c pointers in string array

Question

I'm trying to copy a char array into another char array using pointer arithmetic. It seems to be correct within the copy() but then I don't understand what happens after it gets to main. char a[] does not get the value of char b[] even if I called the function. Am I missing something? Hehehe

    #include <stdio.h>

    void copy(char a[], char b[]){
        int *apoint = &a;
        printf("%d\n", apoint);
        printf("%d\n", &a);
        *apoint = b;
        printf("%d\n", *apoint);
        printf("%s\n", a);
        printf("%s\n", b);
    }

   int main(void){
        char a[100];
        char b[] = "bluhbluh";

        copy(a,b);
        printf("%d\n", a);
   }

sorry!! :( It was the initial answer that followed my chain of thought. Didn't really know pointers don't work like this haha — crispyfriedchicken
– crispyfriedchicken, Commented Nov 14, 2012 at 11:38

ams · Accepted Answer · 2012-11-14 11:19:52Z

2

That's really not very meaningful C. I suspect it didn't even compile without warnings?

First you take the address of array a, then you take the address of array b, convert it to an integer (without a cast!), and then write that integer into the array a. Then, at the end of main, you try to print the char array as an integer.

Pointer arithmetic doesn't quite work like that. :)

I think you really mean to do something like this:

void copy(char a[], char b[]){
    char *ptr_a = a;
    char *ptr_b = b;

    while (*ptr_b != '\0') {
        *ptr_a = *ptr_b;
        ptr_a++;
        ptr_b++;
    }
    *ptr_a = '\0';
}

int main(void){
    char a[100];
    char b[] = "bluhbluh";

    copy(a,b);
    printf("%s\n", a);
}

Of course, the proper way to do it is like this:

#include <string.h>

int main(void){
    char a[100];
    char b[] = "bluhbluh";

    strncpy(a, b, 100);
    printf("%s\n", a);
}

answered Nov 14, 2012 at 11:19

ams

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1 Comment

glglgl Over a year ago

Be aware that strncpy() doesn't put a \0 if the limit is reached. So better call it with 99 (sizeof a - 1) and set a[sizeof a-1] = '\0'.

Mathew Hall · Accepted Answer · 2012-11-14 11:30:32Z

2

As ams points out in his/her answer, if you want a copy of the array, you need to duplicate each entry. Assuming you don't want to duplicate the array, but instead want a to point at b:

apoint doesn't do anything - it can't change the pointers of a or b. If you wanted to modify the address that a or b pointed to, you'd have to pass the address of them to your function. This won't work, however, as arrays are not pointers - it's why you can't reassign arrays in C. You can modify pointers though - the following is an example of modifying a pointer to a to point at b:

#include <stdio.h>

void redirect(char** a, char* b){
    *a = b;
}

int main(void){
    char a[100];
    char b[] = "bluhbluh";

    char* apt = a;
    redirect(&apt,b);
    printf("a=%s, b=%s", apt,b);

}

edited Nov 14, 2012 at 11:30

answered Nov 14, 2012 at 11:25

Mathew Hall

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Comments

pagra · Accepted Answer · 2012-11-14 11:09:38Z

0

apoint should be a char*, don't you have a warning about this pointer assignment? Moreover you're not iterating on all chars of the string...

answered Nov 14, 2012 at 11:09

pagra

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Comments

graziano governatori · Accepted Answer · 2012-11-14 11:12:58Z

0

Maybe the line int *apoint = &a is wrong, &a is pointer to pointer to char, while apoint is int *. Don't you have compile errors?

answered Nov 14, 2012 at 11:12

graziano governatori

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Comments

SajjadHashmi · Accepted Answer · 2012-11-14 11:21:12Z

0

You have some major errors in your code specially this line is incorrect int *apoint = &a; it should be char* *apoint = &a;

//Corrected Code:
#include <stdio.h>

void copy(char a[], char b[]){
    char* *apoint = &a;
    printf("%d\n", apoint);
    printf("%d\n", &a);
    *apoint = b;
    printf("%d\n", *apoint);
    printf("%s\n", a);
    printf("%s\n", b);
}

int main(void){
    char a[100];
    char b[] = "bluhbluh";

    copy(a,b);
    printf("%d\n", a);
}

This will solve the major code issues though I don’t think your copy function would still give your desire results as approach doesn’t seem very right.

answered Nov 14, 2012 at 11:21

SajjadHashmi

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Comments

Omkant · Accepted Answer · 2012-11-14 11:21:21Z

0

Method of copying the char is not good if you enable -Wall or -Werror in gcc you will see so many errors.

The thing you want, which is possible just by copying the pointers in main no need to call copy (a,b);

char *a;
char b[]="bluhbluh"
a=b;

now a started pointing to string "bluhbluh"

But I think that you dont want you want each and every character of array b to be copied into array a

you will have to do like this

void copy(char *a,char *b)
{
  while(*b!='\0')
    *a++=*b++;
 }

int main()
{
  char a[100];
  char b[]="bluhbluh";
  copy(a,b);
  printf("%s\n",a);
}

edited Nov 14, 2012 at 11:21

answered Nov 14, 2012 at 11:15

Omkant

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