用C制作外壳时正确的管道连接方法是什么 [英] what is the proper way to pipe when making a shell in C
问题描述
我正在尝试创建自己的外壳，我相信我已经正确完成了分叉，但是我无法弄清楚如何正确地进行管道连接.任何帮助或提示将不胜感激.
基本上，我的管道无法正常工作，我花了很长时间尝试找出如何使它们正确地在进程之间传输数据.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <unistd.h>
#include "ourhdr.h" // from Steven's book Advanced programing in the UNIX Enviroment
extern int makeargv(char *, char * , char ***);
int main()
{ char **argp;
int i,j,vpret;
char buf[80];
pid_t pid;
int pnum;
//pipe number
int ploc[16];
//pipe location
//(this has a max of 12 possible pipes)
int targ;
int fdleft[2], fdright[2];
printf(" <(^_^)> \nHello \n I am your console and I am here to help you \n");
printf(" If you dont need me anymore just say \"bye\" ");
fflush(stdout);
write(1,"\n(>^_^)> ",8);
while(strcmp(fgets(buf, 80, stdin), "bye\n")!=0)
{
j=makeargv(buf," \n",&argp); //this breaks the line up and returns the number of commands
pnum = 0;
ploc[0] = 0;
if (j > 16) j = 16;
for (i=0;i<j;i++)
{
if ( strcmp(argp[i], "|") == 0)
{
argp[i]= NULL;
ploc[pnum+1] = (i+1);
pnum++;
}
}
for (i = 0; i < (pnum+1); i++)
{
pipe(fdright);
if (i != 0)
{
dup2(fdright[1], fdleft[0]);
}
pid = fork();
switch (pid)
{
case -1:
err_sys("fork failed");
break;
case 0: // child
if (i != pnum)
{
dup2(fdright[1],1);
}
if ( i != 0);
{
dup2(fdright[0],0);
}
//printf("(^o^) running pipe[%i]\n" , i);
targ =(ploc[i]) ;
execvp(argp[targ],&argp[targ]);
write(1,"(-_-) I'm sorry the exec failed \n",33);
exit(1);
default:
dup2(fdleft[1], fdright[1]);
waitpid(pid,NULL, 0);
close(fdright[0]);
close(fdright[1]);
//waitpid(pid,NULL, 0);
}
}
//waitpid(pid, NULL, 0);
write(1,"\n(>^_^)> ",8);
}
printf(" v(^o^)^ BYE BYE!\n");
}
谢谢
各种评论:
-
while (strcmp(fgets(buf, 80, stdin), "bye\n")!=0)如果给外壳提供EOF而不是
bye，则崩溃.不要像这样组合两个函数调用.如果您希望所有内容都处于一个循环条件下，请使用:while (fgets(buf, sizeof(buf), stdin) != 0 && strcmp(buf, "bye\n") != 0)我们将再次讨论命令行长度的限制.
-
由于您没有提供
makeargv()进行查看，因此我们必须假定它可以正常工作. -
将内容分解为命令和管道的循环是:
pnum = 0; ploc[0] = 0; if (j > 16) j = 16; for (i = 0; i < j; i++) { if (strcmp(argp[i], "|") == 0) { argp[i] = NULL; ploc[pnum+1] = (i+1); pnum++; } }假设我们有一个命令行输入:
ls -l | grep lemon.看来您的makeargv()将返回5并按如下所示设置argp:argp[0] = "ls"; argp[1] = "-l"; argp[2] = "|"; argp[3] = "grep"; argp[4] = "lemon"; argp[5] = 0; // Inferred - things will crash sooner or later if wrong您的这个循环会给您:
ploc[0] = 0; ploc[1] = 3; pnum = 1; -
您的代码在
fdleft数组中有一对文件描述符，但是即使在对dup2()的调用中使用它，也永远不会初始化该数组(例如，调用pipe()). /p> -
然后，主
for循环必须运行两次，每个命令一次.对于管道中三个或三个以上命令的一般情况(例如，who | grep me | sort)，您的第一个命令(who)需要其标准输入保持不变，但其标准输出将进入连接who和grep. 中间"命令(第二至第二个命令，在示例中为grep me)每个都需要其标准输入来自上一个管道，并且需要创建一个新管道以供其标准输出使用.最后一个命令(在本示例中，第三个命令sort)需要其标准输入来自最后一个管道，并且其标准输出不变.您的代码无法执行此操作，或者使您无法找到任何地方.
-
使用
pipe()然后使用dup()或dup2()将任何描述符映射到标准I/O描述符时，需要关闭管道的 两端.您根本就没有足够的close()呼叫. -
您的父进程必须依次启动每个子进程，并且仅在所有子进程启动后等待它们退出.有多种组织流程的方式.父母可以分叉一次；孩子可能负责在管道中启动领先的命令，最后执行最后一个命令本身.父级只有一个直子(其他直孙)，因此只需要等待一个命令完成即可.另一种选择是，父级知道管道中的每个进程，并等待它们全部完成.
-
如果父进程在启动其余进程之前等待管道中的每个命令完成，则可能会导致死锁.一个孩子向其管道中写入了太多数据，以致它被内核阻止，直到某个进程从管道中读取数据为止，但尚未启动从管道中读取的过程，而父级正在等待该孩子退出.您还将失去多处理和并发的好处.
-
在案例0"中，您有一个多余的分号.我的编译器就此警告过我.如果您没有警告您，则需要使用更多的编译警告或获得更好的编译器.
case 0: // child if (i != pnum) { dup2(fdright[1], 1); } if (i != 0); // Unwanted semi-colon! { dup2(fdright[0], 0); } -
关于SO上的微型外壳中的管道有很多问题，包括:
对于13636252的回答几乎是通用的.唯一的问题是它使用
char ***，这很容易造成混淆.紧密地编写，它具有相互递归的功能，并且重复次数最少. OTOH，它可以正常工作，并且您的makeargv()函数还使用了char ***参数.
重做的代码
在这里您的代码经过了重新设计，使其可以正常工作.它包括err_sys()和makeargv()的实现.我的makeargv()只是假设命令行中少于32个单词.从我的所有想象中来看，它不是一个强大的命令行解析器.它确实允许您键入ls | wc并给出正确的答案；它还允许who | grep me | sort并给出正确答案；它还允许ls并给出正确的答案.管道符号周围的空格至关重要，不过(在普通外壳中，它们是可选的，因此who|grep me|sort也应适用，但此代码将不起作用.
#include <assert.h>
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/stat.h>
#include <sys/wait.h>
#include <unistd.h>
extern int makeargv(char *line, char *seps, char ***args);
extern void err_sys(const char *msg);
static void dump_argv(char **argv);
static void dump_fds(void);
int main(void)
{
char buf[80];
printf(" <(^_^)> \nHello \n I am your console and I am here to help you\n");
printf(" If you don't need me anymore just say \"bye\"\n");
fflush(stdout);
dump_fds();
printf("(>^_^)> ");
while (fgets(buf, sizeof(buf), stdin) != 0 && strcmp(buf, "bye\n") != 0)
{
pid_t pid;
char **argp;
int fdleft[2] = { -1, -1 };
int fdright[2] = { -1, -1 };
int pnum = 0;
int ploc[16];
int j = makeargv(buf, " \n", &argp);
ploc[0] = 0;
if (j > 16)
j = 16;
for (int i = 0; i < j; i++)
{
if (strcmp(argp[i], "|") == 0)
{
argp[i] = NULL;
ploc[++pnum] = i+1;
}
}
printf("pnum = %d\n", pnum);
for (int k = 0; k < pnum+1; k++)
printf("ploc[%d] = %d\n", k, ploc[k]);
for (int i = 0; i < pnum+1; i++)
{
if (i != pnum)
{
if (pnum > 0)
{
if (pipe(fdright) != 0)
err_sys("pipe");
//printf("%d: fdright = { %d, %d }\n", i, fdright[0], fdright[1]);
//dump_fds();
}
}
if ((pid = fork()) < 0)
err_sys("fork failed");
else if (pid == 0)
{
/* Child */
int targ;
//dump_fds();
if (i != pnum)
{
dup2(fdright[1], 1);
close(fdright[0]);
close(fdright[1]);
}
if (i != 0)
{
dup2(fdleft[0], 0);
close(fdleft[0]);
close(fdleft[1]);
}
targ = ploc[i];
dump_argv(&argp[targ]);
dump_fds();
execvp(argp[targ], &argp[targ]);
fprintf(stderr, "(-_-) I'm sorry the exec failed\n");
exit(1);
}
if (i != 0)
{
//dump_fds();
//printf("%d: fdleft = { %d, %d }\n", i, fdleft[0], fdleft[1]);
assert(fdleft[0] != -1 && fdleft[1] != -1);
close(fdleft[0]);
close(fdleft[1]);
//dump_fds();
}
printf("PID %d launched\n", pid);
fdleft[0] = fdright[0];
fdleft[1] = fdright[1];
}
//dump_fds();
//printf("%d: fdleft = { %d, %d }\n", -1, fdleft[0], fdleft[1]);
close(fdleft[0]);
close(fdleft[1]);
free(argp);
//dump_fds();
int corpse;
int status;
while ((corpse = waitpid(0, &status, 0)) > 0)
printf(":-( PID %d status 0x%.4X\n", corpse, status);
printf("\n(>^_^)> ");
}
printf(" v(^o^)^ BYE BYE!\n");
}
static void dump_argv(char **argv)
{
int n = 0;
char **args;
args = argv;
while (*args++ != 0)
n++;
fprintf(stderr, "%d: %d args\n", getpid(), n);
args = argv;
while (*args != 0)
fprintf(stderr, "[%s]\n", *args++);
fprintf(stderr, "EOA\n");
}
/* Report on open file descriptors (0..19) in process */
static void dump_fds(void)
{
struct stat b;
char buffer[32];
sprintf(buffer, "%d: ", getpid());
char *str = buffer + strlen(buffer);
for (int i = 0; i < 20; i++)
*str++ = (fstat(i, &b) == 0) ? 'o' : '-';
*str++ = '\n';
*str = '\0';
fputs(buffer, stderr);
}
int makeargv(char *line, char *seps, char ***args)
{
enum { MAX_ARGS = 32 };
char **argv = malloc(32 * sizeof(char *)); // Lazy!
if (argv == 0)
err_sys("out of memory in makeargv()");
int n;
char **argp = argv;
char *str = line;
for (n = 0; n < MAX_ARGS - 1; n++)
{
str += strspn(str, seps);
if (*str == '\0')
break;
*argp++ = str;
int len = strcspn(str, seps);
if (len == 0)
break;
str[len] = '\0';
str += len + 1;
}
*argp = 0;
dump_argv(argv);
*args = argv;
return(n);
}
void err_sys(const char *msg)
{
int errnum = errno;
char *errmsg = strerror(errnum);
fprintf(stderr, "%s (%d: %s)\n", msg, errnum, errmsg);
exit(1);
}
示例输出:
$ ./pipes-15673333
<(^_^)>
Hello
I am your console and I am here to help you
If you don't need me anymore just say "bye"
29191: ooo-----------------
(>^_^)> who | grep jl | sort
29191: 6 args
[who]
[|]
[grep]
[jl]
[|]
[sort]
EOA
pnum = 2
ploc[0] = 0
ploc[1] = 2
ploc[2] = 5
PID 29194 launched
PID 29195 launched
29194: 1 args
[who]
EOA
PID 29196 launched
29194: ooo-----------------
29195: 2 args
[grep]
[jl]
EOA
29195: ooo-----------------
29196: 1 args
[sort]
EOA
29196: ooo-----------------
:-( PID 29194 status 0x0000
jleffler console Mar 27 15:11
jleffler ttys000 Mar 27 16:26
jleffler ttys001 Mar 27 16:26
jleffler ttys002 Mar 27 16:26
jleffler ttys003 Mar 27 16:26
jleffler ttys004 Mar 27 16:26
jleffler ttys005 Mar 27 16:26
:-( PID 29195 status 0x0000
:-( PID 29196 status 0x0000
(>^_^)> ls
29191: 1 args
[ls]
EOA
pnum = 0
ploc[0] = 0
PID 29197 launched
29197: 1 args
[ls]
EOA
29197: ooo-----------------
bash.getopts.update makefile pipeline.c pthread-1.c shuntzeroes.c timezeromoves.c
cmpfltint.c mda.c pipeline.dSYM pthread-2.c so.14304827 uint128.c
const-stuff.c mq-saurabh pipes-13905948.c pthread-3.c so.367309 uname.c
dupdata.sql mqp-saurabh pipes-14312939.c quine.c so.6964747 unwrap.c
fifocircle.c multi-pipe-sort.c pipes-15673333 ranges.sql so.6965001 xxx.sql
idsdb00246324.ec multiopts.sh pipes-15673333.c recv.c so.8854855.sql yyy.sql
incunabulum.c nextpipe.c pipes-15673333.dSYM regress.c strandsort.c
madump.c pipeline powa.c send.c streplace.c
:-( PID 29197 status 0x0000
(>^_^)> ls -C | wc
29191: 4 args
[ls]
[-C]
[|]
[wc]
EOA
pnum = 1
ploc[0] = 0
ploc[1] = 3
PID 29200 launched
PID 29201 launched
29200: 2 args
29201: 1 args
[ls]
[wc]
[-C]
EOA
EOA
29201: ooo-----------------
29200: ooo-----------------
:-( PID 29200 status 0x0000
16 46 581
:-( PID 29201 status 0x0000
(>^_^)> bye
v(^o^)^ BYE BYE!
$
I’m attempting to create my own shell I believe i have the forking done correctly but i cannot figure out how to pipe correctly. Any help or tips would be appreciated.
basically my pipes aren’t working and i have spent ages attempting to figure out how to get them to properly transmit the data between the processes.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <unistd.h>
#include "ourhdr.h" // from Steven's book Advanced programing in the UNIX Enviroment
extern int makeargv(char *, char * , char ***);
int main()
{ char **argp;
int i,j,vpret;
char buf[80];
pid_t pid;
int pnum;
//pipe number
int ploc[16];
//pipe location
//(this has a max of 12 possible pipes)
int targ;
int fdleft[2], fdright[2];
printf(" <(^_^)> \nHello \n I am your console and I am here to help you \n");
printf(" If you dont need me anymore just say \"bye\" ");
fflush(stdout);
write(1,"\n(>^_^)> ",8);
while(strcmp(fgets(buf, 80, stdin), "bye\n")!=0)
{
j=makeargv(buf," \n",&argp); //this breaks the line up and returns the number of commands
pnum = 0;
ploc[0] = 0;
if (j > 16) j = 16;
for (i=0;i<j;i++)
{
if ( strcmp(argp[i], "|") == 0)
{
argp[i]= NULL;
ploc[pnum+1] = (i+1);
pnum++;
}
}
for (i = 0; i < (pnum+1); i++)
{
pipe(fdright);
if (i != 0)
{
dup2(fdright[1], fdleft[0]);
}
pid = fork();
switch (pid)
{
case -1:
err_sys("fork failed");
break;
case 0: // child
if (i != pnum)
{
dup2(fdright[1],1);
}
if ( i != 0);
{
dup2(fdright[0],0);
}
//printf("(^o^) running pipe[%i]\n" , i);
targ =(ploc[i]) ;
execvp(argp[targ],&argp[targ]);
write(1,"(-_-) I'm sorry the exec failed \n",33);
exit(1);
default:
dup2(fdleft[1], fdright[1]);
waitpid(pid,NULL, 0);
close(fdright[0]);
close(fdright[1]);
//waitpid(pid,NULL, 0);
}
}
//waitpid(pid, NULL, 0);
write(1,"\n(>^_^)> ",8);
}
printf(" v(^o^)^ BYE BYE!\n");
}
thanks
Various comments:
while (strcmp(fgets(buf, 80, stdin), "bye\n")!=0)This crashes if the shell is given EOF instead of
bye. Don't combine the two function calls like that. If you want it all in one loop condition, then use:while (fgets(buf, sizeof(buf), stdin) != 0 && strcmp(buf, "bye\n") != 0)We'll discuss limits on the length of the command line another time.
Since you've not supplied
makeargv()to look at, we have to assume it does its job OK.Your loop that splits things up into commands and pipes is:
pnum = 0; ploc[0] = 0; if (j > 16) j = 16; for (i = 0; i < j; i++) { if (strcmp(argp[i], "|") == 0) { argp[i] = NULL; ploc[pnum+1] = (i+1); pnum++; } }Let's assume we have a command line input:
ls -l | grep lemon. It appears that yourmakeargv()would return 5 and setargpas follows:argp[0] = "ls"; argp[1] = "-l"; argp[2] = "|"; argp[3] = "grep"; argp[4] = "lemon"; argp[5] = 0; // Inferred - things will crash sooner or later if wrongThis loop of yours would give you:
ploc[0] = 0; ploc[1] = 3; pnum = 1;Your code has a pair of file descriptors in the
fdleftarray, but you never initialize the array (callpipe(), for example), even though you use it in calls todup2().The main
forloop then has to run twice, once for each command. For the general case of three or more commands in a pipeline (who | grep me | sort, for example), your first command (who) needs its standard input untouched, but its standard output going to the pipe that connectswhoandgrep. The 'middle' commands (second to penultimate commands, orgrep mein the example) each needs its standard input to come from the previous pipe and needs to create a new pipe for its standard output to go to. The last command (in this example, the third command,sort), needs its standard input to come from the last pipe and its standard output unchanged.Your code doesn't do that, or get anywhere close.
When you use
pipe()and thendup()ordup2()to map any of the descriptors to standard I/O descriptors, you need to close both ends of the pipe. You simply don't have enoughclose()calls around either.Your parent process has to launch each of the children in turn, and only waits for them to exit after it has launched them all. There are different ways of organizing the processes. The parent could fork once; the child could be responsible for launching the leading commands in the pipeline, finally executing the last command itself. The parent has just one direct child (the others are grandchildren), so it only has to wait for the one command to finish. The alternative is that the parent knows about each of the processes in the pipeline and it waits for them all to finish.
If your parent processes waits for each command in a pipeline to complete before launching the remainder, you can end up with a form of deadlock. One child writes so much data to its pipe that it is blocked by the kernel until some process reads from the pipe, but the process to read from the pipe is not yet launched and the parent is waiting for the child to exit. You also lose the benefits of multi-processing and concurrency.
In your 'case 0', you have an extraneous semi-colon. My compiler warned me about it. If yours didn't warn you about it, you need to use more compilation warnings or get a better compiler.
case 0: // child if (i != pnum) { dup2(fdright[1], 1); } if (i != 0); // Unwanted semi-colon! { dup2(fdright[0], 0); }There are many questions about pipe lines in mini-shells on SO, including:
The answer for 13636252 is most nearly generic. The only snag is it uses
char ***which is apt to be confusing, and it is so tightly written it has mutually recursive functions with minimal repetition. OTOH, it works correctly, and yourmakeargv()function also uses achar ***argument.
Reworked code
Here is your code reworked so that it works. It includes implementations of err_sys() and makeargv(). My makeargv() simply assumes that there will be fewer than 32 words in the command line. It is not, by any stretch of my imagination, a robust command line parser. It does allow you to type ls | wc and gives the correct answer; it also allows who | grep me | sort and gives the correct answer; it also allows ls and gives the correct answer. The spaces around the pipe symbols are crucial, though (in a normal shell, they're optional, so who|grep me|sort should work too, but it won't with this code.
#include <assert.h>
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/stat.h>
#include <sys/wait.h>
#include <unistd.h>
extern int makeargv(char *line, char *seps, char ***args);
extern void err_sys(const char *msg);
static void dump_argv(char **argv);
static void dump_fds(void);
int main(void)
{
char buf[80];
printf(" <(^_^)> \nHello \n I am your console and I am here to help you\n");
printf(" If you don't need me anymore just say \"bye\"\n");
fflush(stdout);
dump_fds();
printf("(>^_^)> ");
while (fgets(buf, sizeof(buf), stdin) != 0 && strcmp(buf, "bye\n") != 0)
{
pid_t pid;
char **argp;
int fdleft[2] = { -1, -1 };
int fdright[2] = { -1, -1 };
int pnum = 0;
int ploc[16];
int j = makeargv(buf, " \n", &argp);
ploc[0] = 0;
if (j > 16)
j = 16;
for (int i = 0; i < j; i++)
{
if (strcmp(argp[i], "|") == 0)
{
argp[i] = NULL;
ploc[++pnum] = i+1;
}
}
printf("pnum = %d\n", pnum);
for (int k = 0; k < pnum+1; k++)
printf("ploc[%d] = %d\n", k, ploc[k]);
for (int i = 0; i < pnum+1; i++)
{
if (i != pnum)
{
if (pnum > 0)
{
if (pipe(fdright) != 0)
err_sys("pipe");
//printf("%d: fdright = { %d, %d }\n", i, fdright[0], fdright[1]);
//dump_fds();
}
}
if ((pid = fork()) < 0)
err_sys("fork failed");
else if (pid == 0)
{
/* Child */
int targ;
//dump_fds();
if (i != pnum)
{
dup2(fdright[1], 1);
close(fdright[0]);
close(fdright[1]);
}
if (i != 0)
{
dup2(fdleft[0], 0);
close(fdleft[0]);
close(fdleft[1]);
}
targ = ploc[i];
dump_argv(&argp[targ]);
dump_fds();
execvp(argp[targ], &argp[targ]);
fprintf(stderr, "(-_-) I'm sorry the exec failed\n");
exit(1);
}
if (i != 0)
{
//dump_fds();
//printf("%d: fdleft = { %d, %d }\n", i, fdleft[0], fdleft[1]);
assert(fdleft[0] != -1 && fdleft[1] != -1);
close(fdleft[0]);
close(fdleft[1]);
//dump_fds();
}
printf("PID %d launched\n", pid);
fdleft[0] = fdright[0];
fdleft[1] = fdright[1];
}
//dump_fds();
//printf("%d: fdleft = { %d, %d }\n", -1, fdleft[0], fdleft[1]);
close(fdleft[0]);
close(fdleft[1]);
free(argp);
//dump_fds();
int corpse;
int status;
while ((corpse = waitpid(0, &status, 0)) > 0)
printf(":-( PID %d status 0x%.4X\n", corpse, status);
printf("\n(>^_^)> ");
}
printf(" v(^o^)^ BYE BYE!\n");
}
static void dump_argv(char **argv)
{
int n = 0;
char **args;
args = argv;
while (*args++ != 0)
n++;
fprintf(stderr, "%d: %d args\n", getpid(), n);
args = argv;
while (*args != 0)
fprintf(stderr, "[%s]\n", *args++);
fprintf(stderr, "EOA\n");
}
/* Report on open file descriptors (0..19) in process */
static void dump_fds(void)
{
struct stat b;
char buffer[32];
sprintf(buffer, "%d: ", getpid());
char *str = buffer + strlen(buffer);
for (int i = 0; i < 20; i++)
*str++ = (fstat(i, &b) == 0) ? 'o' : '-';
*str++ = '\n';
*str = '\0';
fputs(buffer, stderr);
}
int makeargv(char *line, char *seps, char ***args)
{
enum { MAX_ARGS = 32 };
char **argv = malloc(32 * sizeof(char *)); // Lazy!
if (argv == 0)
err_sys("out of memory in makeargv()");
int n;
char **argp = argv;
char *str = line;
for (n = 0; n < MAX_ARGS - 1; n++)
{
str += strspn(str, seps);
if (*str == '\0')
break;
*argp++ = str;
int len = strcspn(str, seps);
if (len == 0)
break;
str[len] = '\0';
str += len + 1;
}
*argp = 0;
dump_argv(argv);
*args = argv;
return(n);
}
void err_sys(const char *msg)
{
int errnum = errno;
char *errmsg = strerror(errnum);
fprintf(stderr, "%s (%d: %s)\n", msg, errnum, errmsg);
exit(1);
}
Sample output:
$ ./pipes-15673333
<(^_^)>
Hello
I am your console and I am here to help you
If you don't need me anymore just say "bye"
29191: ooo-----------------
(>^_^)> who | grep jl | sort
29191: 6 args
[who]
[|]
[grep]
[jl]
[|]
[sort]
EOA
pnum = 2
ploc[0] = 0
ploc[1] = 2
ploc[2] = 5
PID 29194 launched
PID 29195 launched
29194: 1 args
[who]
EOA
PID 29196 launched
29194: ooo-----------------
29195: 2 args
[grep]
[jl]
EOA
29195: ooo-----------------
29196: 1 args
[sort]
EOA
29196: ooo-----------------
:-( PID 29194 status 0x0000
jleffler console Mar 27 15:11
jleffler ttys000 Mar 27 16:26
jleffler ttys001 Mar 27 16:26
jleffler ttys002 Mar 27 16:26
jleffler ttys003 Mar 27 16:26
jleffler ttys004 Mar 27 16:26
jleffler ttys005 Mar 27 16:26
:-( PID 29195 status 0x0000
:-( PID 29196 status 0x0000
(>^_^)> ls
29191: 1 args
[ls]
EOA
pnum = 0
ploc[0] = 0
PID 29197 launched
29197: 1 args
[ls]
EOA
29197: ooo-----------------
bash.getopts.update makefile pipeline.c pthread-1.c shuntzeroes.c timezeromoves.c
cmpfltint.c mda.c pipeline.dSYM pthread-2.c so.14304827 uint128.c
const-stuff.c mq-saurabh pipes-13905948.c pthread-3.c so.367309 uname.c
dupdata.sql mqp-saurabh pipes-14312939.c quine.c so.6964747 unwrap.c
fifocircle.c multi-pipe-sort.c pipes-15673333 ranges.sql so.6965001 xxx.sql
idsdb00246324.ec multiopts.sh pipes-15673333.c recv.c so.8854855.sql yyy.sql
incunabulum.c nextpipe.c pipes-15673333.dSYM regress.c strandsort.c
madump.c pipeline powa.c send.c streplace.c
:-( PID 29197 status 0x0000
(>^_^)> ls -C | wc
29191: 4 args
[ls]
[-C]
[|]
[wc]
EOA
pnum = 1
ploc[0] = 0
ploc[1] = 3
PID 29200 launched
PID 29201 launched
29200: 2 args
29201: 1 args
[ls]
[wc]
[-C]
EOA
EOA
29201: ooo-----------------
29200: ooo-----------------
:-( PID 29200 status 0x0000
16 46 581
:-( PID 29201 status 0x0000
(>^_^)> bye
v(^o^)^ BYE BYE!
$
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