Pass by reference crashed

[audinue]

I'm trying to implement setLength procedure in C...

But why it crashed when I call strlen??

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

void setLength(char *string, int length)
{
   string = (char *)malloc((length+ 1) * sizeof(char));

   int i = 0;

   for(; i<length; i++)
      string[i] = ' ';

   string[length] = '\0';
   
   //Doesn't crash!
   //printf("[%d]\n", strlen(str));
}

int main()
{
   char *str;

   setLength(str, 10);

   //Crash!
   printf("[%d]\n", strlen(str));

   printf("[%s]", str);

   free(str);

   return 0;
}

Something wrong?

[MacGyver]

When you pass a variable to a function, you're really passing a copy. This means when you want to change a variable by passing it to another function, you must actually pass its address.

So because you want to change the content of str, you must pass the address of str. And yes, that is a char **.

[vlrk]

The following is the working code

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

void setLength(char *string, int length)
{
   string = (char *)malloc((length+ 1) * sizeof(char));

   int i = 0;

   for(; i<length; i++)
      string[i] = ' ';

   string[length] = '\0';
   
   //Doesn't crash!
   printf("in side function [%d]\n", strlen(string));
}

int main()
{
   char *str;

   setLength(str, 10);

   //Crash!
   printf("Lenght :[%d]\n", strlen(str));

   printf("[%s]", str);

//   free(str);

   return 0;
}


in side function [10]
Lenght :[0]
[]

i think problem here is you are trying to free memory which you assigned in the fucntion.

btw i donot understand one thing why do want to put spaces inside the string in function.

[MacGyver]

That is "working code"? By what definition? It's certainly wrong.

[audinue]

To: MacGyver

So because you want to change the content of str, you must pass the address of str. And yes, that is a char **.

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

//Neither this function:

void setLength(char **string, int length)
{
   string = (char **)malloc((length+ 1) * sizeof(char));

   int i = 0;

   for(; i<length; i++)
      string[i] = (char *)' ';

   string[length] = (char *)'\0';
}

//Nor this function:

void setLength2(char **string, int length)
{
   *string = (char *)malloc((length+ 1) * sizeof(char));

   int i = 0;

   for(; i<length; i++)
      *string[i] = ' ';

   *string[length] = '\0';
}

//Still doesn't work T_T

int main()
{
   char *str;

   setLength((char **)&str, 10);

   //Crashed >_<!
   printf("Length :[%d]\n", strlen(str));

   printf("[%s]", str);

   //Doesn't matter with this...
   free(str);

   return 0;
}

To: vlrk

btw i donot understand one thing why do want to put spaces inside the string in function.

have you tried SetLength procedure in Delphi?

[MacGyver]

SetLength2() is correct. Try this syntax in your for loop:

(*string)[i]

And of course the same change for adding the '\0'.

[vlrk]

To: MacGyver

i was just seeing on why the crash is happening ..!

I have not checked the program in terms intended functionality ..


I donot have delphi experience..

[audinue]

To: Everybody

Finally, the correct code is:

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

void setLength(char **string, int length)
{
   *string = (char *)malloc((length+ 1) * sizeof(char));

   int i = 0;

   for(; i<length; i++)
      (*string)[i] = ' ';

   (*string)[length] = '\0';
}

int main()
{
   char *str;

   setLength(&str, 10);

   printf("Length: &#37;d\n", strlen(str));
   printf("[%s]", str);

   free(str);

   return 0;
}

The problem about passing arguments by value and by reference from pointer solved...

Thanks a lot for vlrk and MacGyver for the cooperation in this thread.

You are good guys! Thanks!

[MacGyver]

You have one error worth pointing out:

free(&str);

Lose the & in this case, as it may cause program crashes in certain (perhaps most?) cases.

[audinue]

void setLength(char **string, int length)
{
   free(*string);

   *string = (char *)malloc((length + 1) * sizeof(char));

   int i = 0;

   for(; i<length; i++)
      (*string)[i] = ' ';

   (*string)[length] = '\0';
}

free(*string);

By the way, is it safe to do this?
Or is there a way to dectect whether a variable is already allocated or not?

[QuantumPete]

By the way, is it safe to do this?
Or is there a way to dectect whether a variable is already allocated or not?

There is no way to detect whether a pointer already has memory allocated. The usual convention is to set a pointer to NULL (in C) or to 0 (in C++) to indicate that the pointer does not reference any allocated memory.

QuantumPete
Edit:
So you want something like this:

if (*string != NULL) {
   free (*string);
   *string = NULL;
}

Edit Edit:
You should also test that the return value from malloc is not NULL (which indicates a memory problem) and don't cast the return type of malloc.

[Elysia]

The usual convention is to set a pointer to NULL (in C) or to 0 (in C++) to indicate that the pointer does not reference any allocated memory.

I would have to disagree. It's a usual convention to set a pointer to NULL in both C and C++ to indicate that it does not reference to any allocated memory. Whether or not NULL is defined to 0 in C++ is irrelevant. Pointers are set to NULL and it's a good think to think like that since the next standard will introduce a keyword which will act as a null pointer.

[audinue]

You should also test that the return value from malloc is not NULL (which indicates a memory problem) and don't cast the return type of malloc.

I don't get it.
Any example, please?

[Elysia]

*string = (char *)malloc((length + 1) * sizeof(char));
if (*string == NULL)
{
    /* Do something */
}

Remove part in red.
Add part in green.
If malloc fails, it returns NULL. And if it returns NULL and you try to use that memory, you will mostly likely crash your program.

[audinue]

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

//Simple macro :P

#define $free(var) free(var); var = NULL;

void setLength(char **string, int length)
{
   if(*string != NULL)
   {
      $free(*string);
   }

   *string = malloc((length + 1) * sizeof(char));

   if(*string == NULL)
   {
      puts("ERROR: Could not allocated memory space.");
      exit(EXIT_FAILURE);
   }

   int i = 0;

   for(; i<length; i++)
      (*string)[i] = ' ';

   (*string)[length] = '\0';
}

int main()
{
   char *str = NULL;

   int i = 0;

   for(; i<=16; i++)
   {
      setLength(&str, i);

      printf("Length: %d\n", strlen(str));
      printf("[%s]\n", str);

      $free(str);
   }

   return 0;
}