实现矢量的char * insted [英] implementing char* insted of vector

问题描述

你好大家请查看我的代码我试图取一串数字然后压缩它

之前在.hpp我声明了一个字符向量（向量< char> nume）到存储新的压缩数字字符串

现在您可以看到我将其更改为char *和int size

你能推荐一个新的类实现吗？新的字母*和大小而不是矢量



.hpp文件

hello everyone please check out my code i'm trying to take a string of numbers and compress it
before in the .hpp i declared a vector of chars (vector<char> nume) to store the new string of compressed numbers
and now as you can see i changed it to a char* and int size
can you recommend a new implementation of the class using this new char* and size instead of a vector

.hpp file

#include "iostream"
using namespace std;
#include "string"
#include "vector"


class Mint{
public:
	Mint();
	Mint(int);
	Mint(const char*);
	bool operator<(const Mint&);
	bool operator>(const Mint&);
	void afficher();
	void size();
private:
	char* num;
	int size;
};

.cpp文件

.cpp file

#include "Mint.h"
#include "string.h"





Mint::Mint()
{
	taille = 1;
	num = new char[taille];
	*num = '\0';
}

Mint::Mint(const char* s)
{
	if(strlen(s)%2)
		num.push_back(s[0]-'0');
	unsigned int i;
	for(i=strlen(s)%2;i<strlen(s);i+=2)
	{
		int left = s[i] - '0';
		int right = s[i+1]-'0';
		num.push_back(0);
		num.back() = left << 4;
		num.back() |= right;
	}
}
bool Mint::operator<(const Mint& rhs)
{
	unsigned int i;
	if (num.size()<rhs.num.size())
			return true;
	if (num.size()==rhs.num.size())
	for(i=0;i<num.size();i++)
	{
		if(num[i] < rhs.num[i])
			return true;
	}
	return false;
}
bool Mint::operator>(const Mint& rhs)
{
	if (*this < rhs)
	 return false;
	return true;
}
void Mint::afficher()
{

	unsigned int i;
	for (i=0;i<num.size();i++)
	{
		int first_digit = (num[i] & '\xF0')>>4;
		int second_digit = (num[i] & '\x0F');
		if (i || first_digit) cout << first_digit;
		cout << second_digit;
	}
}
void Mint::size()
{
	cout << num.size();
}

推荐答案

正如其他人所发现的，制作标准库容器的版本通常没有意义，因为容器写得很好，就是它们是

• 可靠
• 高效
• 有严格的界面
• ......

As others spotted, it is usually pointless to make you own version of a standard library container, because such containers are well written, that is they are

• Reliable
• Efficient
• Have a rigorous interface
• ...

这篇关于实现矢量的char * insted的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！