自定义C类型字符串类的问题
Problems with Custom C type string class
我有作业,我需要使用C类型字符串创建一个自定义字符串类。大多数看起来还好,但我得到一个运行时错误,只要我测试它,它没有执行我所有的测试,就像它应该。特别是我的+=运算符似乎有问题,但我不明白是什么。
我不能添加或更改任何原型,我应该在可能的情况下使用c++结构而不是C类型。
谢谢!
#include <iostream>
#include <string>
#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] != ' ') {
++i;
}
strSize = i;
strPtr = new char [i+1];
for (i=0; i < strSize; ++i) {
strPtr[i] = str[i];
}
}
//Copy constructor
TStr::TStr(const TStr& str) {
strPtr = new char[str.strSize];
strcpy(strPtr, str.strPtr);
}
//Destructor
TStr::~TStr() {
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) {
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;
}
//overload two relational operators as member functions
bool TStr::operator == (const TStr& str) const {
int value = strcmp(strPtr, str.strPtr);
if (value == 0) {
return true;
} else {
return false;
}
/*int counter=0;
for (int i=0; i < strSize; ++i) {
if (strPtr[i] == str.strPtr[i]) {
++counter;
}
}
if (counter == strSize) {
return true;
} else {
return false;
}
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;
}
//Overload the stream insertion and extraction operators.
ostream& operator << (ostream& out, const TStr& str) {
int size = str.strSize;
for (int i=0; i < size; ++i) {
out << str[i];
}
return out;
}
istream& operator >> (istream& in, TStr& str) {
return in;
}
//overload two other relational operators as global functions
bool operator != (const TStr& S1, const TStr& S2) {
return !(S1 == S2);
}
bool operator <= (const TStr& S1, const TStr& S2) {
return (S1 < S2 || S1 == S2);
}
bool operator > (const TStr& S1, const TStr& S2) {
return !(S1 < S2);
}
bool operator >= (const TStr& S1, const TStr& S2) {
return !(S1 < S2 || S1 == S2);
}
//overload the concatenation operator as a global function
TStr operator + (const TStr& str1, const TStr& str2) {
//return (str1 += str2);
//return (str1 + str2);
}
我要测试的是:
int main() {
authors();
TStr str1 = "VENI"; //initialize str1 using
//the assignment operator
const TStr str2("VEDI"); //initialize str2 using the
//conversion constructor
TStr str3; //initialize str3 to null
TStr str4; //initialize str4 to null
cout << "nTest 1: str1: " << str1 << " str2 " << str2
<< " str3 " << str3 << " ###.n" ; //Test 1
if (str1 <= str2) //Test 2
cout << "nTest 2: " << str1 << " is less "
<< "than " << str2 << endl;
else
cout << "nTest 2: " << str2 << " is less "
<< "than " << str1 << endl;
str1=" Pride is what we have.";
str3 = str1; //Test 3
cout << "nTest 3: The new value of str3 = "
<< str3 << endl;
str3 += " Vanity is what others have. ";
cout<<"nTest 4: The str3: '" << str3<<"' nhas "; //test 4
cout<< countVowels(str3)<<" vowels "<< endl;
/*TStr str5 = str1 + str2 + str3;
cout<<"nTest 5: The str5: '" << str5<<"' nhas "; //test 5
cout<<countVowels(str5)<<" vowels n";
cout<<"nTest 6: The str3 again: " << str3 <<endl; //test 6
cout<<"nn Bye, Bye!";*/
return 0;
}
编辑:我应该补充说,我需要让测试工作作为家庭作业的一部分。
编辑2:这是我到目前为止改变的功能。//constructor; conversion from the char string
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];
strPtr[i + 1] = ' '; //<- this
}
}
//Copy constructor
TStr::TStr(const TStr& str) {
strPtr = new char[str.strSize + 1]; //<-this
strcpy(strPtr, str.strPtr);
}
//overload the concatenation oprerator
TStr TStr::operator += (const TStr& str) {
char *buffer = new char[strSize + str.strSize + 1]; //<- this
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 + 1]; //<- this
assert(strPtr);
for (int i=0; i<strSize; ++i) {
strPtr[i] = str.strPtr[i];
}
}
return *this;
}
编辑3:不幸的是,把它改成下面没有什么真正的。可能是=操作符导致了问题,因为它现在在测试3之后就失败了。
//constructor; conversion from the char string
TStr::TStr(const char *str) {
int i=0;
while (str[i] != ' ') {
++i;
}
strSize = i;
strPtr = new char [i+1];
strcpy(strPtr, str);
}
您在几个函数中涉及到null终止符的几个问题(其中一些问题您已经在编辑中修复了,但无论如何我都将讨论它们,因此这是一个全面的答案)。
-
在从const char *
转换的构造函数中,您不会将空终止符复制到strPtr。> li>在复制构造函数中,没有为空结束符分配足够的空间。> li>在赋值操作符中,没有为空结束符分配足够的空间。(修复…不完全是,见#5) - 在您的赋值操作符中,您没有复制空终止符。
- 当你修复分配时,你在赋值运算符中添加了一个新问题。你现在设置strSize的大小错误。
有几个地方需要+1
为字符串保留结尾的