C语言无法使用引用，一定要使用怎么办？ ------指针的指针做参数

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void fun1(char** s);
void fun2(char** s);
void fun3(char** s);

void fun1(char** s)
{
	char* _s = *s;
	*s = (char*)malloc(sizeof(char)*100); 

	strcpy( *s,"11" );

	//*s = (char*)("11");  不可以这样赋值，会造成内存泄露
	printf( "%s\n",*s );
	fun2(s);

}

void fun2(char** s)
{
	strcpy( *s,"22" );
	printf( "%s\n",*s );
	fun3(s);
}

void fun3(char** s)
{
	strcpy( *s,"33" );
	printf( "%s\n",*s );
}

int main()
{
	char *_s = NULL;
	char** s = &_s;
	fun1(s);

	printf( "final ; %s\n",*s );
	free( *s );
	return 0;
}