I wanted to be really able to explain to a fair amount of detail how does the program ls actually work right from the moment you type the command name and hit ENTER. What goes on in user space and and in kernel space? This is my attempt and what I have learned so far on Linux (Fedora 19, 3.x kernel).
How does the shell find the location of 'ls' ?
| public class CustomMatchers { | |
| public static Matcher<View> withBackground(final int resourceId) { | |
| return new TypeSafeMatcher<View>() { | |
| @Override | |
| public boolean matchesSafely(View view) { | |
| return sameBitmap(view.getContext(), view.getBackground(), resourceId); | |
| } | |
| @Override |
| #!/bin/bash | |
| # Disclaimer - make backups, use at your own risk. | |
| # | |
| # Based on this comment: http://stackoverflow.com/a/13944924/843067 | |
| # Views and stored procedures have to be done separately. | |
| OLDDB="old_db_name" | |
| NEWDB="new_db_name" | |
| MYSQL="mysql -u root -pyour_password " |
