Pointers and Character Array

Question

I have a pointer to a character array:

space=(char**)calloc(100000,sizeof(char*));
for(i=0;i<100000;i++){
    space[i]=(char*)calloc(1,sizeof(char));
}

such that when I use the following command

printf("%s\n",space[0]);

I get "a b c d e"

I want to assign "a b c d e" to

char c[10];

such that

printf("%s",c) yields

"a b c d e"

but when I try

c=space[0]

I get the following error:

incompatible types in assignment

What am I doing wrong?

You can't assign to an array. Use a char * instead, or use strcpy to copy the contents of space[0] into c. — Wug
– Wug, Commented Aug 17, 2012 at 19:46
I don't see how you can get "a b c d e" from that printf with the above code. And I don't see what exactly you're trying to do. — netcoder
– netcoder, Commented Aug 17, 2012 at 20:29

mocj · Accepted Answer · 2012-08-17 20:17:53Z

2

First, you need to allocate enough space to hold the entire string at location space[0]. Currently you are allocating a single character. After you do that you would use strcpy() to copy the string into the newly allocated buffer.

space=(char**)calloc(100000,sizeof(char*));
...
space[0]=(char*)calloc(10,sizeof(char));
strcpy(space[0], "a b c d e");

PS: Don't forget to free() any prior allocated strings at space[0] (like the one you create in the for() loop) before you allocate for the new string. Or you could use realloc() instead.

edited Aug 17, 2012 at 20:17

answered Aug 17, 2012 at 19:48

mocj

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Comments

user836218 · Accepted Answer · 2012-08-18 04:17:33Z

0

c[0] is of the type char *. You either need to do c[0][0] or (char) c.

answered Aug 18, 2012 at 4:17

user836218

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