execv

If I used fork() and execv() to spawn several child processes running in the background and I wanted to bring one of them to the foreground, how could I do that? I am trying to write a shell that can start processes either in the foreground or background. "Background" and "foreground" are not terms used generically for processes, but rather only apply to shells which can wait for jobs on demand. Matt Joiner Complimentarily to Ignacio Vazquez-Abram's answer , I suggest that you emulate the shell foreground/background model. As far as I can tell, backgrounding a process means suspending it. The

I am working on MacOS-X Lion with GCC 4.2. This code works, but I get a warning I would like fix: #include <unistd.h> main() { char *args[] = {"/bin/ls", "-r", "-t", "-l", (char *) 0 }; execv("/bin/ls", args); } warning: deprecated conversion from string constant to 'char*' I do not want the warning to be suppressed, I want not to have it at all. It is C++ code, not C. Using a char *const (so exactly the type required by execv()) still produces the warning. Thank you. This seems to be ok: #include <unistd.h> main() { char const *args[] = {"/bin/ls", "-r", "-t", "-l", NULL }; execv("/bin/ls",

I'm supposed to write an interpreter program which is more like a command prompt. This is some background info: General flow of basic interpreter 1. Prompt for user request. 2. Carry out the user request. 3. Unless user terminates the program, go to step 1. Run a command. Format: R command_path [arg1 to arg4] //a single character ‘R’ which stands for 'Run', followed by the path of the command //the user can supply up to 4 command line arguments Behavior: a. If command exist, run the command in a child process with the supplied command line arguments  Wait until the child process is done b.

i've written two simple programs: int main(int ac, char **argv ) { execv( "/home/me/Desktop/execvtest2", argv ); } and int main(int ac, char **argv ) { execv( "/home/me/Desktop/execvtest1", argv ); } I've compiled them with gcc -g to the according outputfiles. I'm running Ubuntu 10.10 using gcc (Ubuntu/Linaro 4.4.4-14ubuntu5.1) 4.4.5. When I'm debuging the first program with GNU gdb (GDB) 7.2-ubuntu, I can step until the first execv statement, but then the two files just keep running. Even if I set the follow-exec-mode to new, I can't step into the second program. When I set catch exec, gdb

First, I wrote a c++ code as follows: #include <cstdio> int main() { int a,b; while(scanf("%d %d",&a,&b) == 2) printf("%d\n",a+b); return 0; } I use g++ -o a a.cpp to complie it. Afterwards, I wrote python code as follows: import os,sys sys.stdin = open("./data.in","r") sys.stdout = open("./data.out","w") pid = os.fork() if pid == 0: cmd = ["./a","./a"] os.execv(cmd[0],cmd) However, the data.out file contains nothing. That is to say, the child process did not inherit stdin and stdout from his parent process. But when I wrote a c++ code as follows: #include<unistd.h> #include<cstdio> int main()

I hope I haven't missed a similar question. I'm trying to code a mini-shell of my own, using primitive C functions. I got something that should work, but I have a pointer that makes everything bug. My adrCmd pointer should get the command-path string from the searchCmd() function and keep the same value in the main function. In fact: it points to the right value on searchCmd() , but not in the main() . Here's the code: int searchCmd(char* cmd, char* adrCmd){ char* path = getenv("PATH"); if(debug)printf("PATH : %s\n", path); int nbPath = (compteLettre(path, ':')+1); char** pathTab = malloc