can c let me pass string references??

I want to pass a reference to a string variable (char[]) to a funtion and have the function change the original value of that string - but I CANNOT get this to work. I am going mad - any help?

See following little prog which I think SHOULD work but doesn't!! It changes the string variable to some random char??

#include <stdio.h>

void changeString(char *str[]);

void main()
{
char string[]="NO";

printf("\nString: %s\n\n", string);
changeString(&string);
printf("Changed String: %s\n\n", string);

}

void changeString(char *str[])
{
*str="YES";
printf("Change to: %s\n\n", *str);
}

Take out the star *, a string is passed as a pointer. Either use char *str or else char str[]. I would choose the later. This is wrong char *str[]

Thanks for reply - as this is driving me nuts!

But both your suggestions cause my exe to crash when I run it - giving an application error saying something "the memory cant be read".

Any more suggestions? Would really appreciate your help asap?

That's because you probably have to change your code! You want me to write it for you!! I am not wrong.

Code:

---


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

void changeString(char str[]);

void main()
{
char string[]="NO";

printf("\nString: %s\n\n", string);
changeString(string);
printf("Changed String: %s\n\n", string);

}

void changeString(char str[])
{
        strcpy(str,"YES");
        printf("Change to: %s\n\n", str);
}

---

Congratulations, you just found the pointer equivalent to a boogeyman.

This involves talking about the difference between a pointer ad an array... if you were to change
char string[]="NO"
to
char * string = "NO"
Then the program would work... (well, kinda).

This is because the first declares string as an array, and the second declares string as a pointer. The problem happens when you get to this line...

changeString(&string);

Clearly, you're going to change the value of string with this command. Now, when you & a pointer variable, you get the memory address of that pointer, which is how & is supposed to work.
When you & an array (a pointer constant), since it's a constant, it doesn't have any memory address; it's hardcoded into your program. Typically, trying to do this is meaningless. I mean, it's like saying pop (&1); you just can't use & on a constant. With arrays however, C allows it (it shouldn't). When you use & on an array, it just acts as if you had passed the array. Now let's look into your changeString function...

Code:

---

void changeString(char *str[])
{
*str="YES";
printf("Change to: %s\n\n", *str);
}

---

Let's run through this code, with letting str have the same value as string in main. "YES" returns a pointer to a string that has been temporarily placed in memory. We store this pointer in *str. Since str is the same thing as string (if we used string as an array), the memory that's being pointed to is our string itself. So, whatever was in our string ("NO") gets overwritten by a pointer representing wherever the string "YES" is located in memory.

printf("Change to: %s\n\n", *str);
Now, *str contains a pointer, so this line has no problem following that pointer to print the string "YES"

But then we leave the function, and try this...
printf("Changed String: %s\n\n", string);
Again, *string contains a pointer, so when we try to point the chars in *string, we end up printing arbitrary characters.

Something to point out however, is that if you change string to a char *, then although your program will work, it's not gauranteed to work. When it hits the "YES" your program goes to a static buffer, puts "YES" in it, and returns a pointer to that buffer. The program may, however, change the contents of that buffer as it sees fit. The technically correct way to go about this would be to use the strcpy function.

Also, main is supposed to return an int, not void :)

> strcpy(str,"YES");
Oops - but str has only been allocated 3 bytes, and this copies 4 bytes

In main, you should have at least
char string[4]="NO";

Okay, that is true, I wasn't counting!

How about this then ?

#include <stdio.h>

void changeString(char *str);

int main()
{
char string[4]="NO";

printf("\nString: %s\n\n", string);
changeString(string);
printf("Changed String: %s\n\n", string);
return 0;
}

void changeString(char *str)
{
*str='Y';
str++;
*str='E';
str++;
*str='S';
str++;
*str='\0';
}


You know, Im just started learning C and had the same question about passing arrays and was about to post when I read this. Started playing and now I (hopefully) understand (althoug i would use the cpystr too... Honest!! :) )

Sure you can use pointer arithmetic if you want since the pointer is pointing to a sizeof(char) address space. Should not be a problem.