On Tue, 2005-11-22 at 14:01 +1100, Benno wrote:
> On Tue Nov 22, 2005 at 13:26:24 +1100, ashley maher wrote:
> >G'day,
> >
> >I know this is not even C 101 level but would some kind soul please
> >explain to me why this is not even close to working.
>
>
> So you are basicalyl missing what pointers mean. char * is
> a pointer to some memory that can hold a bunch of chars. It
> is not an actual string. I try to explain further:
>
> >
> >int somefunction(char *string1)
> >{
>
> So this is some function that gets pointer to some memory. It
> gets a *copy* of an agument of the stack, and shoves it in a
> local variable called string1. string1 is a pointer to some memory.
>
> > char *string2 = "some words\0";
>
> Here you allocate variable string2, and point it to some memory with
> the data "some words".
> (b.t.w: you don't need \0).
>
> > string1 = (char *)calloc(strlen(string2 + 1), sizeof (char));
>
> Here you allocate zome memory from the heap, specifically enough to a
> copy of the memory pointed to by string2. In doing so it redefines
> what
> ever value was passed in to the function. Oh wait, even more subtle
> than that,
> I missed it. You actually allocate 2 less than need to store string2,
> becuase
> you actually take the length of string2 + 1, which is the length of
> the
> string "ome words", which is 2 less than need to store null terminated
> string2.
>
> > strcpy(string1, string2);
>
> It now copies the contents of the memory pointed to by string2 into
> the new memory pointed to by string1. (Note, no need to use calloc,
> you could have just used malloc, since you just overwrote all that
> memory
> anyway.)
>
> > return 0;
>
> Up to here you haven't done anything that is visible globally or to
> the
> calling function, exception allocate some memory on the heap. And now
> since you return you have created a memory leak, because there is no
> longer any reference to string1.
>
> >}
> >
> >
> >int main ()
> >{
> > char *string;
>
> Ok, this decalres a string point. That is a pointer to some memory.
> But since you didn't initialise it, it currently point to some random
> place in memory.
>
> > somefunction(string);
>
> You call this function that doesn't modify the string at all.
>
> > printf ("\n\nString is: %s\n\n", string);
>
> And then you print out some random data.
>
> > free (string);
>
> And then you try and free something not previously allocated by
> malloc, so it dies in a heap.
>
> > return 0;
> >}
> >
>
> I suggest compiling with -Wall, which will point out a lot of these
> errors before you execute the code.
>
> Another style point is that it is generally consider bad to allocate
> memory
> in one function and free it in another.
>
> I suspect the closest thing to what you wanted to do here would be:
>
> int somefunction(char **string1)
> {
> /* a function that takes a pointer to a pointer to memory */
> char *string2 = "some words";
>
> *string1 = malloc(strlen(string2) + 1);
>
> /* allocate memory and assign it to the char pointer, to which
> string1 points
> to. Known as dereferencing. */
>
> if ((*string1) == NULL) /*SHould at least check return value!
> */
> return -1;
>
> strcpy(*string1, string2);
> /* Copy the string */
>
> return 0;
> }
>
>
> int main(void) {
> char * string;
> int r;
>
> r = somefunction(&string);
> /* Pass the address of string to somefunction */
>
> if (r != 0) {
> fprintf(stderr, "Couldn't allocate memory\n");
> return 1;
> }
>
> /* now we know that string actually points somewhere */
> printf("String is %s\n", string);
>
>
> free(string);
> return 0;
> }
>
>
> Of course thatisn't really the correct way to write it either, that
> would depend
> on what you are actually trying to achieve which I can't grok from the
> code.
>
> Benno
>
thanks Benno
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