在c++中创建自定义C类型字符串类

我有一个分配，我需要在c++中创建一个自定义的C类型字符串类。我用这个有点麻烦。目前，我的代码在开始时出现运行时错误而崩溃。我也知道我的许多函数是错误的，但我想在我继续修复其他函数之前，把成员函数排序。请记住，所有的函数原型都是给我们的，我不能更改它们。可以说，我需要写"胆量"。

我的构造函数有什么问题?

#include <iostream>
#include "tstr.h"
using namespace std;
//Default constructor to initialize the string to null
TStr::TStr() {
    strPtr = 0;
    strSize = 0;
}  
//constructor; conversion from the char string
TStr::TStr(const char *str) {
    int i=0;
    while (str[i] != '/0') {
        strPtr = new char [strlen(str)+1];
        for (i=0; i <strSize;++i) {
            strPtr[i] = str[i];
        }
        ++i;
    }
    strSize = i;
} 
//Copy constructor
TStr::TStr(const TStr&) {
}
//Destructor
TStr::~TStr() {
    if (strPtr) {
        delete[] strPtr;
    }
}
//subscript operators-checks for range
char& TStr::operator [] (int i) {
    assert (i >= 0 && i < strSize);
    return strPtr[i];
}
const char& TStr::operator [] (int i) const {
    assert (i >= 0 && i < strSize);
    return strPtr[i];
}
//overload the concatenation oprerator
TStr TStr::operator += (const TStr& str) {
    //this->strPtr += str.strPtr;
    //this->strSize += str.strSize;
    return *this;
}
//overload the assignment operator
const TStr& TStr::operator = (const TStr& str) {
    if (this != &str) {
        delete[] strPtr;
        strPtr = new char[strSize = str.strSize];
        assert(strPtr);
        for (int i=0; i<strSize; ++i) {
            strPtr[i] = str.strPtr[i];
        }
    }
    return *this;
}
//overload two relational operators as member functions
bool TStr::operator == (const TStr& str) const {
    return (strPtr == str.strPtr && strSize == str.strSize);
}
bool TStr::operator < (const TStr& str) const {
    return (strPtr < str.strPtr && strSize < str.strSize);
}
//the length of the string
int TStr::size() {
    return strSize;
}

感谢您的回复/帮助!:)

编辑1:好吧，构造函数现在正在工作，但我仍然得到一个运行时错误，我90%肯定这是与我的重载+=操作符。它看起来很好，编译也很好。我错过了什么?

(注意:上面的代码只做了很小的改变，但是如果你想看到全部，请告诉我。)

//overload the concatenation oprerator
TStr TStr::operator += (const TStr& str) {
    for(int i = 0; i < strSize; ++i) {
        strPtr[i] += str.strPtr[i];
    }
    return *this;
}

编辑2:好吧，这就是我现在有的。编译很好，但实际上没有像它应该的那样用+=将两个字符串加在一起。有人有什么想法吗?

//overload the concatenation oprerator
TStr TStr::operator += (const TStr& str) {
    char *buffer = new char[strSize + str.strSize + 1];
    strcpy(buffer, strPtr);
    strcat(buffer, str.strPtr);
    delete [] strPtr;
    strPtr = buffer;
    return *this;
}
//overload the assignment operator
const TStr& TStr::operator = (const TStr& str) {
    if (this != &str) {
        delete[] strPtr;
        strPtr = new char[strSize = str.strSize];
        assert(strPtr);
        for (int i=0; i<strSize; ++i) {
            strPtr[i] = str.strPtr[i];
        }
    }
    return *this;
}

// as James perfectly illustrated
TStr::TStr(const char *str) 
{
  int i = 0;
  while (str[i] != '') 
       ++i;
  strSize = i;
  strPtr = new char [i+1];
  while (*strPtr++ = *str++); // with a bit of k&R
} 
//overload the concatenation oprerator
TStr TStr::operator += (const TStr& str) {
    for(int i = 0; i < strSize; ++i) {
        strPtr[i] += str.strPtr[i];
    }
    return *this;
}

//overload the concatenation oprerator
TStr TStr::operator += (const TStr& str) {
    unsigned int i = 0;
    while (str.strPtr[i] != '') 
       ++i;
    // allocate the new buffer
    char* newStr = new char[i + strSize + 1];
    // copy the old string
    unsigned int j = 0;
    for (; j < strSize; ++j) 
    {
        newStr[j] = strPtr[j];
    }
    // update the size
    strSize += i;
    // release the old buffer
    delete[] strPtr;
    // finally concatenate
    char* copyPtr = newStr + j;
    while(*copyPtr++ = *(str.strPtr)++);
    // and swap the pointers
    strPtr = newStr;     
    return *this;
    }

TStr::TStr(const char *str) 
{
  strSize = strlen(str);
  strPtr = new char [strSize+1];
  strcpy(strPtr, str);
}

TStr::TStr(const char *str) 
{
  int i = 0;
  while (str[i] != '') 
       ++i;
  strSize = i;
  strPtr = new char [i+1];
  for (i=0; i < strSize;++i)
        strPtr[i] = str[i];
} 

//constructor; conversion from the char string
TStr::TStr(const char *str) {
    strSize = strlen(str);
    strPtr = new char [strSize+1];
    for (int i=0; i <strSize; ++i) {
        strPtr[i] = str[i];
    }
    strPtr[strSize] = '';
}