damn strings

hi all, i hope u guys can help me, i had a little experience in C but just in math stuff, now i need to handle string but i cant figure it out, i check the forum but i didnt find what im looking for, i now how to create "strings" in C, the problem is when i want to concatenate strings or handle more than one, i want to receive 2 strings and divide them in parts, i receive dates (i know i can make it with integers but my teacher wanted to be with strings, and there are better ways to do this, but my point is why my strings act like that) and then calculate the days between this to dates, when i divide the second string my dummy string fill with trash or values from the other 2 strings, here is the part of my file when i divide the 2 strings

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<alloc.h>
void main(void)
{

int dia1,dia2,mes1,mes2,ano1,ano2,dum,i,totdias;
int mes[12];
char *fecha1=malloc(sizeof(char)*9);
char *fecha2=malloc(sizeof(char)*9);
char *dummy=malloc(sizeof(char)*9);
mes[0]=0,mes[1]=31,mes[2]=28,mes[3]=31,mes[4]=30,mes[5]=31,mes[6]=30,
mes[7]=31,mes[8]=31,mes[9]=30,mes[10]=31,mes[11]=30,mes[12]=31;
totdias=0;
clrscr();
gotoxy(5,7);
printf("Format yyyyymmdd (yearmonthday)");
gotoxy(5,11);
printf("First date: ");
scanf("%s",fecha1);

strncpy(dummy,fecha1,4);
ano1=atoi(dummy);
strncpy(dummy,fecha1,6);
dum=atoi(dummy);
mes1=dum-(ano1*100);
strncpy(dummy,fecha1,8);
dum=atoi(dummy);
dia1=dum-((ano1*10000)+(mes1*100));
gotoxy(5,13);
printf("Second date: ");
scanf("%s",fecha2);
free(dummy); //liberate dummy
strncpy(dummy,fecha2,4);
ano2=atoi(dummy);
strncat(dummy,fecha2,6);
dum=atoi(dummy);
mes2=dum-(ano2*100);
strncat(dummy,fecha2,8);
dum=atoi(dummy);
dia2=dum-((ano2*10000)+(mes2*100));
}

i hope u can help me out, im programming on turbo c++ 3 on windows xp, thanks a lot.

Juan

P.S. Sorry for my bad english

Some tips:

1. mes[12]=31;

This is not going to work because your array is 12 long (0-11)

2. void main(void)

This is wrong, use int main(void);

3. int mes[12];

You can also fill the array when declaring it:
int mes[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

4. char *fecha1=malloc(sizeof(char)*9);

If you allocate something it's always nice to free the allocated memmory (same for fecha2).
Why do you use dynamic memory anyway??:

Code:

---

char fecha1[9];
char fecha2[9];
char dummy[9];

---

> i hope u can help me out, im programming on turbo c++ 3 on windows xp
This mix of the very old and very new cannot be good in the long term

Perhaps something more with the times - at least something which is ANSI-C
http://www.thefreecountry.com/developercity/index.html

In addition to what Monster said

> strncpy(dummy,fecha1,4);
> ano1=atoi(dummy);
strncpy doesn't add a \0 to end the string - so atoi could be in for a long run through lots of memory you don't own

strncpy(dummy,fecha1,4);
dummy[4] = '\0';
ano1=atoi(dummy);

Quote:

---

Originally posted by Salem
strncpy doesn't add a \0 to end the string - so atoi could be in for a long run through lots of memory you don't own

---

Small correction: the strncpy function doesn't add a \0 to the end if the size argument (3rd argument) is less than or equal to the length of the source buffer.

But Salem it right, in your case it doesn't add a null character so you have to add it yourself.

And something else:

Code:

---

strncpy(dummy,fecha1,4);
ano1=atoi(dummy);
strncpy(dummy,fecha1,6);
dum=atoi(dummy);
mes1=dum-(ano1*100);
strncpy(dummy,fecha1,8);
dum=atoi(dummy);
dia1=dum-((ano1*10000)+(mes1*100));

---

You can change this by:

Code:

---

if(strlen(dummy) == 8) /* or >= 8 */
{
  strncpy(dummy,fecha1,4);
  dummy[4] = '\0';
  ano1=atoi(dummy);
  strncpy(dummy+4,fecha1,2);
  dummy[2] = '\0';
  mes1=atoi(dummy);
  strncpy(dummy+6,fecha1,2);
  dummy[2] = '\0';
  dia1=atoi(dummy);
}

---

B.t.w. what happens if I enter more than 9 characters when I enter the date ???

My suggestion would be to modify the strncpy function:

Code:

---



char *Strncpy( char *target, char *source, int max)
{
char *t;
char *s;

 for(s = source, t = target; (s != NULL) && (*s != 0) && (m < max - 1);  s++, t++, m++)
 {
  *t = *s;
 }

*t = 0;

 t = target;

 return t;
}

---

Sebastiani, what's the m variable about, you've forgotten to declare it. ;)

Also, there's a few efficiencies you could make. Here's an alternative version. It doesn't check for source and target being NULL, that's upto the caller. max is the number of characters that will be placed in target (not including the null terminator).

Code:

---

char *Strncpy(char *target, char *source, int max)
{
        char        *t = target;

        while (*source && max)
        {
                *target = *source;
                source++; target++; max--;
        }

        *target = '\0';
       
        return(t);
}

---

I haven't read the whole thread in detail, so maybe I'm off base here!

Not to sound nit-picky, but :

You attached "t" to "target" but then increment "target". You also try to increment "source". You can't do that. Run the function and see for yourself. The compiler treats the char* parameters as if they were the names of arrays, an the copy will fail.

When max = 0, the chance to guarantee that null term is lost...so your loop should be
while(max > 0)

Finally, your point that it is the callers responsibility to check for NULL I have one argument: if perchance "source" is NOT null termed, then checking for a non-NULL pointer is a last-ditch attempt to end the copy loop, which may otherwise run infinitely...

Quote:

---

Small correction: the strncpy function doesn't add a \0 to the end if the size argument (3rd argument) is less than or equal to the length of the source buffer.

---

What? Your 'correction' makes no sense. So if argument 3 (the size) is greater than the source buffer, you say it terminates it? Really? Where at? What point does it decide to null terminate it? What's your source of this information?

Actually I'll correct you both, unless my source is wrong:

According to this...

Quote:

---

The strncpy() function is similar, except that not more
than n bytes of src are copied. Thus, if there is no null
byte among the first n bytes of src, the result wil not be
null-terminated.

In the case where the length of src is less than that of
n, the remainder of dest will be padded with nulls.

---

See? It will copy the null. Provided that N is long enough to count it. It is possible that my source is implementation specific.

Quzah.

thanks all for your input, now is working, i need to work on my logic but i dont hate strings anymore.




Juan Zamudio

Originally posted by Sebastiani
>You attached "t" to "target" but then increment "target".
Yes, but it doesn't matter which you do. You could just as easily put these the other way round, like you imply. You need to keep a pointer to the original target so you can return it at the end of the function.

>You also try to increment "source". You can't do that. Run the function and see for yourself. The compiler treats the char* parameters as if they were the names of arrays, an the copy will fail.
The parameters are treated exactly as defined, pointers to chars. This means pointer arithmatic works fine. Here's an example of strcpy() from K&R2 to prove it.

Code:

---

void strcpy(char *s, char *t)
{
    while (*s++ = *t++)
        ;
}

---

>When max = 0, the chance to guarantee that null term is lost...so your loop should be
>while(max > 0)
max>0 is tested (well, kind of) in my loop. I can't be bothered to research this too far, what I wrote compiled/ran as expected (or so I found).

>Finally, your point that it is the callers responsibility to check for NULL I have one argument: if perchance "source" is NOT null termed, then checking for a non-NULL pointer is a last-ditch attempt to end the copy loop, which may otherwise run infinitely...
These are two different things.. am I understanding you here? The pointer being NULL will result in my function crashing. This is why the caller should ensure it's a good pointer. The null term of the string should always be there, else we'll probably go onto memory that's not ours. I think we're saying the same thing ;)

>Not to sound nit-picky, but...
Please do, it's better that way :)

Quote:

---

Originally posted by quzah


What? Your 'correction' makes no sense. So if argument 3 (the size) is greater than the source buffer, you say it terminates it? Really? Where at? What point does it decide to null terminate it? What's your source of this information?

---

My correction does make sense. If the size argument is greater than the source buffer, the destination buffer will be padded with null characters (MSDN Library).

Code:

---

int main()
{
  char dest[10];
  strncpy(dest, "Hello", 6); /* string will be null terminated */
  return 0;
}

---