bit of C help

[Martin]

so I haven't done C in about 3 years, so I was wondering if I could get a bit of help.

I need to make a function that returns a "char **" type. in my function I have

char[size] result;
// fill the array with stuff
return &result;

I seem to remember that the array's first element is a pointer, so this should return the pointer to that? or is my memory failing me?

 

[sao123]

first of all... all elements of the array are the same...
so... if the first element of the array is a pointer, then all of them are.



return &array;
returns the address of the first element.

so this function...

type* foo()
{
type array[20];
//do something
return &array;
}


would be used like this...

type* array_pointer;
array pointer = foo();


however...
char** is a pointer to a pointer... a highly nontypical scenerio.

 

[aCynic2]

Remember, anything allocated on the stack is freed when the function returns and it's contents are not reliable.

Since you're saying C, not C++, you can malloc for your pointer and return that. So... (keep in mind, it's been perhaps 10 year since I programmed).

char *func()
{
char *p;

if ((p = (char *) malloc (P_SIZE + 1)) = NULL)
return p;

/* do what you need */

return p;
}


Regardless of success you return p, and test it in the receiving function for NULL/not NULL.

 

[Matthias99]

however...
char** is a pointer to a pointer... a highly nontypical scenerio.

...unless you want an array of (char *) pointers -- for instance, a table of variable-length strings. It's also what gets passed into your 'main' function in the 'argv' variable, so it's pretty common to see.

e.g.

char character_array[20];

produces an array of 20 chars -- 'character_array' or '&character_array' or '&(character_array[0])' are of type (char *) or (char[ ]).

char * character_pointer_array[20];

produces an array of 20 (char *) pointers -- 'character_pointer_array' or '&character_pointer_array' or '&(character_pointer_array[0])' are of type ((char *)*) or ((char *)[ ]).