how to pass array used in function to the dynamically allocated array of main?

Question

#include<iostream>

#include<string>    
using namespace std;

char *fun(const char* a)
{
        int size=strlen(a);

        char* str=new char[12];

        for(int i=0;i<size-1;i++)
        {
                str[i]=a[size-i-1];
        }

        str[size-1]='\0';

        return str;
}

main()
{
        int i=0;

        char a[11]={'a','y','u','s','h','r','i','k','h','r','a'};

        char *p=new char[11];

        p=fun(a);

        for(int i=0;i<11;i++)
        {
                cout<<p[i]<<" ";
        }

        delete(ptr);
}

//having some troubles using array p in the main here please help.

Answer 1

There are many issues:

---

Here the a array is not NUL terminated, therefore strlen cannot be used on it in fun:

char a[11]={'a','y','u','s','h','r','i','k','h','r','a'};

---

Here you are allocating a fixed size of 12, but you need a dynamic size depending on the length of the string:

char* str=new char[12];

---

Then:

str[size-1]='\0';

should be:

str[size] = '\0';

otherwise your resulting string will miss the last character.

---

In main:

char *p=new char[11];

is pointless because you assign p right after with p=fun(a);.

---

You call delete(ptr);, but there is no ptr.

---

And finally you should also #include <string.h> be sure that strlen is defined, but one some platforms it compiles also without #include <string.h>.

---

You probably want this:

#include <iostream>
#include <string>
#include <string.h>

using namespace std;

char *fun(const char* a)
{
  int size = strlen(a);
  char* str = new char[size + 1];

  for (int i = 0; i < size; i++)
  {
    str[i] = a[size - i - 1];
  }

  str[size] = '\0';
  return str;
}

int main()
{
  char a[] = "ABCDE";    
  char *p = fun(a);    
  int size = strlen(p);

  for (int i = 0; i < size; i++)
  {
    cout << p[i] << " ";
  }

  delete[] p;
}