Question

c语言 结构体程序求修改

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

struct human
{
int blood;
bool state;//状态
int rate;
};

#define ID 1 //角色ID
#define ID1 2　//角色ID
#define ID2 3 //角色ID
#define ID3 4　　//角色ID

int num=0; //回合数
struct human one={400,0,50};
struct human two={400,0,50};
struct human three={400,0,50};
struct human four={400,0,50};

void show(human &h)
{
printf("第%d回合\n角色%d 体力%d %s 回复率%d%%\n",++num,ID,h.blood, h.state?"中毒":"正常",h.rate);
}
void show1(human &h1)
{
printf("\n角色%d 体力%d %s 回复率%d%%\n",ID1,h1.blood, h1.state?"中毒":"正常",h1.rate);
}
void show2(human &h2)
{
printf("\n角色%d 体力%d %s 回复率%d%%\n",ID2,h2.blood, h2.state?"中毒":"正常",h2.rate);
}
void show3(human &h3)
{
printf("\n角色%d 体力%d %s 回复率%d%%\n",ID3,h3.blood, h3.state?"中毒":"正常",h3.rate);
}

int main()
{
srand( (unsigned)time( NULL ) );
char ch;

show(one);
show1(two);
show2(three);
show3(four);

while(ch=getchar())
{
if(ch==10)
{
int r=rand();
if(r>16384)
{
one.state=1;
one.blood=one.blood-r/100;
if(one.blood<=0)
{
one.blood=0;

}
}
else one.state=0;
show(one);

if(r<16384)
{
two.state=1;
two.blood=two.blood-r/130;
if(two.blood<=0)
{
two.blood=0;

}
}
else two.state=0;
show1(two);

if(r>16384)
{
three.state=1;
three.blood=three.blood-r/105;
if(three.blood<=0)
{
three.blood=0;

}
}
else three.state=0;
show2(three);

if(r<16384)
{
four.state=1;
four.blood=four.blood-r/150;
if(four.blood<=0)
{
four.blood=0;
show3(four);
return 0;
}
}
else four.state=0;
show3(four);

}
}
return 0;
}
问题就一个
在上面程序输出格式不改变的情况下 状态 就是正常 和 中毒 采用函数指针形式来判 定

Answer 1

#include<time.h>
#include <stdio.h>
#include <stdlib.h>
#define ID 1
#define ID1 2
#define ID2 3
#define ID3 4
struct human
{
int blood;
int state;//状态
int rate;
};
int num=0; //回合数
struct human one={400,0,50};
struct human two={400,0,50};
struct human three={400,0,50};
struct human four={400,0,50};
void show(struct human h)
{
printf("第%d回合\n角色%d 体力%d %s 回复率%d%\n",++num,ID,h.blood,h.state?"正常":"中毒",h.rate);
}
void show1(struct human h1)
{
printf("\n角色%d 体力%d %s 回复率%d%%\n",ID1,h1.blood,h1.state?"正常":"中毒",h1.rate);
}
void show2( struct human h2)
{
printf("\n角色%d 体力%d %s 回复率%d%%\n",ID2,h2.blood, (h2.state ? "中毒":"正常"),h2.rate);
}
void show3(struct human h3)
{
printf("\n角色%d 体力%d %s 回复率%d%%\n",ID3,h3.blood,h3.state?"正常":"中毒",h3.rate);
}
int main()
{
char ch;
srand( (unsigned)time( NULL ) );
show(one);
show1(two);
show2(three);
show3(four);
while(ch=getchar())
{
if(ch==10)
{
int r=rand();
if(r>16384)
{
one.state=1;
one.blood=one.blood-r/100;
if(one.blood<=0)
{
one.blood=0;
}
}
else one.state=0;
show(one);
if(r<16384)
{
two.state=1;
two.blood=two.blood-r/130;
if(two.blood<=0)
{
two.blood=0;
}
}
else two.state=0;
show1(two);

if(r>16384)
{
three.state=1;
three.blood=three.blood-r/105;
if(three.blood<=0)
{
three.blood=0;

}
}
else three.state=0;
show2(three);

if(r<16384)
{
four.state=1;
four.blood=four.blood-r/150;
if(four.blood<=0)
{
four.blood=0;
show3(four);
return 0;
}
}
else four.state=0;
show3(four);

}
}
return 0;
}
你的程序我基本上看懂了，上面的代码能够正常执行
1.c语言不支持bool类型，所以定义state时，应该定义成int state
2.char ch,应该写在函数前面，因为c语言不支持变量的随定义随引用（即如果要使用一个变量，只需先写一定义变量的语句即可，并不需要在函数开头定义）
3.c语言不支持引用类型，引用是c++增加的特性
void show(human &h)
应该写成void show( struct human h)
4.较低版本的c语言不支持直接使用自定义的结构体类型定义变量，结构体类型前还需要加struct关键字 如human one是不行的，必须是struct human one

追问

大神 我想问的是 如果状态的判定用 指针来做怎么做好 请大神赐教

追答

我认为你的程序很不错了，用结构体就很好了
我不知道你为什么一定要用指针
如果想学指针的话，有很多的例子程序，你的这个并不是最好的例子...

追问

我程序有个问题如果4个角色其中一个采用r大于16384最后结果这个角色死亡时状态是正常

追答

if(three.blood<=0)
{
three.blood=0;

}
改成这样：if(three.blood<=0)
{
three.blood=0;
three.state=0;
}
其他的也都增加一句...
不知道你是不是这个意思？

Answer 2

指针木有初始化。。