memory dellocation

[c.arnet]

Hi there,

I've written a function that 'converts' a structure of type timeval to
a c-string:

char *tv2Str(struct timeval tvX)
{
char caStr[20];
char *cpResult;

sprintf(caStr, "%ld.%ld", tvX.tv_sec, tvX.tv_usec);

cpResult = (char *)malloc(sizeof(char)*strlen(caStr));
strcpy(cpResult, caStr);

return cpResult;
}

I use it like this:

printf("%s", tv2Str((struct timeval){123, 456}) );

Question:
How (if at all) does the compiler (gcc) ensure that the memory
allocated in the above function is deallocated after the printf call ?

Thanks,
Cornel

[Gordon Burditt]

>char *tv2Str(struct timeval tvX)

{
char caStr[20];
char *cpResult;

sprintf(caStr, "%ld.%ld", tvX.tv_sec, tvX.tv_usec);

cpResult = (char *)malloc(sizeof(char)*strlen(caStr));
strcpy(cpResult, caStr);

return cpResult;
}

I use it like this:

printf("%s", tv2Str((struct timeval){123, 456}) );

Question:
How (if at all) does the compiler (gcc) ensure that the memory
allocated in the above function is deallocated after the printf call ?

It does not. That's your responsibility, and you blew it.
You have a memory leak.

Gordon L. Burditt

[Artie Gold]

Gordon Burditt wrote:

char *tv2Str(struct timeval tvX)
{
char caStr[20];
char *cpResult;

sprintf(caStr, "%ld.%ld", tvX.tv_sec, tvX.tv_usec);

cpResult = (char *)malloc(sizeof(char)*strlen(caStr));
1) Lose the cast. It's not necessary -- and can possibly hide errors
(see the FAQ for further details).

2) sizeof(char) is _defined_ to be 1 -- so you don't need/want this term

3) If you're copying a C-string, always allocate strlen(orig) + 1, to
account for the trailing null.
strcpy(cpResult, caStr);

return cpResult;
}

I use it like this:

printf("%s", tv2Str((struct timeval){123, 456}) );

Question:
How (if at all) does the compiler (gcc) ensure that the memory
allocated in the above function is deallocated after the printf call ?

It does not. That's your responsibility, and you blew it.
You have a memory leak.

Yup. You do!

HTH,
--ag


--
Artie Gold -- Austin, Texas

[Eric Sosman]

Gordon Burditt wrote:

[attribution to c.*****@bluewin.ch re-inserted]:

char *tv2Str(struct timeval tvX)
{
char caStr[20];
char *cpResult;

sprintf(caStr, "%ld.%ld", tvX.tv_sec, tvX.tv_usec);

cpResult = (char *)malloc(sizeof(char)*strlen(caStr));
strcpy(cpResult, caStr);

return cpResult;
}

I use it like this:

printf("%s", tv2Str((struct timeval){123, 456}) );

Question:
How (if at all) does the compiler (gcc) ensure that the memory
allocated in the above function is deallocated after the printf call ?

It does not. That's your responsibility, and you blew it.
You have a memory leak.

Worse: he has undefined behavior because the memory
allocated to `cpResult' is one byte too small to hold the
data copied into it by strcpy(). And that's even ignoring
the possibility of malloc() failure, for which he's written
no test.

--
Er*********@sun.com

[Burne C]

<c.*****@bluewin.ch> wrote in message news:16**************************@posting.google.c om...

Hi there,

I've written a function that 'converts' a structure of type timeval to
a c-string:

char *tv2Str(struct timeval tvX)
{
char caStr[20];
char *cpResult;

sprintf(caStr, "%ld.%ld", tvX.tv_sec, tvX.tv_usec);

cpResult = (char *)malloc(sizeof(char)*strlen(caStr));
strcpy(cpResult, caStr);

return cpResult;
}

I use it like this:

printf("%s", tv2Str((struct timeval){123, 456}) );

Question:
How (if at all) does the compiler (gcc) ensure that the memory
allocated in the above function is deallocated after the printf call ?

Thanks,
Cornel

No, it doesn't. It is memory leak

I think it is not common to allocate a block of memory and return it (because you can't free it).
If you want to retrun a string result, you can request a char* in the function argument and fill it.

char *tv2Str(struct timeval tvX, char *strResult)
{
sprintf(strResult, "%ld.%ld", tvX.tv_sec, tvX.tv_usec);

return strResult;
}

You have to provide the buffer to the call

char somebuffer[50];
printf("%s", tv2Str((struct timeval){123, 456}, somebuffer) );
--
BC