Mistake with storing strings in char pointer array

Question

I know this question might sound quite silly, but I somehow found myself stuck and need help. I have a char* variable char* address="/a/asdasd/c/sdfsdf/adsd"; and I declared an array of char pointer char* store[5]; . I'm trying to divide the content in the variable address by tracing the slash(/) and trying to store each part in the char pointer variable store by doing the following

char* store[5];
char* address="/a/asdasd/c/sdfsdf/adsd";
int k=0;
int j=0;
char* b=NULL;
for(int i=0;i<5;i++)
{
    if(b==0)
    {
        b=strchr(address,'/');
    }
    else
    {
        b=strchr(b,'/');
    }
    j=b-address;
    strncpy(store[i],address+k,j-k);
    k=j;
}

But I see that in the code strncpy(store[i],address+k,j-k) there's an error. The compiler doesn't seem to understand that store[i] is a char pointer, it rather thinks it is a char. Could you see how I can solve the problem?

Thanks for all the help. I've solved it. Solution code is as below:

char* address="/a/asdasd/c/sdfsdf/adsd/asfsd";

     char store[5][100];
     char* b=NULL;
     int k=0;
     int j=0;
     for(int i=0;i<5;i++)
     {
         if(b==0)
         {
             b=strchr(address+1,'/');
         }
         else
         {
             b=strchr(b+1,'/');
         }
         j=strlen(address)-strlen(b);
         strncpy(store[i],address+k+1,j-k-1);
         store[i][j-k-1]='\0';
         printf("%s\n",store[i],j-k);
         k=j;
     }

should there be a int b = 0; somewhere? what about a? store? — Levon
– Levon, Commented Jun 14, 2012 at 11:12
No, there should have been char* b=NULL. I edited it. Thanks. — the_naive
– the_naive, Commented Jun 14, 2012 at 11:13
Your code block should really stand on its own, it's not a good idea to have part of your declarations hidden inside the text of your question. I edited your code to add those missing declarations. Just a suggestion - otherwise it makes it hard to compile/run your code. — Levon
– Levon, Commented Jun 14, 2012 at 11:19

Jay · Accepted Answer · 2012-06-14 11:16:58Z

3

char *store[5]

This is just an array of char pointers. To store strings in each element of this array, you need malloc memory and assign it to the respective pointer.

For Ex, you can change your code to

store[i] = malloc ((j-k)+ 1); // +1 is for the null terminator. Pls check return value of malloc also.
strncpy(store[i],address+k,j-k);

answered Jun 14, 2012 at 11:16

Jay

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2 Comments

the_naive Over a year ago

I knew about using malloc, but I wanted to avoid it because this code is going to be used for project where we don't to use malloc because of memory issues. Isn't there any other way to solve it?

Jay Over a year ago

Can you make it a fixed array like char store[5][100]? and in strncpy, pass the third parameter as (100-1) = 99?

unwind · Accepted Answer · 2012-06-14 11:24:38Z

1

If you want to copy a pointer, you shouldn't be calling strncpy(), since that copies characters.

You want:

store[i] = address + (j - k);

assuming address + (j - k) is the desired starting point for the part.

edited Jun 14, 2012 at 11:24

answered Jun 14, 2012 at 11:08

unwind

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Comments

Whoami · Accepted Answer · 2012-06-14 11:35:44Z

0

If you don't want to have a copy of the string tokens, if you like only to retain the pointers, then just store the address in store[i] as @unwind pointed out. Or else, you could explore strtok () also. only think is that you need to have separate array to keep each length of the string according to your code. Hope this helps.

answered Jun 14, 2012 at 11:35

Whoami

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