Thread: Function returns string

So do you only want one string? Or do you need an array of strings?

Your function has a memory leak. Count the malloc()'s and then count the free()'s.

>Pretty much . . . but what happens if the first call to malloc fails (formula==NULL)? Boom . . . segfault.

Code:

else{
   return 0;
}

?


>You're aware that declarations in the middle of a block is C++/C99 and not C89?
I wasn't. To be honest I thought the formula = malloc(20 * sizeof *formula); line belonged to the declarations and that the whole malloc sequence (starting with if ( formula != NULL )) belonged to that line... which is - looking at it now - stupid.


>Try running gdb from the command line. [edit] Like this: http://cboard.cprogramming.com/showp...79&postcount=7 [/edit]
Afaik I don't have a command line function in DevC++ (I'm aware that this might be a very dumb answer). I'm looking into it now though, something with MinGW I think...

Thanks so much for your help.

>So do you only want one string? Or do you need an array of strings?
Yes I only want one string.

>Your function has a memory leak. Count the malloc()'s and then count the free()'s
Oh, then I'm guessing free(string); in the main has no use?

Click on Start->All Programs->Accessories->Command Prompt (for windows xp). That will give you a command prompt.

Once you have a command prompt open, you can use cd to change directories:

Code:

C:\Documents and Settings\User>cd ..

C:\Documents and Settings>cd "User\My Documents"

C:\Documents and Settings\User\My Documents>cd programs

C:\Documents and Settings\User\My Documents\programs>

Once you're in the same directory as your program's executable, you can debug it with gdb.

Code:

C:\Documents and Settings\User\My Documents\programs>gdb executable.exe
Welcome to GDB ...
(gdb) run
program running . . .
Segmentation fault
(gdb) backtrace
the location of the segfault...
(gdb) list
code listing...
(gdb) quit
The program is running. Exit anyway? y

C:\Documents and Settings\User\My Documents\programs>

Originally Posted by rkooij

>So do you only want one string? Or do you need an array of strings?
Yes I only want one string.

Well, if you only want one string, you only need one malloc statement. Something like this:

Code:

char *full_name(char *first_name, char*lastname) {
    char *name = malloc(strlen(first_name) + strlen(lastname) + 1);
    if(!name) return 0;

    sprintf(name, "%s %s", first_name, last_name);

    return name;
}

void call_name(void) {
    char *name = full_name("John", "Smith");

    puts(name);
    free(name);
}

Output:

Code:

John Smith

Your function has a memory leak. Count the malloc()'s and then count the free()'s
Oh, then I'm guessing free(string); in the main has no use?

Count your malloc statements. 1 + 20. Count your free statements. 1.

Well, if you only want one string, you only need one malloc statement. Something like this:
> Hmmmmm ok!! So that's how malloc works, hehe.... excellent example, thank you very much!

>Count your malloc statements. 1 + 20. Count your free statements. 1.
I count only 1 malloc... ok ok that's after I changed the malloc request according to your tips above

Damn how I wish there was a way to thank you guys other than just putting down the 6 letter word.

I can't express how grateful I am, hope one day I can pay you back.

So what's the latest code - do you only have one malloc and one free?

I count only 1 malloc

Apparently.

Yes, according to dwks' example, only 1 malloc;

Code:

/************************ Floating point converter ****************************/
//* Function:   converts a decimal value to a floating point
//* Parameters: flt: decimal value to convert
//* Returns:    formula: floating point (string)
//*
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;
   
   formula = malloc(21);
   if (!formula)
   {
       return 0;
   }

   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;
}
/******************************************************************************/

Code:

         case '7':
              
              string = floating_point(0.012);
              puts(string);
              free(string);
              
         break;

> char **formula;
Should be char *formula;
to match your return result

> sprintf(*formula,"%d\t%d\t%d\t%d",decimal[0],decimal[1],decimal[2],decimal[3]);
Drop the *, you want to write the string to where the pointer points
sprintf(formula,"%d\t%d\t%d\t%d",decimal[0],decimal[1],decimal[2],decimal[3]);

Ditto with the puts(formula)

> return *formula;
And here as well
return formula;

Originally Posted by Salem

> char **formula;
Should be char *formula;
to match your return result

> sprintf(*formula,"%d\t%d\t%d\t%d",decimal[0],decimal[1],decimal[2],decimal[3]);
Drop the *, you want to write the string to where the pointer points
sprintf(formula,"%d\t%d\t%d\t%d",decimal[0],decimal[1],decimal[2],decimal[3]);

Ditto with the puts(formula)

> return *formula;
And here as well
return formula;

Aren't

> char **formula;
Should be char *formula;
to match your return result

and

> return *formula;
And here as well
return formula;

contradictive?

And I thought I couldn't return a string but had to return a pointer... as it looks from return formula; I'm returning a string now? Or does the asterisk in char* floating_point(float flt) cover that?

contradictive?

Nope. In one place you are delcaring a char* in the other you are using the variable that happens to be a char*

the return type of your function is a char*

int a;
char* formula;

a is an int
formula is a char*

so to return a char* you just return formula.

As said, the same goes for sprintf and puts which expect a char*

And I thought I couldn't return a string but had to return a pointer... as it looks from return formula; I'm returning a string now? Or does the asterisk in char* floating_point(float flt) cover that?

formula is a pointer because that's how you declared it.
The declaration char* formula; is what "covers" it

Basiclly, if you have a string like this,

Code:

char *str;

When you want to use the string, use just plain str, such as for puts() calls or whatever. *str is the same as str[0], and so you should only use it when you want to access the first character in the string.

Given the char*str above, if your function also returns a char*, then you can just use this

Code:

return str;

because "str" is a char*. If you returned *str, it would be like returning a character, which isn't what the signature would indicate.

That's great explanation guys, kind of what I already had in mind. Thank you both of you!