What is wrong with this C code?

daYz · 10-05-2007, 11:35 AM

Can someone explain to me why the following C code is not working? Thanks

Code:

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


int main ()
{


char string[] = "string";

char *var;

int i;



for(i=0; i<=6 ;i++){


int six = 6;

var = calloc(six, sizeof(char));

*(var + i) = string[i];


}


printf("%c", *(var + 0));
printf("%c", *(var + 1));
printf("%c", *(var + 2));
printf("%c", *(var + 3));
printf("%c", *(var + 4));
printf("%c", *(var + 5));


return 0;

}

kaz2100 · 10-05-2007, 11:43 AM

Hya,

I am not sure what you want to program, however, if I guess you want to print out string, following is the story:

You allocate memory to *var six times, so that only the last memory space allocated is printed out.

Happy Penguins!

swodniw · 10-05-2007, 11:45 AM

It is working, just the way you want it to work is not correct

Think about it, if you allocate and assign in the first loop to index 0 then continue to the next loop and allocate and assign to index 1, what is the value at index 0?

daYz · 10-05-2007, 11:56 AM

Thanks guys

I get it now.

kaz2100 · 10-05-2007, 12:02 PM

Hya,

index 0 is "zero" because memory is allocated by calloc.

Happy Penguins!

PTrenholme · 10-05-2007, 12:10 PM

You might also want to consider the termination test in your first for loop: You seem to be assigning seven values, but only using six of them.

daYz · 10-05-2007, 12:28 PM

Quote:

Hya,

index 0 is "zero" because memory is allocated by calloc.

Happy Penguins!

Thanks

Quote:

You might also want to consider the termination test in your first for loop: You seem to be assigning seven values, but only using six of them.

I understand what you mean, but can you show me how you would do that?

rsashok · 10-05-2007, 01:02 PM

Is it what you want?

Code:

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

int main ()
{
char *string = "string";
char *var, size;

size = strlen(string);

if (var = calloc(size, sizeof(char)))
{
   memcpy(var, string, size);

   while(size--)
   {
       printf("%c", *var++); 
   }
}
else
{
   printf("Cannot allocate memory\n");
}

return 0;
}

daYz · 10-05-2007, 01:29 PM

I don't understand this part:

Code:

while(size--)
   {
       printf("%c", *var++); 
   }

Would you mind telling me what it does exactly?

Thanks

Fidori · 10-05-2007, 04:24 PM

The expression "variable++" increments the variable by one and returns the value that the variable had before incrementing. "variable--" works similarly but decrements the variable.

rsashok · 10-05-2007, 04:39 PM

Quote:

while(size--)
{
printf("%c", *var++);
}

var - is the pointer (e.g. address of a memory location)
*var - is content of a memory pointed by "var"
*var++ - read memory location at address "var" and increment value of "var" by 1.

Let say, the initial value of "var" is 100, then "*var++" will return value stored at location 100, and increments "var" by 1, so next time location at address 101 will be read.

There are few things to point out:

Quote:

char *string = "string"

will have terminating zero at the end, because it is declared as a string, but after your program execution "var" have only six bytes of "string" without terminating zero. In other words there is a difference between these declarations:

Quote:

char string[] = "string"; /* have zero at the end */
and
char string[] = {'s', 't', 'r', 'i', 'n', 'g'}; /* no zero at the end*/

PTrenholme · 10-05-2007, 08:28 PM

Quote:

<snip>I understand what you mean, but can you show me how you would do that?</snip>

Sheesh! If you "understand" what I meant, why ask?

What I hope you understood, is that the "=" is not needed in this statement:

for(i=0; i<=6 ;i++){

<edit>
I don't think that the order of operation makes any difference in the increment of i, but, as rsasock pointed out, there is a difference between ++i and i++
</edit>

daYz · 10-06-2007, 11:03 AM

Thanks guys, it is all clear to me now.

Quote:

<snip>I understand what you mean, but can you show me how you would do that?</snip>
Sheesh! If you "understand" what I meant, why ask?

I wasn't aware you meant the

Code:

i<=6

part when your where talking about the "termination test". I am trying to learn C, and you can't expect me to understand the language all at once.

Regards,

Ben

kennithwang · 10-11-2007, 01:09 AM

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

int main ()
{

char string[] = "string";

char *var;

//int i;
int j=0;

{

int six = 6;

var = (char *)calloc(six+1, sizeof(char));
for (j = 0; j < 6; j++)
*(var + j) = string[j];
*(var + 6) = '\0';

}

printf("%c\n", *(var + 0));
printf("%c\n", *(var + 1));
printf("%c\n", *(var + 2));
printf("%c\n", *(var + 3));
printf("%c\n", *(var + 4));
printf("%c\n", *(var + 5));

return 0;

}

daYz · 10-11-2007, 07:15 AM

xie xie kennithwang

Tai Che