利用兩個管道進行進程間雙向通信在第一篇練習已經大致作出說明，下面將進行一個更為綜合的練習

首先看題目：

設有二元函數f(x,y) = f(x) + f(y)
其中： f(x) = f(x-1) * x (x >1)
f(x)=1 (x=1)
f(y) = f(y-1) + f(y-2) (y> 2)
f(y)=1 (y=1,2)
請編程建立3 個并發協作進程，它們分別完成f(x,y)、f(x)、f(y)

實現的方法很多，這里只用管道實現，大致流程如下：

1.在父進程代碼中初始化四個管道，兩個用于父-子進程1，另外兩個父-子進程2；

2.父進程創建2個子進程，子進程1計算函數f(x)，子進程2計算函數f(y)；

3.父進程向子進程1、2發送數據；

4.子進程1、2均在管道里讀出數據，并進行計算；

5.計算完畢后，子進程1、2向父進程發送結果；

6。父進程接受數據，打印出來

?

下面是實現過程：

[cpp]?view plaincopy

/*pipe3.c*/??

#include?<unistd.h>??

#include?<sys/stat.h>??

#include?<sys/types.h>??

#include?<stdio.h>??

#include?<fcntl.h>??

#define?MAXLINE?1024??

#define?READ????0??

#define?WRITE???1??

/*函數x*/??

int?functionx(int?nx);??

/*函數y*/??

int?functiony(int?ny);??

main(void)??

{??

????pid_t?pid_x,pid_y;??

????int?fdx1[2],fdy1[2],fdx2[2],fdy2[2];??

????/*初始化管道*/??

????pipe(fdx1);??

????pipe(fdy1);??

????pipe(fdx2);??

????pipe(fdy2);??

????/*創建子進程1*/??

????pid_x?=?fork();??

????if(pid_x?<?0)??

????{??

????????printf("Create?process?error!/n");??

????????exit(0);??

????}??

????if(pid_x?==?0)??

????{??

????????int?numx,funx;??

????????printf("childx?process?ID:%d/n",getpid());??

????????close(fdx1[WRITE]);??

????????close(fdx2[READ]);??

????????/*從管道讀出x*/??

????????read(fdx1[READ],&numx,sizeof(int));??

????????/*函數計算*/??

????????funx?=?functionx(numx);??

????????printf("childx??x=%d/n",funx);??

????????/*向管道發送*/??

????????write(fdx2[WRITE],&funx,sizeof(int));??

????????close(fdx1[READ]);??

????????close(fdx2[WRITE]);??

????}??

????if(pid_x?>?0)??

????{??

????????/*創建子進程2*/??

????????pid_y?=fork();??

????????if(pid_y?<?0)??

????????{??

????????????printf("Create?process?error!/n");??

????????????exit(0);??

????????}??

????????if(pid_y?==?0)??

????????{??

????????????int?numy,funy;??

????????????printf("childy?process?ID:%d/n",getpid());??

????????????close(fdy1[WRITE]);??

????????????close(fdy2[READ]);??

????????????/*從管道讀出x*/??

????????????read(fdy1[READ],&numy,sizeof(int));??

????????????/*函數計算*/??

????????????funy?=?functiony(numy);??

????????????printf("childy??y=%d/n",funy);??

????????????/*向管道發送*/??

????????????write(fdy2[WRITE],&funy,sizeof(int));??

????????}??

????????if(pid_y?>?0)??

????????{?????

????????????int?x,y,funxy;??

????????????int?fx,fy;??

????????????sleep(1);??

????????????printf("parentxy?process?ID:%d/n%",getpid());??

????????????/*參數輸入*/??

????????????printf("enter?x,y/n");??

????????????scanf("%d,%d",&x,&y);??

????????????close(fdx2[WRITE]);??

????????????close(fdx1[READ]);??

????????????close(fdy2[WRITE]);??

????????????close(fdy1[READ]);??

????????????/*管道發送*/??

????????????write(fdx1[WRITE],&x,sizeof(int));??

????????????write(fdy1[WRITE],&y,sizeof(int));??

????????????/*等待子進程計算*/??

????????????sleep(1);??

????????????/*管道讀入*/??

????????????read(fdx2[READ],&fx,sizeof(int));??

????????????read(fdy2[READ],&fy,sizeof(int));??

????????????funxy?=?fx+fy;??

????????????printf("f(x)?=?%d/nf(y)?=?%d/nfun(x,y)?=?%d",fx,fy,funxy);??

????????????waitpid(pid_x,NULL,0);??

????????????waitpid(pid_y,NULL,0);??

????????}??

????}?????

}??

int?functionx(int?nx)??

{??

????int?sum?=?1;??

????int?i?=?1;??

????if(nx?<?0)??

????{??

????????printf("errorx!/n");??

????????exit(0);??

????}??

????while(i?<=?nx)??

????{??

????????sum?*=i++;??

????}??

????return?sum;??

}??

int?functiony(int?ny)??

{??

????int?f1=1,f2=1,f3;??

????int?i?=3;??

????if(ny?<=?2)??

????????return?f1;??

????while(i?<=?ny)??

????{??

????????f3?=?f1+f2;??

????????f1?=?f2;??

????????f2?=?f3;??

????????i++;??

????}??

????return?f3;??

}??

?

編譯：

$ gcc pipe3.c -o pipe3

運行：

$ ./pipe

?

我們輸入測試數據x=3,y=4

輸出為f(x)=6,f(y)=3,f(x,y)=9

成功

總結

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