it is wrong.
   array = myfun( s );

The reason assigning a value to array is incorrect in this place is because we're trying to assign 'array' a new value, rather than use it's currently allocated memory block.

With an array, you cannot just use 'realloc' or something to give it new space. This is the only real way they differ.

Quzah.

Ah, thank you for clearing that up Quzah. I knew I wasn't getting it completely right and was hoping someone else would pop in and fill in the blanks.

-Prelude