strcpy and strcat return an error in c

Question

   char first[]="aa";               
   strcat(first,ar[0]);

   char *second[10];   
   strcpy(second[0],first);

I want to use strcat and strcpy. strcat works but there is an error in strcpy. how can I modified it?

`char first[]="aa";` has allocated 3 bytes for the array. How can `strcat()` append to that? There is no room. — Weather Vane, Nov 28 '15 at 17:23
In the second case, you have not allocated any memory to your pointer array. — Weather Vane, Nov 28 '15 at 17:24
@esrtr it might work today, but it will fail at the very moment you proudly demonstrate your project. — Weather Vane, Nov 28 '15 at 17:25
By allocating enough memory to hold `"aa"` and `ar[0]` and a terminating `nul` byte. We don't know what `ar[0]` is, but you could try, for example, `char first[100] = "aa";` — Weather Vane, Nov 28 '15 at 17:28
I tried to initialize. I tried char *second[10]=NULL; but it returns an error @AugustKarlstrom — esrtr, Nov 28 '15 at 17:30
In the second case you are now (in comment) assigning a `NULL` pointer, which you cannot dereference (copy to or from). You still have to allocate memory, please research library function `malloc()`. However, even when you *have* allocated memory, it *still* needs to be initialised to use as a string. Copying a string to it is one way to do this. — Weather Vane, Nov 28 '15 at 17:34

August Karlstrom · Answer 1 · 2016-01-03T15:28:52.880

In C there are basically three kind of string/character array variables:

string constant pointers
character arrays where the length is set at compile time
dynamic character arrays where the length is set at run-time.

For a dynamic array, the memory it uses is reserved when the program is run. The program below illustrates the three different cases. It also provided a convenient macro for memory allocation.

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

#define NEW_ARRAY(ptr, n) \
    { \
        (ptr) = malloc((n) * sizeof (ptr)[0]); \
        if ((ptr) == NULL) { \
            fprintf(stderr, "error: Memory exhausted\n"); \
            exit(EXIT_FAILURE); \
        } \
    }

int main(void)
{
    const char *s1 = "foo";
    char s2[4] = "bar";
    char *s3;

    NEW_ARRAY(s3, 4);
    strcpy(s3, "baz");

    return 0;
}

strcpy and strcat return an error in c

1 Answers1