Question

所以我们必须创建一个程序来创建3个子进程，这些进程可以相互传递信息。

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上的大写字母来改变该数据（即小写字母为大写，反之亦然）。
第三个进程将信息写入文件。

我的问题是，我似乎无法使用fork()和waitpid()来运行流程。对于waitpid()，老实说，我不知道第二和第三个args中属于什么。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;
}

pipe.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);

}

pipe.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

Answer 1

我看到的第一件事，但是我并不太自信，因为它密集的代码，你应该只等待执行时间最长的孩子。如果您不知道将要进行哪个子进程，最好的办法是等待一个fifo或让每个孩子创建下一个孩子，这样父母只需要等待一个孩子。

是的，让每个孩子都做好自己的工作，然后等待那个孩子，当它退出时，主程序开始创建下一个孩子。没有理由让他们同时做这件事。

Answer 2

在担心waitpid()之前，您应该照顾do_child1()。 do_child0()和do_child1()之间的管道有效，但do_child1()中的行缓冲管理是一团糟。 do_child0()总是每行写40个字符，而do_child1()首先只读取1个字符（“R”），然后是40个字符（它丢弃），然后是另外40个字符（从第2个字符开始）第二行），产生输出“RUT穿过田野和树林”。然后输入无限循环while(j < 40);因此，该计划不会继续。

管道上的这个任务可以帮助我吗？

2 个答案: