C中的fork（）、pipe（）、dup2（）和exec（）有问题
这是我的密码：C中的fork（）、pipe（）、dup2（）和exec（）有问题,c,exec,fork,pipe,dup2,C,Exec,Fork,Pipe,Dup2,这是我的密码： #include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <wait.h> #include <readline/readline.h> #define NUMPIPES 2 int main(int argc, char *argv[]) { char *bBuffer, *sPtr, *aPtr = NULL, *pipeCom
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <wait.h>
#include <readline/readline.h>
#define NUMPIPES 2
int main(int argc, char *argv[]) {
char *bBuffer, *sPtr, *aPtr = NULL, *pipeComms[NUMPIPES], *cmdArgs[10];
int fdPipe[2], pCount, aCount, i, status, lPids[NUMPIPES];
pid_t pid;
pipe(fdPipe);
while(1) {
bBuffer = readline("Shell> ");
if(!strcasecmp(bBuffer, "exit")) {
return 0;
}
sPtr = bBuffer;
pCount = -1;
do {
aPtr = strsep(&sPtr, "|");
pipeComms[++pCount] = aPtr;
} while(aPtr);
for(i = 0; i < pCount; i++) {
aCount = -1;
do {
aPtr = strsep(&pipeComms[i], " ");
cmdArgs[++aCount] = aPtr;
} while(aPtr);
cmdArgs[aCount] = 0;
if(strlen(cmdArgs[0]) > 0) {
pid = fork();
if(pid == 0) {
if(i == 0) {
close(fdPipe[0]);
dup2(fdPipe[1], STDOUT_FILENO);
close(fdPipe[1]);
} else if(i == 1) {
close(fdPipe[1]);
dup2(fdPipe[0], STDIN_FILENO);
close(fdPipe[0]);
}
execvp(cmdArgs[0], cmdArgs);
exit(1);
} else {
lPids[i] = pid;
/*waitpid(pid, &status, 0);
if(WIFEXITED(status)) {
printf("[%d] TERMINATED (Status: %d)\n",
pid, WEXITSTATUS(status));
}*/
}
}
}
for(i = 0; i < pCount; i++) {
waitpid(lPids[i], &status, 0);
if(WIFEXITED(status)) {
printf("[%d] TERMINATED (Status: %d)\n",
lPids[i], WEXITSTATUS(status));
}
}
}
return 0;
}
nazgulled ~/Projects/SO/G08 $ ./a.out
Shell> echo q|sudo fdisk /dev/sda
[4838] TERMINATED (Status: 0)
The number of cylinders for this disk is set to 1305.
There is nothing wrong with that, but this is larger than 1024,
and could in certain setups cause problems with:
1) software that runs at boot time (e.g., old versions of LILO)
2) booting and partitioning software from other OSs
(e.g., DOS FDISK, OS/2 FDISK)
Command (m for help):
[4839] TERMINATED (Status: 0)
那么，我的代码怎么了？我认为您的分叉进程将继续执行 尝试以下任一方法：
- 将其更改为“return execvp”
- 添加“出口（1）；”执行后
- 一个潜在的问题是cmdargs的末尾可能有垃圾。在将数组传递给execvp（）之前，应该使用空指针终止该数组
不过，看起来grep正在接受STDIN，因此这可能不会导致任何问题。在execvp（）之后应该会有一个错误退出-它有时会失败
exit(EXIT_FAILURE);
cmdArgs[aCount] = 0;
我不清楚您是否让两个程序都免费运行—您似乎要求管道中的第一个程序在启动第二个程序之前完成，如果第一个程序因管道已满而阻塞，这不是成功的秘诀。乔纳森的想法是正确的。您依赖第一个进程来分叉所有其他进程。每一个都必须运行到完成，然后才能分叉下一个 相反，像您正在做的那样在循环中分叉进程，但是在内部循环之外等待它们（在大循环的底部等待shell提示）
即使在管道的第一个命令退出（并且thust关闭
stdout=~fdPipe[1]fdPipe[1]stdin=~fdPipe[0]|管道（fdPipe）for cmd in cmds
if there is a next cmd
pipe(new_fds)
fork
if child
if there is a previous cmd
dup2(old_fds[0], 0)
close(old_fds[0])
close(old_fds[1])
if there is a next cmd
close(new_fds[0])
dup2(new_fds[1], 1)
close(new_fds[1])
exec cmd || die
else
if there is a previous cmd
close(old_fds[0])
close(old_fds[1])
if there is a next cmd
old_fds = new_fds
if there are multiple cmds
close(old_fds[0])
close(old_fds[1])
closedup2fdPipe{STDIN\u FILENO，STDOUT\u FILENO}fdPipe1fdPipe2pipe（）exec*（）exit（1）exec*（）loop //for prompt
next prompt
loop //to fork tasks, store the pids
if pid == 0 run command
else store the pid
end loop
loop // on pids
wait
end loop
end loop
for cmd in cmds
if there is a next cmd
pipe(new_fds)
fork
if child
if there is a previous cmd
dup2(old_fds[0], 0)
close(old_fds[0])
close(old_fds[1])
if there is a next cmd
close(new_fds[0])
dup2(new_fds[1], 1)
close(new_fds[1])
exec cmd || die
else
if there is a previous cmd
close(old_fds[0])
close(old_fds[1])
if there is a next cmd
old_fds = new_fds
if there are multiple cmds
close(old_fds[0])
close(old_fds[1])
fdPipe1 fdPipe3
v v
cmd1 | cmd2 | cmd3 | cmd4 | cmd5
^ ^
fdPipe2 fdPipe4
/* suppose stdin and stdout have been closed...
* for example, if your program was started with "./a.out <&- >&-" */
close(0), close(1);
/* then the result you get back from pipe() is {0, 1} or {1, 0}, since
* fd numbers are always allocated from the lowest available */
pipe(fdPipe);
close(0);
dup2(fdPipe[0], 0);