0

所以我们必须创建一个程序来创建 3 个子进程,它们相互传递信息。

  • 第一个进程从文件中读取数据。
  • 第二个过程通过转换每个字母的大写来更改数据(即小写到大写,反之亦然)。
  • 第三个进程将信息写入文件。

我的问题是我似乎无法通过使用fork()and来运行进程waitpid()。因为waitpid(),老实说,我不知道第二个和第三个参数属于什么。这是我的主要内容:

#include <windows.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <sys/types.h>
#include <sys/wait.h>
#include "pipe.c"


int main(void) {

int p, p2, rc, rc2;
rc = _pipe(f_des);
rc2 = _pipe(C1C2);

pid_t pid[3];

pid[0] = fork();
if (pid[0] == 0) {
    puts("create child 1");
    do_child0();
    exit(0);
}
pid[1] = fork();
if (pid[1] == 0) {
    puts("create child 2");
    do_child1();
    exit(0);
}
pid[2] = fork();
if (pid[2] == 0) {
    puts("create child 3");
    do_child2();
    exit(0);
}

waitpid(pid[0], NULL, 0);
waitpid(pid[1], NULL, 0);
waitpid(pid[2], NULL, -1); 
printf("children done, parent goes");

return 0;
}

管道.c:

#include "pipe.h"

void do_child1() {

int c, p2;
int rc, rc2;
char *line2, *line3;
int i = 0;
int j = 0;
int k = 0;
char buffer[80];
//char *line;

close(f_des[1]);
close(C1C2[0]);
printf("Received by child 2 from child 1: ");   

while(k<26) {
for (i = 0; i <= 40; i++) {
    if ((read(f_des[0], buffer, 1)) != '\0') {
//while((rc = read(f_des[0], buffer, 40))) {
    printf("%s", buffer);
    j = 0;
    line2 = (char *)malloc(sizeof(char)*40);
    read(f_des[0], buffer, 40);
    line2 = buffer;
    line3 = (char *)malloc(sizeof(char)*40);
    line2 = line3; 

    rc = read(f_des[0], line2, 40);
    printf("%s\n",line2);
    while(j < 40) {
        if(islower(*(line3+j)))
            *(line3+j) = toupper(*(line3+j));
        if(isupper(*(line2+j)))
            *(line3+j) = tolower(*(line3+j));
        }
    rc2 = write(C1C2[1], line3, 40);
    printf("\nLine 3: %s", line3);

    k++;
}
}
}

printf("\n");
close(C1C2[1]);
close(f_des[0]);
exit(0);
}

void do_child2() {
int c;
int rc;
char *line4;

close(C1C2[1]);

while((rc= read(C1C2[0], line4, 40)>0)) {
    line4 = (char*)malloc(sizeof(char)*40);
    printf("%s\n", line4);
}

FILE *file;
file = fopen("C:\\Users\\apkim_000\\Desktop\\newReluctance.c", "a+");
fprintf(file, "%s", line4);
fclose(file);




}



void do_child0() {
FILE *inf = fopen("C:\\MinGW\\bin\\Reluctance.c", "r");
int c;
int rc; 
int i =0;
char buffer[80];
char *line;
cntr = 0;
close(f_des[0]);

printf("From child 1: \n");
while (1) {
    fgets(buffer, 100, inf); //reads in at most 80 char from a line
    if (feof(inf)) //this checks to see if the special EOF was read
        break;     //if so, break out of while and continue with your main
    line = (char*)malloc(sizeof(char)*40);
    line = strtok(buffer, "\n");
    printf("%s\n", line);
    i++;
    cntr++;
    write(f_des[1], buffer, 40);
}
close(f_des[1]);

exit(0);

}

管道.h:

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


void do_child1();
void do_child2();
void do_child0();

int f_des[2];
int C1C2[2];
int cntr, p3;

编译器输出:

create child 1
From child 1:
Reluctance
OUT through the fields and the woods
And over the walls I have wended;
I have climbed the hills of view
And looked at the world, and descended;
I have come by the highway home, 
And lo, it is ended.
The leaves are all dead on the ground,
Save those that the oak is keeping
To ravel them one by one
And let them go scraping and creeping
Out over the crusted snow,
When others are sleeping.
And the dead leaves lie huddled and still,
No longer blown hither and thither;
The last lone aster is gone;
The flowers of the witch-hazel wither;
The heart is still aching to seek,
But the feet question 'Whither?'
Ah, when to the heart of man
Was it ever less than a treason
To go with the drift of things,
To yield with a grace to reason,
And bow and accept the end
create child 2
Received by child 2 from child 1: RUT through the fields and the woods
create child 3
4

2 回答 2

0

我看到的第一件事,但我不太自信,因为它的密集代码是你应该只等待执行时间最长的孩子。如果您不知道将是哪个子进程,最好的办法是等待一个 fifo 或让每个子进程创建下一个子进程,这样父进程只需等待一个子进程。

是的,让每个孩子做它的工作,等待那个孩子,当它退出时,主程序然后开始创建下一个孩子。没有理由让他们同时这样做。

于 2013-05-10T20:16:36.670 回答
0

在担心之前waitpid(),你应该照顾好do_child1()do_child0()和之间的管道do_child1()可以工作,但是其中的行缓冲区管理do_child1()是一团糟。do_child0()总是每行写入 40 个字符,而do_child1()首先只读取 1 个字符(“R”),然后读取 40 个字符(它丢弃),然后再读取 40 个字符(从第 2 行的第 2 个字符开始),产生输出“RUT穿过田野和树林”。while(j < 40)然后进入一个无限循环;因此,程序不会继续。

于 2014-07-04T11:57:12.160 回答