c++ - 创建没有字符串库的字符串类

Question

我试图在不使用字符串库的情况下创建自己的字符串类。 #include <string>

我可以使用 C 风格的字符串!

我必须制作 iString ctor、复制 ctor、dtor 并重载运算符 >>、<<、+、*

到目前为止，我的问题是，当我尝试编译我的作品时，我的每个 operator+ 都出现了这个错误。和 operator* ...这是:

warning: reference to local variable s3 returned [enabled by default]

s3只是iString对象名称。

这是我目前所拥有的:

.h文件

#ifndef __ISTRING_H__
#define __ISTRING_H__

struct iString{
  char * chars;
  unsigned int length;
  unsigned int capacity;
  iString();  // ctor
  iString(const char *); // copy ctor
  iString(const iString&); // copy ctor
  ~iString(); // dtor
};

std::istream &operator>>(std::istream &in, const iString &s);
std::ostream &operator<<(std::ostream &out, const iString &s);
iString &operator+(const iString &s1, const iString &s2);
iString &operator+(const iString &s1, const char *c_str);
iString &operator*(const iString &s, const int k);
iString &operator*(const int k, const iString &s);


#endif

.cc 文件

#include <iostream>
#include "istring.h"
#include <cstring>

using namespace std;

iString::iString(){  // this is my ctor
  chars = new char[1];
  length = 0;
  capacity = length + 1;
  chars[0] = '\0';
}

iString::iString(const char *word){ // this is my copy ctor from a c-string
  length = strlen(word);
  capacity = length + 1;
  chars = new char[capacity];
  strcpy(chars, word);
}

iString::iString(const iString &s){ // this is my copy ctor from an iString
  length = strlen(s.length);
  capacity = length + 1;
  chars = new char[capacity];
  strcpy(chars, s.chars);
}


iString::~iString() { //dtor
  delete [] this->chars;
}

istream &operator>>(istream &in, const iString &s){  // not done, don't worry

}

ostream &operator<<(ostream &out, const iString &s){
  //out<<s.chars;
  return out<<s.chars;
}

iString &operator+(const iString &s1, const iString &s2){ 
  iString s3;

  char *new_str;// = new char[size];
  new_str = strcat(s1.chars, s2.chars);

  s3 = iString(new_str);
  return s3;
}

// tried a different approach 
iString &operator+(const iString &s1, const char *c_str){ 

  int size = s1.length + strlen(c_str) + 1;
  char *new_str = new char[size];

  strcpy(new_str, s1.chars);
  strcat(new_str, c_str); 

  iString s3(new_str);
  delete [] new_str;
  return s3;
}

iString &operator*(const iString &s, const int k){

  int size = (s.length * k) + 1;
  char *c_str = new char[size];
  for(int i = 0; i < k; i++){
    strcat(c_str, s.chars);
  }

  iString t(c_str);
  delete[] c_str;
  return t;
}

iString &operator*(const int k, const iString &s){
  return s * k;
}

Answer 1

您不应该从 operator+( , ) 返回引用

iString operator+(const iString &s1, const iString &s2){
 ...
}