指针函数,结构体指针

博客围绕角色数据处理展开,包含三方面内容:一是数据中存储角色血值信息,血值类型为int,值为100;二是编写函数返回角色名字信息地址,无则返回0;三是查找id=1且level=8的结构体信息数量。

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1、这一堆数据中存储了角色的血值信息,假设血值的类型为int类型,值为100(10进制)

#include "stdafx.h"
#include<stdio.h>

char data[100] = {
	0x02,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x07,0x09,					
	0x00,0x20,0x10,0x03,0x03,0x0C,0x00,0x00,0x44,0x00,					
	0x00,0x33,0x00,0x47,0x0C,0x0E,0x00,0x0D,0x00,0x11,					
	0x00,0x00,0x00,0x02,0x64,0x00,0x00,0x00,0xAA,0x00,					
	0x00,0x00,0x64,0x10,0x00,0x00,0x00,0x00,0x00,0x00,					
	0x00,0x00,0x02,0x00,0x74,0x0F,0x41,0x00,0x00,0x00,					
	0x01,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x0A,0x00,					
	0x00,0x02,0x74,0x0F,0x41,0x00,0x06,0x08,0x00,0x00,					
	0x00,0x00,0x00,0x64,0x00,0x0F,0x00,0x00,0x0D,0x00,					
	0x00,0x00,0x23,0x00,0x00,0x64,0x00,0x00,0x64,0x00					
};

void Fun()
{
	int* p = (int*)data;
	printf("%x\n",*p);
	char* q = (char*)p;

	for(int i = 0;i < 96;i++)
	{
		if(*p == 0x64)
		{
			printf("%x\n",p);
		}
		q = q + 1;
		p = (int*)q;
	}
}

int main(int argc, char* argv[])
{
	Fun();
	return 0;
}

2、编写函数,返回角色名字信息的地址,如果没有返回0

#include "stdafx.h"
#include<stdio.h>
#include<string.h>

char data[100] = {
	0x02,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x07,0x09,					
	0x00,0x20,0x10,0x03,0x03,0x0C,0x00,0x00,0x44,0x00,					
	0x00,0x33,0x00,0x47,0x0C,0x0E,0x00,0x0D,0x00,0x11,					
	0x00,0x00,0x00,0x02,0x64,0x00,0x00,0x00,0xAA,0x00,					
	0x00,0x00,0x64,0x10,0x00,0x00,0x00,0x00,0x00,0x00,					
	0x00,0x00,0x02,0x00,0x74,0x0F,0x41,0x00,0x00,0x00,					
	0x01,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x0A,0x00,					
	0x00,0x02,0x74,0x0F,0x41,0x00,0x06,0x08,0x00,0x00,					
	0x00,0x00,0x00,0x64,0x00,0x0F,0x00,0x00,0x0D,0x00,					
	0x00,0x00,0x23,0x00,0x00,0x64,0x00,0x00,0x64,0x00					
};

char* FindRoleNameAddr(char* pData,char* pRoleName)
{
	for(int i=0;i<100;i++)
	{
		if(strcmp(pData,pRoleName))
		{
			pData++;
		}
		else
		{
			return pData;
		}
	}

	return 0;
}

int main(int argc, char* argv[])
{
	char* pRoleName = "WOW";
	char* pData = data;
	char* p = FindRoleNameAddr(pData,pRoleName);
	printf("%x\n",p);
}

3、查找这些数据中,有几个id=1 level=8的结构体信息

#include "stdafx.h"
#include<stdio.h>

char arr[] = {
	0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x07,0x09,					
	0x00,0x20,0x10,0x03,0x03,0x0C,0x00,0x00,0x44,0x00,					
	0x00,0x33,0x01,0x00,0x00,0x08,0x00,0x00,0x00,0x00,					
	0x00,0x00,0x00,0x02,0x64,0x00,0x00,0x00,0xAA,0x00,					
	0x00,0x00,0x64,0x01,0x00,0x00,0x00,0x08,0x00,0x00,					
	0x00,0x00,0x02,0x00,0x74,0x0F,0x41,0x00,0x00,0x00,					
	0x01,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x0A,0x00,					
	0x00,0x02,0x57,0x4F,0x57,0x00,0x06,0x08,0x00,0x00,					
	0x00,0x00,0x00,0x64,0x00,0x0F,0x00,0x00,0x0D,0x00,					
	0x00,0x00,0x23,0x00,0x00,0x64,0x00,0x00,0x64,0x00			
};

typedef struct TagPlayer
{
	int id;
	int level;
}Player;

void Fun()
{
	Player* p;
	p = (Player*)arr;
	char* q = (char*)p;

	for(int i=0;i<92;i++)
	{
		if(p->id == 0x1 && p->level == 0x8)
		{			
			printf("%x\n",p);
		}		
		q++;
		p = (Player*)q;
	}
}

int main(int argc, char* argv[])
{
	Fun()getchar();
	return 0;
}
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