Build a Compiler Bomb

Question

Introduction

You're probably familiar with zip bombs, XML bombs, etc. Put simply, they are (relatively) small files which produce enormous output when interpreted by naïve software. The challenge here is to abuse a compiler in the same way.

Challenge

Write some source code which occupies 512 bytes or less and which compiles into a file which occupies the most possible space. Largest output file wins!

Rules

OK, so there are a few important clarifications, definitions and restrictions;

• The output of the compilation must be an ELF file, a Windows Portable Executable (.exe), or virtual bytecode for the JVM or .Net's CLR (other types of virtual bytecode are also likely to be OK if asked for). Update: Python's .pyc / .pyo output also counts.
• If your language-of-choice can't be compiled directly into one of those formats, transpilation followed by compilation is also allowed (Update: you can transpile multiple times, just so long as you never use the same language more than once).
• Your source code can consist of multiple files, and even resource files, but the summed size of all these files must not exceed 512 bytes.
• You cannot use any other input than your source file(s) and the standard library of your language-of-choice. Static linking standard libraries is OK when it's supported. Specifically, no third party libraries or OS libraries.
• It must be possible to invoke your compilation using a command or series of commands. If you require specific flags when compiling, these count towards your byte limit (e.g. if your compile line is gcc bomb.c -o bomb -O3 -lm, the -O3 -lm part (7 bytes) will be counted (note the initial leading space isn't counted).
• Preprocessors are permitted only if they are a standard compilation option for your language.
• The environment is up to you, but in the interests of making this verifiable, please stick to recent (i.e. available) compiler versions and operating systems (and obviously specify which you're using).
• It must compile without errors (warnings are OK), and crashing the compiler doesn't count for anything.
• What your program actually does is irrelevant, though it can't be anything malicious. It doesn't even have to be able to start.

Example 1

The C program

main(){return 1;}

Compiled with Apple LLVM version 7.0.2 (clang-700.1.81) on OS X 10.11 (64-bit):

clang bomb.c -o bomb -pg

Produces a file of 9228 bytes. The total source size is 17+3 (for the -pg) = 20 bytes, which is easily within size limit.

Example 2

The Brainfuck program:

++++++[->++++++++++++<]>.----[--<+++>]<-.+++++++..+++.[--->+<]>-----.--
-[-<+++>]<.---[--->++++<]>-.+++.------.--------.-[---<+>]<.[--->+<]>-.

Transpiled with awib to c with:

./awib < bomb.bf > bomb.c

Then compiled with Apple LLVM version 7.0.2 (clang-700.1.81) on OS X 10.11 (64-bit):

clang bomb.c

Produces a file of 8464 bytes. The total input here is 143 bytes (since @lang_c is the default for awib it didn't need to be added to the source file, and there are no special flags on either command).

Also note that in this case, the temporary bomb.c file is 802 bytes, but this counts towards neither the source size nor the output size.

Final Note

If an output of more than 4GB is achieved (perhaps if somebody finds a turing complete preprocessor), the competition will be for the smallest source which produces a file of at least that size (it's just not practical to test submissions which get too big).

Answer 1

C, (14 + 15) = 29 byte source, 17,179,875,837 (16 GB) byte executable

Thanks to @viraptor for 6 bytes off.

Thanks to @hvd for 2 bytes off and executable size x4.

This defines the main function as a large array and initialises its first element. This causes GCC to store the entire array in the resulting executable.

Because this array is bigger than 2GB, we need to provide the -mcmodel=medium flag to GCC. The extra 15 bytes are included in the score, as per the rules.

main[-1u]={1};

Don't expect this code to do anything nice when run.

Compile with:

gcc -mcmodel=medium cbomb.c -o cbomb

---

It took me a while to get round to testing @hvd's suggestion - and to find a machine with enough juice to handle it. Eventually I found a old non-production RedHat 5.6 VM with 10GB RAM, 12GB swap, and /tmp set to a large local partition. GCC version is 4.1.2. Total compile time about 27 minutes.

Due to the CPU and RAM load, I recommend against doing this compile on any remotely production-related machine.

Answer 2

C#, about 1 min to compile, 28MB output binary:

class X<A,B,C,D,E>{class Y:X<Y,Y,Y,Y,Y>{Y.Y.Y.Y.Y.Y.Y.Y.Y y;}}

Adding more Y's will increase the size exponentially.

An explanation by Pharap as per @Odomontois' request:

This answer is abusing inheritance and type parameters to create recursion. To understand what's happening, it's easier to first simplify the problem. Consider class X<A> { class Y : X<Y> { Y y; } }, which generates the generic class X<A>, which has an inner class Y. X<A>.Y inherits X<Y>, hence X<A>.Y also has an inner class Y, which is then X<A>.Y.Y. This then also has an inner class Y, and that inner class Y has an inner class Y etc. This means that you can use scope resolution (.) ad infinitum, and every time you use it, the compiler has to deduce another level of inheritance and type parameterisation.

By adding additional type parameters, the work the compiler has to do at each stage is further increased.

Consider the following cases:
In class X<A> { class Y : X<Y> { Y y;} } type param A has a type of X<A>.Y.
In class X<A> { class Y : X<Y> { Y.Y y;} } type param A has a type of X<X<A>.Y>.Y.
In class X<A> { class Y : X<Y> { Y.Y.Y y;} } type param A has a type of X<X<X<A>.Y>.Y>.Y.
In class X<A,B> { class Y : X<Y,Y> { Y y;} } type param A is X<A,B>.Y and B is X<A,B>.Y.
In class X<A> { class Y : X<Y> { Y.Y y;} } type param A is X<X<A,B>.Y, X<A,B>.Y>.Y and B is X<X<A,B>.Y, X<A,B>.Y>.Y.
In class X<A> { class Y : X<Y> { Y.Y.Y y;} } type param A is X<X<X<A,B>.Y, X<A,B>.Y>.Y, X<X<A,B>.Y, X<A,B>.Y>.Y>.Y and B is X<X<X<A,B>.Y, X<A,B>.Y>.Y, X<X<A,B>.Y, X<A,B>.Y>.Y>.Y.

Following this pattern, one can only imagine¹ the work the compiler would have to do to to deduce what A to E are in Y.Y.Y.Y.Y.Y.Y.Y.Y in the definition class X<A,B,C,D,E>{class Y:X<Y,Y,Y,Y,Y>{Y.Y.Y.Y.Y.Y.Y.Y.Y y;}}.

¹ You could figure it out, but you'd need a lot of patience, and intellisense won't help you out here.

Answer 3

Python 3, 13 byte source, 9,057,900,463 byte (8.5GiB) .pyc-file

(1<<19**8,)*2

Edit: Changed the code to the version above after I realized the rules say output size beyond 4GiB doesn't matter, and the code for this one is ever so slightly shorter; The previous code - and more importantly the explanation - can be found below.

---

Python 3, 16 byte source, >32TB .pyc-file (if you have enough memory, disk space and patience)

(1<<19**8,)*4**7

Explanation: Python 3 does constant folding, and you get big numbers fast with exponentation. The format used by .pyc files stores the length of the integer representation using 4 bytes, though, and in reality the limit seems to be more like 2**31, so using just exponentation to generate one big number, the limit seems to be generating a 2GB .pyc file from an 8 byte source. (19**8 is a bit shy of 8*2**31, so 1<<19**8 has a binary representation just under 2GB; the multiplication by eight is because we want bytes, not bits)

However, tuples are also immutable and multiplying a tuple is also constant folded, so we can duplicate that 2GB blob as many times as we want, up to at least 2**31 times, probably. The 4**7 to get to 32TB was chosen just because it was the first exponent I could find that beat the previous 16TB answer.

Unfortunately, with the memory I have on my own computer, I could test this only up to a multiplier of 2, ie. (1<<19**8,)*2, which generated a 8.5GB file, which I hope demonstrates that the answer is realistic (ie. the file size isn't limited to 2**32=4GB).

Also, I have no idea why the file size I got when testing was 8.5GB instead of the 4GB-ish I expected, and the file is big enough that I don't feel like poking around it at the moment.

Answer 4

If an output of more than 4GB is achieved (perhaps if somebody finds a turing complete preprocessor), the competition will be for the smallest source which produces a file of at least that size (it's just not practical to test submissions which get too big).

"Template Haskell" allows Haskell code to be generated at compile-time using Haskell, and is hence a turing complete pre-processor.

Here's my attempt, parameterised by an arbitrary numerical expression FOO:

import Language.Haskell.TH;main=print $(ListE .replicate FOO<$>[|0|])

The magic is the code inside the "splice" $(...). This will be executed at compile time, to generate a Haskell AST, which is grafted on to the program's AST in place of the splice.

In this case, we make a simple AST representing the literal 0, we replicate this FOO times to make a list, then we use ListE from the Language.Haskell.TH module to turn this list of ASTs into one big AST, representing the literal [0, 0, 0, 0, 0, ...].

The resulting program is equivalent to main = print [0, 0, 0, ...] with FOO repetitions of 0.

To compile to ELF:

$ ghc -XTemplateHaskell big.hs
[1 of 1] Compiling Main             ( big.hs, big.o )
Linking big ...
$ file big
big: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /nix/store/mibabdfiaznqaxqiy4bqhj3m9gaj45km-glibc-2.21/lib/ld-linux.so.2, for GNU/Linux 2.6.32, not stripped

This weighs in at 83 bytes (66 for the Haskell code and 17 for the -XTemplateHaskell argument), plus the length of FOO.

We can avoid the compiler argument and just compile with ghc, but we have to put {-# LANGUAGE TemplateHaskell#-} at the beginning, which bumps the code up to 97 bytes.

Here are a few example expressions for FOO, and the size of the resulting binary:

FOO         FOO size    Total size    Binary size
-------------------------------------------------
(2^10)      6B          89B           1.1MB
(2^15)      6B          89B           3.6MB
(2^17)      6B          89B           12MB
(2^18)      6B          89B           23MB
(2^19)      6B          89B           44MB

I ran out of RAM compiling with (2^20).

We can also make an infinite list, using repeat instead of replicate FOO, but that prevents the compiler from halting ;)

Answer 5

C++, 250 + 26 = 276 bytes

template<int A,int B>struct a{static const int n;};
template<int A,int B>const int a<A,B>::n=a<A-1,a<A,B-1>::n>::n;
template<int A>struct a<A,0>{static const int n=a<A-1,1>::n;};
template<int B>struct a<0,B>{static const int n=B+1;};
int h=a<4,2>::n;

This is the Ackermann function implemented in templates. I'm not able to compile with h=a<4,2>::n; on my little (6GB) machine, but I did manage h=a<3,14> for a 26M output file. You can tune the constants to hit your platform's limits - see the linked Wikipedia article for guidance.

Requires -g flag to GCC (because it's all the debug symbols that actually consume any space), and a larger-than-default template depth. My compile line ended up as

g++ -ftemplate-depth=999999 -g -c -o 69189.o 69189.cpp

Platform information

g++ (Ubuntu 4.8.2-19ubuntu1) 4.8.2
Linux 3.13.0-46-generic #79-Ubuntu SMP x86_64 GNU/Linux

Answer 6

ASM, 61 bytes (29 bytes source, 32 bytes for flags), 4,294,975,320 bytes executable

.globl main
main:
.zero 1<<32

Compile with gcc the_file.s -mcmodel=large -Wl,-fuse-ld=gold

Answer 7

Here's my C answer from 2005. Would produce a 16TB binary if you had 16TB RAM (you don't).

struct indblock{
   uint32_t blocks[4096];
};

struct dindblock {
    struct indblock blocks[4096];
};

struct tindblock {
    struct dindblock blocks[4096];
};

struct inode {
    char data[52]; /* not bothering to retype the details */
    struct indblock ind;
    struct dindblock dint;
    struct tindblock tind;
};

struct inode bbtinode;

int main(){}

Answer 8

Plain old C preprocessor: 214 bytes input, 5MB output

Inspired by my real-world preprocessor fail here.

#define A B+B+B+B+B+B+B+B+B+B
#define B C+C+C+C+C+C+C+C+C+C
#define C D+D+D+D+D+D+D+D+D+D
#define D E+E+E+E+E+E+E+E+E+E
#define E F+F+F+F+F+F+F+F+F+F
#define F x+x+x+x+x+x+x+x+x+x

int main(void) { int x, y = A; }

Experiments show that each level of #defines will (as expected) make the output approximately ten times larger. But since this example took more than an hour to compile, I never went on to "G".

Answer 9

Java, 450 + 22 = 472 bytes source, ~1GB class file

B.java (golfed version, warning during compilation)

import javax.annotation.processing.*;@SupportedAnnotationTypes("java.lang.Override")public class B extends AbstractProcessor{@Override public boolean process(java.util.Set a,RoundEnvironment r){if(a.size()>0){try(java.io.Writer w=processingEnv.getFiler().createSourceFile("C").openWriter()){w.write("class C{int ");for(int i=0;i<16380;++i){for(int j=0;j<65500;++j){w.write("i");}w.write(i+";int ");}w.write("i;}");}catch(Exception e){}}return true;}}

B.java (ungolfed version)

import java.io.Writer;
import java.util.Set;

import javax.annotation.processing.AbstractProcessor;
import javax.annotation.processing.RoundEnvironment;
import javax.annotation.processing.SupportedAnnotationTypes;
import javax.annotation.processing.SupportedSourceVersion;
import javax.lang.model.SourceVersion;
import javax.lang.model.element.TypeElement;

@SupportedAnnotationTypes("java.lang.Override")
@SupportedSourceVersion(SourceVersion.RELEASE_8)
public class B extends AbstractProcessor {
    @Override
    public boolean process(Set<? extends TypeElement> annotations, RoundEnvironment roundEnv) {
        if (annotations.size() > 0) {
            try (Writer writer = processingEnv.getFiler().createSourceFile("C").openWriter()) {
                writer.write("class C{int ");
                for (int i = 0; i < 16380; ++i) {
                    for (int j = 0; j < 65500; ++j) {
                        writer.write("i");
                    }
                    writer.write(i + ";int ");
                }
                writer.write("i;}");
            } catch (Exception e) {
            }
        }
        return true;
    }
}

Compilation

javac B.java
javac -J-Xmx16G -processor B B.java

Explanation

This bomb uses Annotation Processors. It needs 2 compile passes. The first pass builds the processor class B. During the second pass the processor creates a new source file C.java, and compiles it to a C.class with a size of 1,073,141,162 bytes.

There are several limitations when trying to create a big class file:

• Creating identifiers longer than about 64k results in: error: UTF8 representation for string "iiiiiiiiiiiiiiiiiiii..." is too long for the constant pool.
• Creating more than about 64k variables/functions results in: error: too many constants
• There is also a limit of about 64k for the code size of a function.
• There seems to be a general limit (bug?) in the java compiler of about 1GB for the .class file. If I increase 16380 to 16390 in the above code the compiler never returns.
• There is also a limit of about 1GB for the .java file. Increasing 16380 to 16400 in the above code results in: An exception has occurred in the compiler (1.8.0_66). Please file a bug ... followed by a java.lang.IllegalArgumentException.

Answer 10

C, 26 byte source, 2,139,103,367 byte output, valid program

const main[255<<21]={195};

Compiled using: gcc cbomb.c -o cbomb (gcc version 4.6.3, Ubuntu 12.04, ~77 seconds)

I thought I'd try to see how large I could make a valid program without using any command line options. I got the idea from this answer: https://codegolf.stackexchange.com/a/69193/44946 by Digital Trauma. See the comments there as to why this compiles.

How it works: The const removes the write flag from the pages in the segment, so main can be executed. The 195 is the Intel machine code for a return. And since the Intel architecture is little-endian, this is the first byte. The program will exit with whatever the start up code put in the eax register, likely 0.

It's only about 2 gig because the linker is using 32 bit signed values for offsets. It's 8 meg smaller than 2 gig because the compiler/linker needs some space to work and this is the largest I could get it without linker errors - ymmv.

Answer 11

Boo, 71 bytes. Compile time: 9 minutes. 134,222,236 byte executable

macro R(e as int):
 for i in range(2**e):yield R.Body
x = 0
R 25:++x

Uses a macro R (for Repeat) to cause the compiler to multiply the increment statement an arbitrary number of times. No special compiler flags are needed; simply save the file as bomb.boo and invoke the compiler with booc bomb.boo to build it.

Answer 12

Kotlin, 90 bytes source, 177416 bytes (173 KB) compiled JVM binary

inline fun a(x:(Int)->Any){x(0);x(1)}
fun b()=a{a{a{a{a{a{a{a{a{a{a{println(it)}}}}}}}}}}}

Technically, you could make this even longer by nesting the expression further. However, the compiler crashes with a StackOverflow error if you increase the recursion.

Answer 13

C++, 214 bytes (no special compile options needed)

#define Z struct X
#define T template<int N
T,int M=N>Z;struct Y{static int f(){return 0;}};T>Z<N,0>:Y{};T>Z<0,N>:Y{};T,int M>Z{static int f(){static int x[99999]={X<N-1,M>::f()+X<N,M-1>::f()};}};int x=X<80>::f();

It's a fairly straightforward two-dimensional template recursion (recursion depth goes as the square-root of total templates emitted, so won't exceed platform limits), with a small amount of static data in each one.

Generated object file with g++ 4.9.3 x86_64-pc-cygwin is 2567355421 bytes (2.4GiB).

Increasing the initial value above 80 breaks the cygwin gcc assembler (too many segments).

Also, 99999 can be replaced by 9<<19 or similar for increased size without changing the source code... but I don't think I need to use any more disk space than I already am ;)

Answer 14

Scala - 70 byte source, 22980842 byte result (after jar)

import scala.{specialized => s}
class X[@s A, @s B, @s C, @s D, @s E]

This produces 9⁵ (about 59,000) specialized class files, which pack into a jar of about 23 MB. You can in principle keep going if you have a filesystem that can handle that many files and enough memory.

(If the jar command must be included, it's 82 bytes.)

Answer 15

C, 284 bytes + 2 for the -c in gcc bomb.c -o bomb.o -c; output: 2 147 484 052 bytes

#define a 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
#define b a,a,a,a,a,a,a,a,a,a,a,a,a,a,a,a
#define c b,b,b,b,b,b,b,b,b,b,b,b,b,b,b,b
#define d c,c,c,c,c,c,c,c,c,c,c,c,c,c,c,c
#define e d,d,d,d,d,d,d,d,d,d,d,d,d,d,d,d
#define f e,e,e,e,e,e,e,e,e,e,e,e,e,e,e,e
__int128 x[]={f,f,f,f,f,f,f,f};

Answer 16

Boo, way more than you can expect from this

macro R(e as int):for i in range(9**e):yield R.Body
x = 0
R 99:++x

Answer 17

Python 3:

9**9**9**9**9

Tetration bomb