Thread: working with srings adresses and pointers,

Originally Posted by Lired

but i think im creating a new memory space for this wich is not what is intended.

You are not - your movefunction() acts on the m1 and m2 you pass it (i.e. in this case, the ones declared in your main() function), so you're okay on that front.

However, the exercise states that "this function must work even if m1 and m2 share memory space", i.e. if they overlap; however, yours will not - think about if you did e.g. "movefunction(&m1[1], &m1[0], n)". Before you fix this, I suggest you write a small test case to show you that it's broken. To fix, you need to allocate a new array, copy all the contents of the source buffer in there, then copy them all to the destination buffer, then free the array. If you're allowed to use variable-length arrays, you can just declare a new array "char tmp[n]" in your function - otherwise you'll have to use malloc()/free().

Originally Posted by JohnGraham

You are not - your movefunction() acts on the m1 and m2 you pass it (i.e. in this case, the ones declared in your main() function), so you're okay on that front.

However, the exercise states that "this function must work even if m1 and m2 share memory space", i.e. if they overlap; however, yours will not - think about if you did e.g. "movefunction(&m1[1], &m1[0], n)". Before you fix this, I suggest you write a small test case to show you that it's broken. To fix, you need to allocate a new array, copy all the contents of the source buffer in there, then copy them all to the destination buffer, then free the array. If you're allowed to use variable-length arrays, you can just declare a new array "char tmp[n]" in your function - otherwise you'll have to use malloc()/free().

well i dont think i quite understand. are u saying to use another string equal to m2 that, because it is a diferent variable has a different memory location? for example, like this:

Code:

void movefunction (char m1[], char m2[], int n){
    char temp[1000];
    int i;

    if(((m1-m2)<0)||((m1-m2)<n)){
        for (i=0;i!=n;i++){
                temp[i] = m2[i];
            }
        for (i=0;i!=n;i++){
                m1[i] = temp[i];
            }
    }
    else {
        for (i=0;i!=n;i++){
                   m1[i] = m2[i];
            }
    }
}

and also the example where you were saying it didnt work, was it, for example,

m1="sl"
m2="o"
n=2
returns "o"

?

i guess here it should return "ol" but it doesn't...

Originally Posted by anduril462

It can be done without the temporary array, but it takes some thought.

Not if you mean using pointer comparisons (other than equality/inequality). The behaviour of comparison operations other than equality and inequality are undefined if the two pointers don't point to parts of the same object (or array with "one past its end").

Personally, I'd use memmove(), but I suppose that's not the point of the exercise.

Damn! I was hoping you had another approach to do it, which didn't involve such pointer comparisons.

In fact you can do smth like this:

Code:

#include <stdio.h>
#include <string.h>
#define MIN(a,b) ((a)<(b)) ? a:b;
void movefunction(char m1[], char m2[], int n);
 
int main() 
{
 
char m1[1000];
char m2[1000];
int n;
 
printf("write string 1 \n");    
scanf("%s", m1);
printf("write string 2 \n");
scanf("%s",m2);    
printf("write the number of bytes to move\n");
scanf("%d",&n);
 
movefunction(m1, m2, n);
 
printf("%s", m1);


return 0;
 
}
 
 
void movefunction(char *s1, char *s2, int n){
   int i,move;
   move=MIN(strlen(s1),n);
   move=MIN(strlen(s2),move);
   for (i=0; i<move; i++,s1++,s2++)
          *s1 = *s2;
}

So you will use pointers and you will not run out of strings lengths

That will not work if the strings overlap, Omen - among other things.

Originally Posted by Omen

Maybe, I didn't get the problem... I thought I have to change first n bytes of string one in the first n bytes of string two. No?

No. Look at post #2.

Also, to reiterate a reply I gave to your other post, it is against site policy to just give complete code solutions.