Samrat Man Singh

Email: mail@samrat.me

Hi! I’m Samrat, a programmer from Nepal.

Piping output to a pager

2019-10-28

When you run git diff on a project where you’ve made more than a screenful of changes, the output is automatically redirected to a pager and you can browse through your changes using the familiar less interface. This post shows a minimal example piping output to less.

To achieve this, git creates a pipe and spawns a new process, whose purpose is to exec the pager.

int fds[2];
if (pipe(fds) == -1) {
  perror_die("pipe");
}
int child_pid = fork();

Let’s start by looking at the parent process, even though in the code this case is handled last. The STDOUT of the parent process is aliased to the write end of the pipe, so any further printing in the parent process writes stuff to fds[WRITE_END]. Once we’re done, we close the write end of the pipe to signal EOF to the child process.

We wait for the child process to exit since otherwise control returns to the shell.

switch (child_pid)
// ...
  default: { /* Parent */
    /* STDOUT is now fds[WRITE_END] */
    dup2(fds[WRITE_END], STDOUT_FILENO);
    /* parent doesn't read from pipe */
    close(fds[READ_END]);

    /* "Business" logic which determines what to actually print */
    int num_lines = 1024;
    if (argc > 1) {
      num_lines = atoi(argv[1]);
    }
    print_n_lines(num_lines);


    fflush(stdout);

    /* Signal EOF to the pager process */
    close(STDOUT_FILENO);

    int stat_loc;
    waitpid(child_pid, &stat_loc, 0);
    break;
  }

The STDIN of the new process is aliased to the read end of the pipe. Anything being printed in the parent process is thus now an input to the pager process. After setting up STDIN, this process runs less.

switch (child_pid) {
case 0: {    /* Child(pager) */
    /* Pager process doesn't write to pipe */
    close(fds[WRITE_END]);

    /* Make READ_END of pipe pager's STDIN */
    dup2(fds[READ_END], STDIN_FILENO);

    /* F -> quit-if-one-screen */
    /* R -> preserve color formatting */
    /* X -> don't send some special instructions eg. to clear terminal screen before starting */
    char *less_argv[] = {"less", "-FRX", NULL};
    int exec_status = execvp(less_argv[0], less_argv);

    fprintf(stderr,
            "execvp failed with status: %d and errno: %d\n", exec_status, errno);
    break;
  }
  
// ...
} // switch

The full example can be found in this gist. You can also check out the Ruby and Rust implementations of this.

Myers' diff algorithm in Clojure

2019-10-21

If you’re a programmer, you’re probably fairly comfortable with understanding and interpreting diff patches of the kind that git and other version control systems use. In this post we’ll walk through a Clojure implementation of the Myers’ diff algorithm, which is the algorithm used by Git.

I have been implementing Git following along James Coglan’s book “Building Git”– you can check out the project here. The project itself is in Rust but I thought it might be an interesting exercise to implement the algorithm in Clojure.

Edit graph

Say we want to turn a file with the lines A,B,C into A,C,E. The Myers’ diff algorithm finds out the shortest edit script that does so. The algorithm works on an edit graph as shown above– a rightward traversal from one vertex to another indicates deletion of a line from the old file; a downward traversal indicates insertion of a line from the new file.

Also notice that in our example ABC -> ACE there are two lines which are the same(A & C). Myers’ algorithm leverages this to produce the shortest edit script. In our edit graph above, you can see that where lines are equal we have drawn dotted lines to indicate diagonal traversals which require neither insertion nor deletion. The Myers’ algorithm maximizes the number of diagonal traversals in the edit script it produces1.

Let’s walk through the output Myers’ algorithm produces for our example above:

The first character is the same in both files so take the first diagonal down. Then delete B. The third character in the old file(C) is the same as the second in the new file so keep it and do a diagonal traversal again. Then insert E. This series of edits on the old file has produced our desired new file, and so we stop.

Finally, before we dive into the algorithm, we have one more addition to the edit graph. We will draw downward-diagonals across the graph, also called k-lines. Each k-line has the equation k = x - y.

Implementation

The maximum number of operations(insertions or deletions) the edit script can have is len(old) + len(new)– in this case we would be deleting every line in the old file and inserting every line from the new file.

We iterate through 0 to maximum number of operations and see what the farthest point along each k-line we can go. In the reduce, we can see that our accumulator is a tuple with v and trace.

v maps each k to the farthest x we have found so far along that k-line. We can discard the y values since we can easily compute them(y = x - k).

trace is a vector of the vs we’ve seen so far.

We can see that the function ultimately returns trace. This will be used to reconstruct the path that we took along the graph.

(defn shortest-edit [a b]
  (let [max (+ (count a) (count b))
        [_ trace] (reduce (fn [[v trace] d]
                            (let [[found-end? farthest-xs] (get-farthest-xs-for-d v a b d)
                                  [new-v new-trace] [(merge v farthest-xs)
                                                     (conj trace v)]]
                              (if found-end?
                                (reduced [new-v new-trace])
                                [new-v new-trace])))
                          [{1 0} []]
                          (range 0 (inc max)))]
    trace))

To get to a line k, we can either go down from k+1 or go right from line k-1. We use the move-down? function to decide whether to move down or right:

(defn move-down?
  [v d k]
  (or (= k (- d))
      (and (not= k d)
           (< (get v (dec k))
              (get v (inc k))))))

Next, in get-farthest-xy-along-k* we figure out what the farthest (x,y) we can reach in this round d along the k line given the v map we’ve built so far. a and b contain the lines that we want to diff. This simply uses move-down?as described above; then if possible walks down any diagonals.

(defn get-farthest-xy-along-k*
  [v d a b k]
  (let [x (if (move-down? v d k)
            (get v (inc k))
            (inc (get v (dec k))))
        y (- x k)
        ;; walk down diagonals
        [x y] (loop [[x y] [x y]]
                (if (and (< x (count a))
                         (< y (count b))
                         (= (get a x) (get b y)))
                  (recur [(inc x) (inc y)])
                  [x y]))]
    [x y]))

get-farthest-xs-for-d iterates from -d to d(both inclusive), 2 steps at a time2 and stores the farthest x for each k, returning early if we reach the bottom-right corner in our edit graph.

(defn get-farthest-xs-for-d
  [v a b d]
  (let [get-farthest-xy-along-k (partial get-farthest-xy-along-k* v d a b)]
    (reduce (fn [[found-end? k->x] k]
              (let [[x y] (get-farthest-xy-along-k k)]
                (if (and (>= x (count a))
                         (>= y (count b)))
                  (reduced [true k->x])
                  [found-end? (assoc k->x k x)])))
            [false {}]
            (range (- d) (inc d) 2))))

So far, we’ve built up functions that give us a trace:

myers.core> (shortest-edit (clojure.string/split "ABCABBA" #"")
                  (clojure.string/split "CBABAC" #""))
[{1 0}
 {1 0, 0 0}
 {1 1, 0 0, -1 0}
 {1 1, 0 2, -1 0, -2 2, 2 3}
 {1 5, 0 2, -1 4, -2 2, 2 3, -3 3, 3 5}
 {1 5, 0 5, -1 4, -2 4, 2 7, -3 3, 3 5, -4 3, 4 7}]
myers.core>

Now, let’s look at how to turn this into a diff.

Backtracking and producing the diff

This next function will take an input a trace and output an edit sequence. Each element in the edit-sequence is of the form [prev-x prev-y x y].

This time we start from the bottom-right node of the edit graph. In each iteration, we figure out the previous (x,y)– also walking back up diagonals if required.

(defn trace->edit-seq
  [n m trace]
  (let [[_ edit-seq]
        (reduce (fn [[[x y] edit-seq] [d v]]
                  (let [k (- x y)
                        prev-k (if (move-down? v d k)
                                 (inc k)
                                 (dec k))
                        prev-x (get v prev-k)
                        prev-y (- prev-x prev-k)
                        [[x y] edit-seq] (loop [[x y] [x y]
                                                edit-seq edit-seq]
                                           (if (and (> x prev-x) (> y prev-y))
                                             (recur [(dec x) (dec y)]
                                                    (conj edit-seq
                                                          [(dec x) (dec y) x y]))
                                             [[x y] edit-seq]))
                        edit-seq (if (pos? d)
                                   (conj edit-seq [prev-x prev-y x y])
                                   edit-seq)]
                    [[prev-x prev-y] edit-seq]))
                [[n m] []]
                (reverse (map vector (range) trace)))]
    (reverse edit-seq)))

Once we have an edit sequence, printing the diff is pretty straightforward:

(defn diff
  [a b]
  (for [[prev-x prev-y x y] (trace->edit-seq (count a)
                                             (count b)
                                             (shortest-edit a b))]
    (let [a-line (get a prev-x)
          b-line (get b prev-y)]
      (cond
        (= x prev-x) {:op :ins
                      :content b-line}
        (= y prev-y) {:op :del
                      :content a-line}
        :else {:op :eql
               :content a-line}))))

(defn print-diff
  [diff]
  (doseq [{:keys [op content]} diff]
    (let [d-out (case op
                  :ins (format "+%s" content)
                  :del (format "-%s" content)
                  :eql (format " %s" content))]
      (println d-out))))

  1. which is also why finding the shortest edit script and finding the longest common subsequence are equivalent problems.

    [return]

  2. if d is even, you can only reach an even k-line, and vice-versa.

    [return]

Summary of recent reading(July-September 2019)

2019-09-30

Summary of recent reading(April-June 2019)

2019-06-30

Summary of recent reading(January-March 2019)

2019-03-17