I have this piece of code for running a Python program and I expect my shell to run a python program when I enter something like the following : mysh> hello.py
But hello.py is not passed to /usr/bin/python in execvp and I need help figuring out how to correctly pass arguments to execvp.
if (pid==0) // child process
{
if (py_flag==1)
execvp("/usr/bin/python",argv);
else
{
execvp(argv[0],argv);
perror("error");
}
}
And here's the result I am receiving:
./basic_shell
mysh> hello.py
I am a python program
Python 2.6.6 (r266:84292, May 27 2013, 05:35:12)
[GCC 4.4.7 20120313 (Red Hat 4.4.7-3)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>>
While I expect to receive a result like this:
$ python hello.py
sys.argv[0] = hello.py
Please let me know how can this be fixed.
I changed my code to this and now it is working as expected:
if (pid==0) //child process
{
if (py_flag==1)
{
char *new_argv[2];
new_argv[0]="/usr/bin/python";
new_argv[1]=argv[0];
new_argv[2]=0;
//execvp("/usr/bin/python",argv);
execvp(new_argv[0],new_argv);
}
else
{
execvp(argv[0],argv);
perror("error");
}
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.