c++ 自己实现getLine函数，可以接受任意的字符长度的输入

int  getline(char ** iStr) {
	int i = 0;
	char c;
	int count = 5,acNumber=0;
	char* temp = new char[5];

	//保存开始的字符串开始的指针
	char * start = temp;
	*(start + 4) = '
 
 
int  getline(char ** iStr) {
	int i = 0;
	char c;
	int count = 5,acNumber=0;
	char* temp = new char[5];

	//保存开始的字符串开始的指针
	char * start = temp;
	*(start + 4) = '\0';
	cout << "开始输入:\n" << endl;
	while (c = getchar())
	{
		//防止输入有问题造成死循环
		if (c == '\n')break;
		i++;
		cout << "开始读取输入:" <<c<< endl;
		if (i == count) {
			char* tempc = new  char[count+1];
			char * tempStar = start;
			strcpy_s(tempc, count+1, start);
			delete[] tempStar;
			temp = new char[count * 2];
			char * tempct = tempc;
			strcpy_s(temp, count+1, tempc);
			delete[] tempct;
			start = temp;
			count *= 2;
			*(start + count-1) = '\0';
		}
		*(start + acNumber) = c;
		acNumber++;
		cout << "添加一个字符" << endl;
		if (c == ';')break;
	}
	*(start +acNumber) = '\0';
	cout << "start长度为：" << acNumber << endl;
	//申请一个合适的空间
	char* tempc = new char(acNumber+1);
	cout << "开始复制" << endl;
	char * tempstart = start;
	strcpy_s(tempc, acNumber+1 , start);
	cout << "结束复制:" << *start <<endl;
	delete[] tempstart;
	cout << "清除完成" << endl;
	/*for (int i = 0; i < acNumber; i++)
	{
		cout << "输入为：" <<*(++tempc) <<endl;
	}*/
	*iStr = tempc;
	cout << "赋值完成" << endl;
	return acNumber;
}
 
 
    在c++以前得到用户的输入只能提前定义一个固定长度的字符串，感觉不是很方便，因为不知道用户输入的字符串的长度到底是多长。今天我就仿照着c#中的List的实现方法实现了一个可以接受任意长度的getLine()函数，具体代码如下，在使用时不要忘了在前面引入<string.h>。具体代码如下。 
  
  

';
	cout << "开始输入:\n" << endl;
	while (c = getchar())
	{
		//防止输入有问题造成死循环
		if (c == '\n')break;
		i++;
		cout << "开始读取输入:" <<c<< endl;
		if (i == count) {
			char* tempc = new  char[count+1];
			char * tempStar = start;
			strcpy_s(tempc, count+1, start);
			delete[] tempStar;
			temp = new char[count * 2];
			char * tempct = tempc;
			strcpy_s(temp, count+1, tempc);
			delete[] tempct;
			start = temp;
			count *= 2;
			*(start + count-1) = '
 
 
int  getline(char ** iStr) {
	int i = 0;
	char c;
	int count = 5,acNumber=0;
	char* temp = new char[5];

	//保存开始的字符串开始的指针
	char * start = temp;
	*(start + 4) = '\0';
	cout << "开始输入:\n" << endl;
	while (c = getchar())
	{
		//防止输入有问题造成死循环
		if (c == '\n')break;
		i++;
		cout << "开始读取输入:" <<c<< endl;
		if (i == count) {
			char* tempc = new  char[count+1];
			char * tempStar = start;
			strcpy_s(tempc, count+1, start);
			delete[] tempStar;
			temp = new char[count * 2];
			char * tempct = tempc;
			strcpy_s(temp, count+1, tempc);
			delete[] tempct;
			start = temp;
			count *= 2;
			*(start + count-1) = '\0';
		}
		*(start + acNumber) = c;
		acNumber++;
		cout << "添加一个字符" << endl;
		if (c == ';')break;
	}
	*(start +acNumber) = '\0';
	cout << "start长度为：" << acNumber << endl;
	//申请一个合适的空间
	char* tempc = new char(acNumber+1);
	cout << "开始复制" << endl;
	char * tempstart = start;
	strcpy_s(tempc, acNumber+1 , start);
	cout << "结束复制:" << *start <<endl;
	delete[] tempstart;
	cout << "清除完成" << endl;
	/*for (int i = 0; i < acNumber; i++)
	{
		cout << "输入为：" <<*(++tempc) <<endl;
	}*/
	*iStr = tempc;
	cout << "赋值完成" << endl;
	return acNumber;
}
 
 
    在c++以前得到用户的输入只能提前定义一个固定长度的字符串，感觉不是很方便，因为不知道用户输入的字符串的长度到底是多长。今天我就仿照着c#中的List的实现方法实现了一个可以接受任意长度的getLine()函数，具体代码如下，在使用时不要忘了在前面引入<string.h>。具体代码如下。 
  
  

';
		}
		*(start + acNumber) = c;
		acNumber++;
		cout << "添加一个字符" << endl;
		if (c == ';')break;
	}
	*(start +acNumber) = '
 
 
int  getline(char ** iStr) {
	int i = 0;
	char c;
	int count = 5,acNumber=0;
	char* temp = new char[5];

	//保存开始的字符串开始的指针
	char * start = temp;
	*(start + 4) = '\0';
	cout << "开始输入:\n" << endl;
	while (c = getchar())
	{
		//防止输入有问题造成死循环
		if (c == '\n')break;
		i++;
		cout << "开始读取输入:" <<c<< endl;
		if (i == count) {
			char* tempc = new  char[count+1];
			char * tempStar = start;
			strcpy_s(tempc, count+1, start);
			delete[] tempStar;
			temp = new char[count * 2];
			char * tempct = tempc;
			strcpy_s(temp, count+1, tempc);
			delete[] tempct;
			start = temp;
			count *= 2;
			*(start + count-1) = '\0';
		}
		*(start + acNumber) = c;
		acNumber++;
		cout << "添加一个字符" << endl;
		if (c == ';')break;
	}
	*(start +acNumber) = '\0';
	cout << "start长度为：" << acNumber << endl;
	//申请一个合适的空间
	char* tempc = new char(acNumber+1);
	cout << "开始复制" << endl;
	char * tempstart = start;
	strcpy_s(tempc, acNumber+1 , start);
	cout << "结束复制:" << *start <<endl;
	delete[] tempstart;
	cout << "清除完成" << endl;
	/*for (int i = 0; i < acNumber; i++)
	{
		cout << "输入为：" <<*(++tempc) <<endl;
	}*/
	*iStr = tempc;
	cout << "赋值完成" << endl;
	return acNumber;
}
 
 
    在c++以前得到用户的输入只能提前定义一个固定长度的字符串，感觉不是很方便，因为不知道用户输入的字符串的长度到底是多长。今天我就仿照着c#中的List的实现方法实现了一个可以接受任意长度的getLine()函数，具体代码如下，在使用时不要忘了在前面引入<string.h>。具体代码如下。 
  
  

';
	cout << "start长度为：" << acNumber << endl;
	//申请一个合适的空间
	char* tempc = new char(acNumber+1);
	cout << "开始复制" << endl;
	char * tempstart = start;
	strcpy_s(tempc, acNumber+1 , start);
	cout << "结束复制:" << *start <<endl;
	delete[] tempstart;
	cout << "清除完成" << endl;
	/*for (int i = 0; i < acNumber; i++)
	{
		cout << "输入为：" <<*(++tempc) <<endl;
	}*/
	*iStr = tempc;
	cout << "赋值完成" << endl;
	return acNumber;
}int  getline(char ** iStr) {
	int i = 0;
	char