實現字符串(char*)的前N個字符放到後面

/*
實現字符串(char*)的前N個字符放到後面
*/
#include <iostream>

using namespace std;

char* func(char *str, int N)
{
	if (NULL == str)
		return 0;
	int len = strlen(str);
	if (N >= len)
		N = N % len;
	if (len<=0 || N<=0)
		return str;
	char *str1 = new char[N+1];
	char *str2 = new char[len-N+1];
	for (int i=0; i<N; i++)
	{
		str1[i] = str[i];
	}
	str1[N] = '\0';
	for (int i=0; i<len-N; i++)
	{
		str2[i] = str[N+i];
	}
	str2[len-N] = '\0';
	char *str3 = new char[len+1];
	for (int i=0; i<len-N; i++)
	{
		str3[i] = str2[i];
	}
	for (int i=0; i<N; i++)
	{
		str3[i+len-N] = str1[i];
	}
	str3[len] = '\0';

	delete[] str1;
	delete[] str2;
	return str3;
}

int main()
{
	char *str = "abcdefg";
	char *ss = func(str,5666);
	if (ss != NULL)
		cout << ss << endl;

	return 0;
}

用string特性實現請見http://blog.csdn.net/redeagle_gbf/article/details/16823365