输入/输出管道
本质上,我正在尝试替换:
/directory/program < input.txt > output.txt
为了避免使用硬盘驱动器,请在C ++中使用管道。这是我的代码:
//LET THE PLUMBING BEGIN int fd_p2c[2], fd_pFc[2], bytes_read;
// "p2c" = pipe_to_child, "pFc" = pipe_from_child (see above link)
pid_t childpid;
char readbuffer[80];
string program_name;// <---- includes program name + full path
string gulp_command;// <---- includes my line-by-line stdin for program execution
string receive_output = "";
pipe(fd_p2c);//create pipe-to-child
pipe(fd_pFc);//create pipe-from-child
childpid = fork();//create fork
if (childpid < 0)
{
cout << "Fork failed" << endl;
exit(-1);
}
else if (childpid == 0)
{
dup2(0,fd_p2c[0]);//close stdout & make read end of p2c into stdout
close(fd_p2c[0]);//close read end of p2c
close(fd_p2c[1]);//close write end of p2c
dup2(1,fd_pFc[1]);//close stdin & make read end of pFc into stdin
close(fd_pFc[1]);//close write end of pFc
close(fd_pFc[0]);//close read end of pFc
//Execute the required program
execl(program_name.c_str(),program_name.c_str(),(char *) 0);
exit(0);
}
else
{
close(fd_p2c[0]);//close read end of p2c
close(fd_pFc[1]);//close write end of pFc
//"Loop" - send all data to child on write end of p2c
write(fd_p2c[1], gulp_command.c_str(), (strlen(gulp_command.c_str())));
close(fd_p2c[1]);//close write end of p2c
//Loop - receive all data to child on read end of pFc
while (1)
{
bytes_read = read(fd_pFc[0], readbuffer, sizeof(readbuffer));
if (bytes_read <= 0)//if nothing read from buffer...
break;//...break loop
receive_output += readbuffer;//append data to string
}
close(fd_pFc[0]);//close read end of pFc
}
我绝对可以确保上述字符串已正确初始化。但是,发生了两件事对我来说没有意义:
(1)我正在执行的程序报告“输入文件为空”。由于我未使用“ <”调用程序,因此不应期望输入文件。相反,它应该期待键盘输入。此外,它应该阅读“
gulp_command”中包含的文本。
(2)程序的报告(通过标准输出提供)出现在终端中。这很奇怪,因为此管道的目的是将stdout传输到我的字符串“
receive_output”。但是由于它出现在屏幕上,所以向我表明该信息没有通过管道正确传递给变量。如果我在if语句的末尾实现以下内容,
cout << receive_output << endl;
我什么也没得到,好像字符串是空的。我感谢您能给我任何帮助!
编辑:澄清
我的程序当前正在使用文本文件与另一个程序通信。我的程序编写了一个文本文件(例如input.txt),该文件由外部程序读取。然后,该程序将生成output.txt,我的程序会读取该文件。所以是这样的:
my code -> input.txt -> program -> output.txt -> my code
因此,我的代码当前使用的是
system("program < input.txt > output.txt");
我想使用管道替换此过程。我想将我的输入作为标准输入传递给程序,并让我的代码将该程序的标准输出读入字符串。
回答:
您的主要问题是您需要dup2()
反转参数。您需要使用:
dup2(fd_p2c[0], 0); // Duplicate read end of pipe to standard inputdup2(fd_pFc[1], 1); // Duplicate write end of pipe to standard output
我陷入了误读您写为OK的错误的念头,直到我对设置代码进行错误检查并从dup2()
调用中获得了意外的值,这告诉我出了什么问题。如果出现问题,请插入之前跳过的错误检查。
您也没有确保从子级读取的数据为空终止;此代码可以。
工作代码(带有诊断程序),使用cat
最简单的“其他命令”:
#include <unistd.h>#include <string>
#include <iostream>
using namespace std;
int main()
{
int fd_p2c[2], fd_c2p[2], bytes_read;
pid_t childpid;
char readbuffer[80];
string program_name = "/bin/cat";
string gulp_command = "this is the command data sent to the child cat (kitten?)";
string receive_output = "";
if (pipe(fd_p2c) != 0 || pipe(fd_c2p) != 0)
{
cerr << "Failed to pipe\n";
exit(1);
}
childpid = fork();
if (childpid < 0)
{
cout << "Fork failed" << endl;
exit(-1);
}
else if (childpid == 0)
{
if (dup2(fd_p2c[0], 0) != 0 ||
close(fd_p2c[0]) != 0 ||
close(fd_p2c[1]) != 0)
{
cerr << "Child: failed to set up standard input\n";
exit(1);
}
if (dup2(fd_c2p[1], 1) != 1 ||
close(fd_c2p[1]) != 0 ||
close(fd_c2p[0]) != 0)
{
cerr << "Child: failed to set up standard output\n";
exit(1);
}
execl(program_name.c_str(), program_name.c_str(), (char *) 0);
cerr << "Failed to execute " << program_name << endl;
exit(1);
}
else
{
close(fd_p2c[0]);
close(fd_c2p[1]);
cout << "Writing to child: <<" << gulp_command << ">>" << endl;
int nbytes = gulp_command.length();
if (write(fd_p2c[1], gulp_command.c_str(), nbytes) != nbytes)
{
cerr << "Parent: short write to child\n";
exit(1);
}
close(fd_p2c[1]);
while (1)
{
bytes_read = read(fd_c2p[0], readbuffer, sizeof(readbuffer)-1);
if (bytes_read <= 0)
break;
readbuffer[bytes_read] = '\0';
receive_output += readbuffer;
}
close(fd_c2p[0]);
cout << "From child: <<" << receive_output << ">>" << endl;
}
return 0;
}
样本输出:
Writing to child: <<this is the command data sent to the child cat (kitten?)>>From child: <<this is the command data sent to the child cat (kitten?)>>
请注意,您将需要小心以确保不会陷入代码僵局。如果您具有严格的同步协议(因此父级写一条消息并以锁步方式读取响应),则应该没问题,但如果父级正在尝试写一个太大而无法放入子级管道的消息当孩子试图写一条太大的消息而无法放入父级管道中时,则每个消息都将被阻止写入,而等待对方阅读。
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