为什么在这里使用进程替换会导致挂起？

Question

我有一个程序需要启动其他程序，可能用文件和管道替换它们的 stdio。虽然在 "work" 看来，子进程确实从源管道中获取了 I/O，但不幸的是，它也会导致挂起。子流程似乎永远不会得到 EOF.

这里是代码的最小复制，为什么打印后挂起[=13=]？

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>

int main(int argc, char *argv[]) {

    switch (pid_t pid = fork()) {
    case 0: {
        // in child

        // replace the child's stdin with whatever filename is given as argv[1]
        freopen(argv[1], "r+b", stdin);

        // construct an argv array for to exec, no need for anything except 
        // argv[0] since we want it to use stdin
        char path[]  = "/bin/cat";
        char *args[] = {path, NULL};

        // run it!
        execv(args[0], args);
        abort(); // we should never get here!
    }
    case -1:
        // error
        return -1;
    default: {
        // in parent, just wait for the sub-process to terminate
        int status;
        const auto r = waitpid(pid, &status, __WALL);

        if (r == -1) {
            perror("waitpid");
            return -1;
        }
        break;
    }
    }
}

# runs printf creating a pipe, which is then passed as the argv of my test program
./test >(printf "Hello\n")

Answer 1

./test <(printf "Hello\n")

切换到 >(...) 到 <(...) 以从 printf 读取而不是写入它。

freopen(argv[1], "rb", stdin);

不要使用 r+。您只是从文件中读取，所以将其设置为 r.

为什么在这里使用进程替换会导致挂起？

Why does using process substitution here cause a hang?

c++

linux

bash

shell

process-substitution