C++ Combining two zero terminated strings? - c++

Question

So I am doing a question where I have to join two zero terminated strings, the first contains a word, and the second is empty and twice the size of the original array. I was able to get this working using the following code
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
char str1[] = "test";
char str2[(sizeof(str1)-1)*2];
char *p;
int count = 0;
for(p = str1; *p != 0; p++) {
str2[count] = *p;
count++;
}
cout << str2;
}
However I have to use a function with the following prototype
char *combine(char *a);
So I tried this
#include <stdio.h>
#include <iostream>
using namespace std;
char *copy_and_reverse(char *a) {
char str2[8];
int count = 0;
char* b = str2;
for(a; *a != 0; a++) {
str2[count] = *a;
count++;
}
return b;
}
int main()
{
char str1[] = "test";
char *a;
a = str1;
char* b = copy_and_reverse(a);
for(b; *b != 0; b++) {
cout << *b;
}
}
But it does not work (it is printing the string but it's printing a few random characters after it), I'm getting so confused with the pointers, can anyone help me out with this?
Edit: here is the question I am trying to answer
Write a function in C++ that takes as a char * style zero terminated string and returns a char* string twice the length of the input. The first half of the returned string should contain a copy of the contents of the original array. The second half of the string should contain the contents of the original string in reverse order.
The function should have the following prototype:
char *copy_and_reverse(char* a);
Note: you should not use any library functions (e.g from string.h).

Answer 1

There are two big problems in your copy_and_reverse code.
After copying the input string, you are not terminating the result. This means str2 is not a valid string. Fix:
str2[count] = '\0'; // after the loop
copy_and_reverse returns a pointer to a local variable (str2). After the function returns, all its local variables are gone, and main is dealing with an invalid pointer. To fix this, either use static memory (e.g. by declaring str2 as static or making it a global variable) or dynamic memory (allocate storage with new[] (or malloc())). Both approaches have their disadvantages.
Minor stuff:
variable; does nothing (see for (a; ...), for (b; ...)).
str2 isn't big enough for the final result. str1 is 5 bytes long ('t', 'e', 's', 't', '\0'), so char str2[8] is sufficient for now, but in the end you want to allocate length * 2 + 1 bytes for your result.

Answer 2

I believe that this will suit your needs:
#include <stdio.h>
#include <stdlib.h>
static char* copy_and_reverse(char* a);
static int strlen(char *c); // self-implemented
int main(void) {
char *a = "some string";
char *b = copy_and_reverse(a);
printf("%s", b);
free(b);
return 0;
}
static char* copy_and_reverse(char* a) {
int n = strlen(a);
char *b = new char[n * 2 + 1]; // get twice the length of a and one more for \0
for (int i = 0; i < n; ++i) { // does copying and reversing
b[i] = a[i];
b[i+n] = a[n-i-1];
}
b[2 * n] = '\0'; // null out last one
return b;
}
static int strlen(char *c) {
char *s = c;
while( *s++ );
return s-c-1;
}