Zig

XML Path Performance Optimizations

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Discusses some important things to know about using Chilkat paths in the Chilkat XML API. Zig

const std = @import("std");
const chilkat = @import("chilkat");

pub fn main(init: std.process.Init) !void {
    const alloc = init.arena.allocator();

    const xml = try chilkat.Xml.init();
    defer xml.deinit();

    // Let's load XML containing the following:

    // <?xml version="1.0" encoding="utf-8"?>
    // <xyz>
    //     <licenses>
    //         <license>
    //             <id>1234</id>
    //         </license>
    //         <license>
    //             <id>1234</id>
    //         </license>
    // ...
    // My sample XML contains 64,000 "license" nodes ..
    // ...
    //         <license>
    //             <id>1234</id>
    //         </license>
    //         <license>
    //             <id>1234</id>
    //         </license>
    //     </licenses>
    // </xyz>
    //
    xml.loadXmlFile("qa_output/large.xml") catch {
        std.debug.print("{s}\n", .{try xml.getLastErrorText(alloc)});
        return;
    };

    // Iterating over the individual "license" nodes with this code snippet is
    // extremely slow:
    var lic_count: i32 = xml.numChildrenHavingTag("licenses|license");
    std.debug.print("license count = {d}\n", .{lic_count});

    var s: [:0]const u8 = "";
    var i: i32 = 0;
    // If "10" is changed to licCount, then it becomes apparent that this loop gets slower with each iteration.
    while (i < 10) {
        xml.setI(i);
        s = try xml.getChildContent(alloc, "licenses|license[i]|id");
        std.debug.print("{d}: {s}\n", .{ i, s });
        i = i + 1;
    }

    // The reason it is extremely slow is that the "license[i]" part of the path passed to GetChildContent
    // says: find the i'th child of "licenses" having the tag "license".  Chilkat cannot assume that all
    // children of an XML node have the same tag.  Therefore it's not possible to directly access the i'th child.
    // Internally, Chilkat must start at the 1st child and iterate until it reaches the i'th child having the
    // tag "license".

    // For example, imagine if the XML was like this:

    // <?xml version="1.0" encoding="utf-8"?>
    // <xyz>
    //     <licenses>
    //         <license>
    //             <id>1234</id>
    //         </license>
    //         <somethingElse>
    //             <a>abc</a>
    //         </somethingElse>
    //         <license>
    //             <id>1234</id>
    //         </license>
    // ...

    // In the above XML, the 1st "license" is the 1st child of "licenses", but the 2nd "license"
    // is the 3rd child of "licenses".

    // If you already know that all children have the same tag, there is a shortcut that allows
    // for direct access to that child.  Just leave off the tag name, like this:

    i = 0;
    // If "10" is changed to licCount, then we can see the time for each loop is the same, and it's fast.
    while (i < 10) {
        xml.setI(i);
        s = try xml.getChildContent(alloc, "licenses|[i]|id");
        std.debug.print("{d}: {s}\n", .{ i, s });
        i = i + 1;
    }

    // When we pass just the index "[i]", we're saying: Get the i'th child regardless of tag.
    // This is extremely fast because internally we can just access the i'th child directly.

    // Another performance improvement is to call NumChildrenAt rather than NumChildrenHavingTag.
    // For example:
    lic_count = xml.numChildrenAt("licenses");
    std.debug.print("licCount = {d}\n", .{lic_count});

    // NumChildrenAt returns the total number of children at the tag path.  If we already know
    // all children will have the same tag, we can just get the count.
}