Memory allocation problem

Hi all, I am new to C and am having a memory allocation problem. I am creating a char* and initializing to NULL, then passing it to another function which allocates some memory for it, and copies some characters. When the second function returns, my char* returns to NULL. I'm sure I am missing something simple here but cannot figure it out. Any help would be greatly appreciated. See the example code below:

Code:

void functionA() {
  char *text = NULL;
  functionB(text);
  //variable text is still NULL
}

void functionB(char *text){
  char *tempText = "12345";
  size_t len = strlen(tempText);
  text = (char*)malloc(len * sizeof(char) + 1);
  strcpy(text, tempText );
}

Because C only has pass by value.
So if you want functionB() to change a pointer declared in functionA, then you need to pass a pointer to that pointer.

Code:

void functionA() {
  char *text = NULL;
  functionB(&text);
}

void functionB(char **text){
  char *tempText = "12345";
  size_t len = strlen(tempText);
  *text = malloc(len * sizeof(char) + 1);  // cast not needed in C
  strcpy(*text, tempText );
}

Thanks for the reply, that has clarified a lot for me however I do have one more question. If the function that allocates the memory for the char* is another level deep (say functionC() called from functionB()), I am assuming that I will need to pass the char** in functionB() as a reference, and functionC() will pick it up as a char***. I was wondering if there was any way to simply pass the memory address from start to finish? I'm asking as the software I am working on is quite badly written, and may require the allocation to happen in perhaps 3 or 4 nested functions (unless I do a significant re-write). I'd rather not be working with char******...etc

If the function that allocates the memory for the char* is another level deep (say functionC() called from functionB()), I am assuming that I will need to pass the char** in functionB() as a reference, and functionC() will pick it up as a char***

No..(AFAIK), just passing the same pointer should suffice ..(As long as all the allocation take place on the heap...and you don't return local pointers .)

Originally Posted by rtlm

Thanks for the reply, that has clarified a lot for me however I do have one more question. If the function that allocates the memory for the char* is another level deep (say functionC() called from functionB()), I am assuming that I will need to pass the char** in functionB() as a reference, and functionC() will pick it up as a char***. I was wondering if there was any way to simply pass the memory address from start to finish? I'm asking as the software I am working on is quite badly written, and may require the allocation to happen in perhaps 3 or 4 nested functions (unless I do a significant re-write). I'd rather not be working with char******...etc

Originally Posted by manasij7479

No..(AFAIK), just passing the same pointer should suffice ..(As long as all the allocation take place on the heap...and you don't return local pointers .)

Manasij7479 is correct, you are not going to keep adding layers of redirection. Just keeping passing the pointer the same way you did in the first function, where the original pointer is referred to as *text.

Code:

void foo(void){
     char *text=NULL;
     fooA(&text);
}
void fooA(char **textA){
     *textA = <whatever>
     fooB(&*textA);
}
void fooB(char **textB){
     *textB = <whatever>
}
ect...

thanks for all the help guys!

Code:

fooB(&*textA);

A little redundant, no?