Hello to all.
Is this program licking memory?
Code:int main()
{
while(i<100)
{
...
printf("Example %s\n",function());
...
i++;
}
return 0;
}
char* function ()
{
char* data;
...
data = "memory";
....
return data;
}
Thanks!
Printable View
Hello to all.
Is this program licking memory?
Code:int main()
{
while(i<100)
{
...
printf("Example %s\n",function());
...
i++;
}
return 0;
}
char* function ()
{
char* data;
...
data = "memory";
....
return data;
}
Thanks!
No.
If it were to actually compile and run, it would only be using the program stack.
isn't 100,000 out of bounds for an integer? thought it was 16,000, unless i'm the dinosaur;->
For a 2 byte integer, yes. For a 4 byte integer, no.Quote:
Originally Posted by rogster001
The code is flawed.
1) It never calls the function, instead it prints the actual function address which is not a string. You may very well get junk or a crash.
2) The function "function" returns a pointer to local data. I'm not sure if it might just return a proper string, but I wouldn't bet on it.
3) It should be const char*, not char*.
4) i is never defined.
You forgot one.Quote:
Originally Posted by Elysia
RAM is a terrible thing to taste. :DQuote:
Originally Posted by guillermoh
Do I need to malloc()?Quote:
2) The function "function" returns a pointer to local data. I'm not sure if it might just return a proper string, but I wouldn't bet on it.
you need to call function like
function()
Sorry. typing mistakeQuote:
you need to call function like
function()
Most likely, yes. Then you need to free.Quote:
Originally Posted by guillermoh
Why if he needs only to return some const string?Quote:
Originally Posted by Elysia
Does this solve my problems?
Code:int main()
{
gchar *string;
while(i<100)
{
...
string = function();
printf("Example %s\n",string);
free (string);
...
i++;
}
return 0;
}
char* function ()
{
char* data = "";
...
data = concatenate(data,"memory");
....
return data;
}
char * concatenate (char* str1, char* str2)
{
char *result;
if ( str2 != NULL )
{
result = malloc(strlen(str1) + strlen(str2) + 1);
if ( result != NULL )
{
strcpy(result, str1);
strcat(result, str2);
}
}
return result;
}
Depends on what is your problemQuote:
Originally Posted by guillermoh
Currently this code is more complicated without a gain
I'm trying to solve this:
Quote:
1) It never calls the function, instead it prints the actual function address which is not a string. You may very well get junk or a crash.
2) The function "function" returns a pointer to local data. I'm not sure if it might just return a proper string, but I wouldn't bet on it.
For either short or int, -32767 to +32767 is guaranteed.Quote:
Originally Posted by rogster001
For signed, yes. For unsigned, that would be 0 - 65535.
to solve the 1st - just add () to function to indicate the call of the functionQuote:
Originally Posted by guillermoh
About the second - when you return something like
return "some const string";
you returning pointer to const string that will not be destructed, so its OK...
If you are creating the string inside your function - its another story
I understand. So this would be fine
Thanks for you help!Code:
const char *text[] = {"memory","information"};
int main()
{
while(i<100)
{
...
printf("Example %s\n",function());
...
i++;
}
return 0;
}
char* function ()
{
char* data;
...
data = text[0];
....
return data;
}
Yes, it would work fine, but not the best solution. To return constant data, consider making it static to your function instead.
That way you avoid global variables.
You should declare function asQuote:
Originally Posted by guillermoh
const char* function(void);
in this case
Thanks.
What about
static const char* string[] = .... ?
its okQuote:
Originally Posted by guillermoh
Code:const char* sHelloWorld(void)
{
static const char* hello = "Hello, World!";
return hello;
}
> static const char* string[] = .... ?
Did you really want an array of read-only strings?
Probably, because that's what they would be even without const and static - string literals are non-writable in most modern OS's.Quote:
Originally Posted by citizen
--
Mats
but with const you will get warning from compiler when trying to assign pointer to this string to the regular char* pointerQuote:
Originally Posted by matsp
Well aware of that, except that most of the examples up to now have been simple variables and not arrays. The OP might have been confused by vart's example. He demonstrated an entirely different type and its behavior.Quote:
Originally Posted by matsp