Why doesnt this malloc work please?

[rkooij]

Hi,

Another malloc problem (the day I understand malloc you will all get a round of beer on me).

Below is part of the code of a function for converting float values into floating point:

char* floating_point(float flt)
{
   char *formula;
   
   formula = malloc(21);

   return formula;
}

The program exits without warning when it reaches the malloc line in the code above.

To be complete, here is how I call the function:

                    fa_string = floating_point(formula_a);
                    fprintf(fpout,"%s\n", fa_string);
                    free(fa_string);

Pulling my hair out since 9am this morning (its 16:45h now). What is wrong here?

Thanks for any help.


René

[King Mir]

That looks like it will chrash in the next line: fprintf(fpout,"%s\n", fa_string);

malloc(), does allocate 21 bytes for you, but those bytes contain random values. If you try to read them as a null terminating string, you may have to read past the end of the allocated space. This isn't supposed to work, and as a result your program crashes. Not sure why windows doesn't give an error message, but it happends. If you change your other code, then it might show the error message.

There are many solutions, especially for integer types (including char, w_char, int, long, short).

[rkooij]

No it really crashes on that specific line, I've printed characters and getch(); after every code line and the malloc line is where it exited.

[King Mir]

I don't see anything wrong there. The only thing that could be done is

if(!(formula = malloc(21)))
{
     ;//free up resources 
     if(!(formula = malloc(21)))
          exit(-1);
}

[SKeane]

Try

char* floating_point(float flt)
{
   char *formula;
   
   formula = malloc(21);

   if ( formula == NULL) printf("malloc() failed\n");
   else strcpy(formula, "foobar");

   return formula;
}

See if it still fails in the same place. Go on, humour us.

[rkooij]

Tried both of your suggestions, but in both cases it fails like it has done before...

I will soon be left hair less.

[rkooij]

It does work when I run it in command prompt.... but only the first time... when I call the same function for the second time it exits like it did above....

Any ideas?

[whiteflags]

> Below is part of the code of a function for converting float values into floating point:
I simply do not understand what is going on because a floating point number is a floating point number. Are you trying to convert a number to string representation? Use snprintf with an empty string then and keep it simple:

size_t len = 21;
char *str = malloc(len);
if ( str )
{
   int ret = snprintf(str, len, "%f", 3.14159f );
   if ( ret < len )
   { 
      puts(str);
      free(str);
   }
   /* else you need to resize the buffer */
}

Something like that.

[dwks]

You could use FLT_DIGITS (at least I think that's what it's called) to figure out how big your buffer needs to be. It's in <limits.h>.

[Dave_Sinkula]

Below is part of the code of a function for converting float values into floating point:

char* floating_point(float flt)
{
   char *formula;
   
   formula = malloc(21);

   return formula;
}

The program exits without warning when it reaches the malloc line in the code above.

To me, this sounds like you had trashed your dynamic memory elsewhere in the code you didn't post, and you only discover this when you try a subsequent attempt at further dynamic allocation.

[rkooij]

I simply do not understand what is going on because a floating point number is a floating point number. Are you trying to convert a number to string representation? Use snprintf with an empty string then and keep it simple

To me, this sounds like you had trashed your dynamic memory elsewhere in the code you didn't post, and you only discover this when you try a subsequent attempt at further dynamic allocation.

Here's the complete code of the function to make clear what I want it to do:

/******************************************************************************/
char* floating_point(float flt)
{
   char hex[9], hex_2[9], temp[3], *p;
   char *formula;
   int decimal[4], i=0, ii=0;
   short cnt=0, cnt2=0;

   union
   {
      unsigned long i;
      float f;
   }v;

if(!(formula = malloc(21)))
{
     free(formula);//free up resources 
     if(!(formula = malloc(21)))
          exit(-1);
}

   v.f = flt;
   sprintf(hex_2,"%08lx",v.i);
   puts(hex_2);

   cnt=7;
   while(cnt2<=3)
   {

      sprintf(temp,"%c%c",hex_2[cnt-1],hex_2[cnt]);
      printf("%s\t",temp);
      decimal[cnt2] = strtol(temp,&p,16);
      cnt-=2;
      cnt2++;
   }
   printf("\n");
   sprintf(formula,"%d\t%d\t%d\t%d",decimal[0],decimal[1],decimal[2],decimal[3]);
   puts(formula);

   return formula;
}
/******************************************************************************/

I hope this helps, as the problem still occurs...

[rkooij]

Edit: for heaven's sake, It seems to be working now... but I didnt change anything so I think it's a bit dodgy...
Maybe it has to do with a memory leak, but I don't see where it's leaking?

[quzah]

if(!(formula = malloc(21)))
{
     free(formula);//free up resources 
     if(!(formula = malloc(21)))
          exit(-1);
}

If formula is NULL, because malloc just failed, why are you calling free on it?


Quzah.

[rkooij]

if(!(formula = malloc(21)))
{
     free(formula);//free up resources 
     if(!(formula = malloc(21)))
          exit(-1);
}

If formula is NULL, because malloc just failed, why are you calling free on it?


Quzah.

Hmm don't know What else is meant with "free up resources" posted by King Mir somewhere above?

[SKeane]

You can't "free" a resource if you don't have it in the first place.