Thread: memory allocation

I was wondering if anyone can give me an answer to this:
Here is the code

int main(int argv, char *argc[])
{
char **name;
char *tempstring = "this string"

name = (char**)malloc(sizeof(char**) * NoOfInFiles))

for(i=0;i<NoOfFile;i++)
name[i] = strdup(tempstring);

// Do other stuff

free(name);
return(0);
}

ok now my question is this:
If the memory for name is freed, will it free the memory that has been allocted by strdup() or will that memory still be set aside?
The way I think it works is that I would have to do this first :

for(i=0;i<NoOfFiles;i++)
free(name[i]);

before freeing name? is this right or am I just messed up?
And the pointer to pointer has to stay. It is too deep in the original code to remove it.
any help would be appreciated

This is a memory leak.

strdup() uses malloc to allocate memory for the duplicate string. When you call free( name );, you are deallocating the memory for the pointers that name points to, but you are not deallocating the memory strdup() set aside for the individual strings.

What's wrong with this picture?

Code:

name = (char**)malloc(sizeof(char**) * NoOfInFiles)) 

for(i=0;i<NoOfFile;i++)

It's the empty line in between malloc and the start of your loop. Nowhere do you test to see if malloc failed, always test malloc, always.

Code:

name = (char**)malloc(sizeof(char**) * NoOfInFiles)) 
if ( name != NULL ) {
  for(i=0;i<NoOfFile;i++) 
  /* The rest of your code */
}
else {
  /* Handle the error */
}

There are also two problems with this line, not true errors but they have the potential for annoyance.

name = (char**)malloc(sizeof(char**) * NoOfInFiles))

In C, you aren't required to cast malloc, and it's usually best not to because if you cast the result of malloc then that could hide an error, such as forgetting to include stdlib.h. You also make malloc type specific this way when you could write it much more maintenance friendly.

name = malloc( NoOfInFiles * sizeof **name )

This saves you the trouble of having to change several instances of char ** because malloc is no longer bound to the type of the variable that it is allocating memory for. If you change name to int then you don't have to change the above line at all.

-Prelude

Thanks for the help. I know I didn't check malloc. I didnt intend to... this is crap code..just example and I just omitted checking for convience. As for the cast, that is the way a teacher showed our class how to do it. Actually come to think of it, every teacher I had for C told us to cast malloc. I do however see your point

>every teacher I had for C told us to cast malloc
I hate to say it, but your teachers were all wrong. In older C malloc was required to be cast because of how the generic pointer ( char * or void * depending on how far back you go ) behaved of which I'll not bore you, but under the ANSI/ISO C89 standard casting is optional and considered a bad practice for the reasons I stated earlier.

-Prelude

you wont bore me with history... I find the history of C quite interesting. It wouldn't surprise me to know my teachers were wrong. Seems like they will hire anyone to teach. My advanced C teacher didn't know squat about manipulating bit or how to access hardware but she could make a linked list ..... lol

>but she could make a linked list
Sad, but often true. Teachers tend to be one step ahead of the students if you're lucky. Very few instructors are knowledgeable, if they were then they would have high paying jobs and wouldn't need to teach.

>I find the history of C quite interesting
void * only made it into C with the ANSI standard, before that the char * played the role of generic pointer that void * now enjoys and because of a more strict pointer to pointer conversion rule, you had to cast from a generic pointer to anything else before using it, even in assignment. The ANSI standard changed the rule a bit for assignment in that pointers could be assigned to and from void * and compared with them without the use of an explicit cast. Though other pointer to pointer conversions still require you to cast them.

All very interesting if you have the time to figure it out, K&R offers a good explanation of pointer casting and rules in section A6.6 of the reference.

-Prelude

Thanks for that bit of info Prelude. I read about the change from char* to void* but I didnt know about the cast being dropped. I will have to go read me bible again.
As for the code here is something interesting to chew on.
I dropped the cast like you said to do and i get this error:

error C2440: '=' : cannot convert from 'void *' to 'char ** '
Conversion from 'void*' to pointer to non-'void' requires an explicit cast
Error executing cl.exe.

this is the code in question
if(!(name = malloc(sizeof(char**) * NoOfInFiles)))
{
printf("memory error\n");
exit(1);
}

that is when I reallized that I was using a .cpp file extension so I changed it to .C and there we go, it compiles without a complaint. Seems to me that c++ changed this to require an explicit cast. I know this is a C board but though you might find it interesting if you code in both languages(even though I should be using the new operator in C++ anyways).

I also tested this
for(i=0;i<NoOfInFiles;i++)
{
printf("Freeing %s\n",name[i]);
free(name[i]);
}
and it worked fine, so I got that solved at least.

>that is when I reallized that I was using a .cpp file extension
That's one of the traps that C programmers fall into with C++, C doesn't require explicit casts yet C++ does. That's also one of the reasons C++ programmers who migrate to C use explicit casts.

Sometimes it's hard to think of C and C++ as two completely different languages. The syntax is similar but there are some subtle differences since C++ and C both evolved after C++ was created as a superset of C.

-Prelude

Originally posted by Sayeh
It is called 'type coercion'.

Type conversion (K&R Section 2.7)

I suspect the reason C programmers use the term casting when they mean type conversion is also due to K&R, which seems to use the phrases interchangeably. In fact, the index of K&R (second edition) has this to say:

• type conversion operator, see cast

[edit]
It should be noted that the terms 'type conversion' and 'cast' are not interchangeable. A type conversion is done automatically, whereas a cast is explicit.

Code:

int i;
char c;

c = i; /* type conversion */
c = (char) i; /* cast */

I'll admit this seems a little anal, but my original post did not make this clear :)
[/edit]