malloc/free of return value

Bladactania · 02-10-2009, 09:45 AM

I'm writing a function that returns a C-style string (char *). THe function performs several operations on the 2 incoming strings and returns the result. While I'm performing these operations I use a temp variable...

Code:

char * sTemp = malloc((sizeof(s1) * strlen(s1)) + (sizeof(s2) * strlen(s2)));

Then at the end of the function I return sTemp. But what about deallocating the memory I allocated for sTemp? Is is automatically deallocated when the function completes? if not can I deallocate AFTER the return statement?

**laserlight** · 02-10-2009, 09:48 AM

Quote:

Originally Posted by Bladactania

Then at the end of the function I return sTemp. But what about deallocating the memory I allocated for sTemp?

The caller (or the caller's caller, etc) should be responsible for deallocating the memory.

Quote:

Originally Posted by Bladactania

Is is automatically deallocated when the function completes?

No, and that is a Good Thing, otherwise the caller would then have a pointer to deallocated memory.

Quote:

Originally Posted by Bladactania

if not can I deallocate AFTER the return statement?

No, but then as I noted, the caller can deallocate the memory, so it is kind of like deallocating after the return statement.

Bladactania · 02-10-2009, 09:50 AM

Thanks for the quick reply...

How does the caller deallocate the memory for a pointer with function level scope?

Bladactania · 02-10-2009, 09:54 AM

Another thing... If I'm returning a pointer to a string like this, do I need to allocate the memory for the pointer I'm returning the value into?

Code:

test = malloc(.....);
test = mystringfunction(s1, s2);

**laserlight** · 02-10-2009, 10:00 AM

Quote:

Originally Posted by Bladactania

How does the caller deallocate the memory for a pointer with function level scope?

It uses a pointer. The pointer in the function is destroyed when it goes out of scope, so another pointer is needed to store the address of the dynamically allocated object.

Quote:

Originally Posted by Bladactania

If I'm returning a pointer to a string like this, do I need to allocate the memory for the pointer I'm returning the value into?

Well, you probably do not need to dynamically allocate memory for such a pointer as a local variable will do.

Bladactania · 02-10-2009, 10:02 AM

Here's my function Code...

Code:

 char * String_Concat (const char *s1, const char *s2) {
   char *sTemp = malloc((sizeof(s1) * strlen(s1)) + (sizeof(s2) * strlen(s2)));

   strcpy(sTemp, s1);
   strcat(sTemp, s2);

   return sTemp;
 }

In main I call the function like so...

Code:

char   *sDiagnostic_Log_Path; // Full path of diagnostic log file

sDiagnostic_Log_Path = String_Concat(sFlash_Card_Base_Path, sDiagnostic_Log_Name);

Is that sufficient? I'm terribly scared of memory leaks!

Elysia · 02-10-2009, 10:03 AM

Code:

char* foo() { return malloc(10); }
void foo2(char** p) { *p = malloc(10); }
int main()
{
    char* p;
    p = foo();
    free(p);
    foo2(&p);
    free(p);
}

**Salem** · 02-10-2009, 11:29 AM

> char *sTemp = malloc((sizeof(s1) * strlen(s1)) + (sizeof(s2) * strlen(s2)));
Why sizeof(s1) ?
That's the size of the pointer, not a single char.

Cheaply, it's
char *sTemp = malloc( strlen(s1) + strlen(s2) + 1);

Pedantically, in line with what you're doing
char *sTemp = malloc((sizeof(s1[0]) * strlen(s1)) + (sizeof(s2[0]) * strlen(s2)) + 1);

Bladactania · 02-10-2009, 12:03 PM

Could I use sizeof(*s1)? I suppose I could use sizeof(char) since I know it will be a character.

**Salem** · 02-10-2009, 12:04 PM

If you like, it's the same thing.

Bladactania · 02-10-2009, 12:06 PM

when using sizeof(*s1) I get a "Nonportable pointer conversion" warning... Is there a way to avoid this?

Meldreth · 02-10-2009, 12:09 PM

avoid it by not using sizeof(*s1). sizeof(char) is 1 anyway and N*1 is always going to be N so it's redundant.

**Salem** · 02-10-2009, 12:14 PM

What a strange error message from applying sizeof to something.

Bladactania · 02-10-2009, 12:25 PM

I think the warning comes from the strlen(s1) not the sizeof(char).

Bladactania · 02-10-2009, 12:25 PM

I am using Turbo C version 3.0... *sigh*

02-10-2009, 09:45 AM	#1
Bladactania Registered User Join Date: Feb 2009 Posts: 278	malloc/free of return value I'm writing a function that returns a C-style string (char ). THe function performs several operations on the 2 incoming strings and returns the result. While I'm performing these operations I use a temp variable... Code: char sTemp = malloc((sizeof(s1) * strlen(s1)) + (sizeof(s2) * strlen(s2))); Then at the end of the function I return sTemp. But what about deallocating the memory I allocated for sTemp? Is is automatically deallocated when the function completes? if not can I deallocate AFTER the return statement?
Bladactania is offline

02-10-2009, 09:50 AM	#3
Bladactania Registered User Join Date: Feb 2009 Posts: 278	Thanks for the quick reply... How does the caller deallocate the memory for a pointer with function level scope?
Bladactania is offline

02-10-2009, 09:54 AM	#4
Bladactania Registered User Join Date: Feb 2009 Posts: 278	Another thing... If I'm returning a pointer to a string like this, do I need to allocate the memory for the pointer I'm returning the value into? Code: test = malloc(.....); test = mystringfunction(s1, s2);
Bladactania is offline

02-10-2009, 11:29 AM	#8
Salem and the hat of vanishing Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,214	> char sTemp = malloc((sizeof(s1) strlen(s1)) + (sizeof(s2) * strlen(s2))); Why sizeof(s1) ? That's the size of the pointer, not a single char. Cheaply, it's char sTemp = malloc( strlen(s1) + strlen(s2) + 1); Pedantically, in line with what you're doing char sTemp = malloc((sizeof(s1[0]) * strlen(s1)) + (sizeof(s2[0]) * strlen(s2)) + 1); __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Up to 8Mb PlusNet broadband from only £5.99 a month!
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02-10-2009, 12:03 PM	#9
Bladactania Registered User Join Date: Feb 2009 Posts: 278	Could I use sizeof(*s1)? I suppose I could use sizeof(char) since I know it will be a character.
Bladactania is offline

02-10-2009, 12:04 PM	#10
Salem and the hat of vanishing Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,214	If you like, it's the same thing. __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Up to 8Mb PlusNet broadband from only £5.99 a month!
Salem is offline

02-10-2009, 12:06 PM	#11
Bladactania Registered User Join Date: Feb 2009 Posts: 278	when using sizeof(*s1) I get a "Nonportable pointer conversion" warning... Is there a way to avoid this?
Bladactania is offline

02-10-2009, 12:09 PM	#12
Meldreth Registered User Join Date: Feb 2009 Posts: 138	avoid it by not using sizeof(s1). sizeof(char) is 1 anyway and N1 is always going to be N so it's redundant.
Meldreth is offline

02-10-2009, 12:14 PM	#13
Salem and the hat of vanishing Join Date: Aug 2001 Location: The edge of the known universe Posts: 21,214	What a strange error message from applying sizeof to something. __________________ If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut. Up to 8Mb PlusNet broadband from only £5.99 a month!
Salem is offline

02-10-2009, 12:25 PM	#14
Bladactania Registered User Join Date: Feb 2009 Posts: 278	I think the warning comes from the strlen(s1) not the sizeof(char).
Bladactania is offline

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