所以这应该做的是fork，让子进程获取文件的文本，然后让父进程修改该文本并将其写入新文件。我有各种奇怪的东西。整个代码大致是这个。
#include <iostream>
#include <termios.h>
#include <cstdio>
#include <cstdlib>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <string.h>
#include <sys/wait.h>
using namespace std;
int parentPID;
int main(int argc, char* argv[]){
parentPID = getpid();
int pipey[2];
int worked = pipe(pipey);
if( worked == - 1){
cout << "Oops. Didn't make a pipe.";
}
//cout << "About to fork!!!";
fork();
if(getpid() != parentPID){//Only run in the child process
char* argvec1[3] = {"cat", "colorfile.txt", (char*)0};
dup2(pipey[1], 1);
execv("/bin/cat", argvec1);
}
else{//Only run in the parent process.
int someInt;
cout << "In the parent process";
pid_t status = wait(&someInt);
dup2(pipey[0], 0);
creat("newfile.txt", 0777);
chmod("newfile.txt", 0777);
int targetFile = open("newfile.txt", O_WRONLY);
if(targetFile == -1){
cout << "\nOops, couldn't open targetFile, ";
perror("because ");
}
else{
cout << "\nOpened target file.";
}
dup2(targetFile, 1);
//char* argvec2[] = {"sed", "-e", "s/color/colour/g", (char*)0};
//execv("/bin/sed", argvec2);
cout << "something went terribly wrong";
}
}
特别麻烦的是三件事，第一件事，这段代码......
creat("newfile.txt", 0777);
chmod("newfile.txt", 0777);
int targetFile = open("newfile.txt", O_WRONLY);
if(targetFile == -1){
cout << "\nOops, couldn't open targetFile, ";
perror("because ");
}
else{
cout << "\nOpened target file.";
}
dup2(targetFile, 1);
...不会将“打开的目标文件”写入标准输出。相反，它把它放在newfile.txt中，所以dup2正在改变输出命令之前出现的输出？...如果我注释掉dup2，那么它就不会发生，它肯定是特定的dup2打电话让它发生。
第二，这段代码片段......
creat("newfile.txt", 0777);
chmod("newfile.txt", 0777);
int targetFile = open("newfile.txt", O_WRONLY);
if(targetFile == -1){
cout << "\nOops, couldn't open targetFile, ";
perror("because ");
}
else{
cout << "\nOpened target file.";
}
//dup2(targetFile, 1);
char* argvec2[] = {"sed", "-e", "s/color/colour/g", (char*)0};
execv("/bin/sed", argvec2);
cout << "something went terribly wrong";
...根本没有输出关于打开文件的任何成功/失败。它打印出原始文件的内容，经过适当修改，但不会终止。它只是一种存在，直到我使用ctrl-C来杀死当前进程。最后的cout不会显示。
最后，这......
creat("newfile.txt", 0777);
chmod("newfile.txt", 0777);
int targetFile = open("newfile.txt", O_WRONLY);
if(targetFile == -1){
cout << "\nOops, couldn't open targetFile, ";
perror("because ");
}
else{
cout << "\nOpened target file.";
}
dup2(targetFile, 1);
char* argvec2[] = {"sed", "-e", "s/color/colour/g", (char*)0};
execv("/bin/sed", argvec2);
cout << "something went terribly wrong";
...不会给我输出标准输出或newfile.txt。
感觉这些系统调用中的一些只是以他们想要的顺序执行，半独立于我编写它们的顺序，因此很难用它们做任何事情。
答案 0 :(得分：0)
使用名为'colorfile.txt'的输入文件，如：
color
color color
color color color
color color color color
我更新了您的代码，以不同方式执行重复和调试消息。最重要的更改（正如我在您对您的问题的评论中所提到的）是关闭父进程中管道的写入端，以避免该进程挂起。同样重要的（用于调试目的，并处理有关事件顺序的混淆）是使用'endl'来刷新输出。否则，当仍然保持缓冲数据时，会发生dup2调用，并且可以在写入之后更改该数据的目标，因为在基础库刷新输出时（例如，在进程退出时）会发生写入。
请注意，对于管道，我使用术语“写入侧”和“读取侧”来分别引用由 pipe（2）创建的小型文件描述符数组的索引1和0 ：管道的用户写入并从中读取的边。
这些更新针对Linux 3.10.17内核上的g ++（GCC）4.8.2。据推测，其他人的包括等等会有所不同。
#include <iostream>
#include <cstdio>
#include <fcntl.h>
#include <unistd.h>
using namespace std;
int main() {
int pipey[2];
if (pipe(pipey) == -1) {
perror("pipe failed");
return 1;
}
cout << "About to fork!!!" << endl;
pid_t pid = fork();
if (pid == -1) {
perror("fork failed");
return 1;
}
if (pid == 0) { // child process
const char *av[] = { "cat", "colorfile.txt", 0 };
cout << "In the child process" << endl;
cout << "colorful child message before dup2" << endl;
dup2(pipey[1], 1);
close(pipey[1]);
cout << "colorful child message after dup2" << endl;
close(pipey[0]);
execvp(av[0], (char * const *) av);
perror("failed to exec cat");
return 1;
} else { // parent process.
const char *av[] = { "sed", "-e", "s/color/colour/g", 0 };
cout << "In the parent process" << endl;
int targetFd = open("newfile.txt", O_CREAT|O_TRUNC|O_WRONLY, 0644);
if (targetFd == -1){
perror("failed to open newfile.txt");
return 1;
} else {
cout << "Opened target file." << endl;
}
cout << "colorful parent message before dup2s" << endl;
dup2(pipey[0], 0);
close(pipey[0]);
dup2(targetFd, 1);
close(targetFd);
cout << "colorful parent message after dup2s" << endl;
// The soon-to-be exec'd cat process will have the read side of
// the pipe duped to its stdin, and the child process created
// above *will* write data that cat will see, but unless the
// write side is closed by *all* its writers prior to exec (and
// this parent process is one of the writers), cat will see no
// end-of-file on its stdin.
//
// Note how easily this deadlock can be created, within a single
// process.
close(pipey[1]);
execvp(av[0], (char * const *) av);
perror("failed to exec sed");
return 1;
}
}
当我运行时，我看到：
About to fork!!!
In the parent process
In the child process
colorful child message before dup2
Opened target file.
colorful parent message before dup2s
输出文件newfile.txt的内容是：
colorful parent message after dup2s
colourful child message after dup2
colour
colour colour
colour colour colour
colour colour colour colour
如果你理解为什么一条消息是彩色的而另一条消息是丰富多彩的，那么你就会抓住它。