Question

Using x86 platform , I want to start my application named myapp through this method:execl("./myapp","");It’s OK! But failed when I’m using ARM platform + embedded linux. Why ? Any help will be appreciated. Thanks in advance.

Answer 1

If you would like to use execle to pass in the same environment that your calling application had, you can use this:

#include <unistd.h>
extern char **environ;

/* ... */
execle("./myApp","./myApp",NULL,environ);