Утечка при перегрузки оператора "+" - C++ - Ответ 695450

class andString
{
    char *sTring;
    unsigned int nLenght;
    
    
public:
    friend ostream &operator<<(ostream &, andString &);
    friend istream &operator>>(istream &, andString &);
 
    //
    andString &operator=(const andString &);//General
    andString &operator=(const char *);
    andString &operator=(const char);
 
    //
    andString &operator+(const andString &);//General !Здесь то и проблемы, со всеми операторами +!
 
    friend andString &operator+(const andString &,const char *);
///....///тд

#include <iostream>
using namespace std;
 
#include "andString.h"
 
//
andString::andString()
{
    nLenght = 0;
    sTring = NULL;
}
//
ostream &operator<<(ostream &stream,andString &aStr)
{
    stream<<aStr.sTring;
    return stream;
}
//
istream &operator>>(istream &stream,andString &aStr)
{
    char *sTemp=new char[1024];
    stream>>sTemp;
    delete [] aStr.sTring;
    aStr.nLenght = strlen(sTemp)+1;
    aStr.sTring = new char[aStr.nLenght];
    for(int i=0; i<aStr.nLenght ; ++i)
        aStr.sTring[i] = sTemp[i];
    delete [] sTemp;
    return stream;
}
//////..../////
//Add
//!
andString &andString::operator+(const andString &aStr)
{
    andString *asOpTemp = new andString;
    delete [] asOpTemp->sTring;
    asOpTemp->nLenght = nLenght + aStr.nLenght - 1;
    asOpTemp->sTring = new char[asOpTemp->nLenght];
    for(int i = 0 ; i < nLenght-1 ; ++i)
        asOpTemp->sTring[i] = sTring[i];
    for(int i = 0 ; i < aStr.nLenght ; ++i)
        asOpTemp->sTring[i + nLenght -1] = aStr.sTring[i];
    return *asOpTemp;
}
////.....////