colang - CO language compiler

CO is a language designed for use in programming olympiads and contests like IOI and ACM ICPC. It features a simple C-style syntax, opinionated algorithm-oriented standard library, C++ tier performance, and many small bits that make solving tasks easier and more fun.

colang translates CO source code into C source code, so you can use CO on every judge that accepts C or C++.

Status

colang is in early development stage, so it can't do much yet. This section will be updated when most core language features will be implemented, so stay tuned!

Building

If you still want to try it out right now, grab the source, install SBT, and prepare your build environment. You need two things to successfully build colang: CO standard library and GCC binary on your PATH.

CO standard library

colang depends on the CO standard library that must be present at any of the following locations: ~/.colang-libs/, /usr/local/lib/colang, /usr/lib/colang, /lib/colang. The standard library is included in this repo (the stdlib directory), so the most simple and reliable installation method is creating a symlink from ~/.colang-libs/ to the stdlib directory. On Unix-like systems this can be done like this:

ln -s path/to/repo/stdlib ~/.colang-libs

On Windows, run cmd.exe as administrator and in your home directory (C:\Users\<you>\) execute

mklink /D .colang-libs\ path\to\repo\stdlib

Assembly

Once you have GCC and CO stdlib ready, build the compiler with sbt assembly. It will produce an executable standalone JAR file under target/scala-2.11.

Compiling and running programs

You can now compile and execute the solution to the A + B problem:

void main() {
    int a, b
    read(a)
    read(b)
    println(a + b)
}

The syntax is not final: a more convenient I/O interface will be eventually supported.

Let's go for a more fun example. At this point, you can already solve quadratic equations with CO:

double abs(double x) {
    if (x >= 0.0) return x else return -x
}

double sqr(double x) {
    return x * x
}

double EPS = 1.0e-9

//If we have no sqrt(), we can just write our own :)
double sqrt(double x) {
    double r = 10.0

    while (abs(r * r - x) > EPS) {
        r = r - (sqr(r) - x) / (2.0 * r)
    }

    return r
}

void main() {
    double a, b, c
    read(a)
    read(b)
    read(c)

    double det = sqr(b) - 4.0 * a * c
    if (det < 0.0) {
        //Can't print strings yet :(
        println(-1)
        return
    }

    double r1 = (-b - sqrt(det)) / (2.0 * a)
    double r2 = (-b + sqrt(det)) / (2.0 * a)

    println(r1)
    println(r2)
}

For the curious, the generated code looks like this.