实验四 用信号量解决进程互斥与同步问题

#include<stdio.h>

#include<stdlib.h>

#include<math.h>

#include<time.h>

#define N 100

 

 

typedef struct process

{

       int num;

}P;

 

P a[N];

int wordtime=0;

void main(){

int i=0,j,run=1,n,p=0,v=0,num;

printf("输入存放量的大小\n");

scanf("%d",&n);

while(run==1)

{

srand(time(NULL));

j=rand()%2+1;

if(1==j)

{

       if(i<5&&p==0&&v==0)

       {

              p++;

              v++;

              i++;

       printf("生产\n");

       p--;

              v--;

       }

else{

printf("仓库已满无法生产\n");

}     

      

}

else

{

if(i>0&&p==0&&v==0)

{

              p++;

              v++;

       i--;

       printf("消费\n");

              p--;

              v--;

}

else{

printf("无货物无法消费\n");

}     

 

}

printf("是否继续||1.是 2.否\n");

scanf("%d",&run);

 

}

}

转载于:https://www.cnblogs.com/garxiu/p/5603264.html