I want to compile a C program so simple I can explain all of the assembly

I wish someone would dissect JVM in exactly the same way: i.e. clearly explained what needs to be stripped off to have a quickly loading "hello world" implementation eating less than 1MB of RAM and starting in a few microseconds, like a normal, sane process should.

All these wonderful things are being built (Clojure, JRuby) on top of JVM that are of no use outside of web/EE because default JVM is so heavy.

Yes, a VM does a lot more than just bootstrapping your stdlib, like in case of libc, yet I keep thinking there must be plenty of unnecessary fat to strip off. Just look at Microsoft CLR: same feature set, yet none of that sluggish starting, RAM-wasting JVM nonsense.

I'm not sure I understand the revelation here. GCC is often used to target single boards with little resources. You can build gcc for the environment you want to target. Sounds like attempting to use a compiler flag in a way it wasn't intended. If you wanted to reduce the binary size wouldn't you look at the linker?

I think he's talking about this article: http://www.phreedom.org/solar/code/tinype/

It's not very well documented, but BiteScript looks very interesting:

http://blog.headius.com/2009/03/bitescript-001-ruby-dsl-for-...

http://blog.headius.com/2009/05/bitescript-002-scripting-exa...

Lets see a really simple program you can explain all te assembly :)

We need to write our program like this:

$echo 0000000: 55 48 89 e5 b8 ff aa 00 00 c9 c3 |xxd -r > sum2.bin

Here we have the SAME little program in C: $ cat sum.c int sum(void){ return 0x00ff + 0xaa00; }

We can getLook at the results: $ gcc -c sum.c -o sum.o (get the raw opcodes in osx, intel arch )

$ otool sum.o -td|sed -n '3,$p'| awk '{ print $0}'|xxd -r > sum.bin

Now you can look at asm level your code: $ ndisasm -b 32 sum.bin 00000000 55 push ebp 00000001 48 dec eax 00000002 89E5 mov ebp,esp 00000004 B8FFAA0000 mov eax,0xaaff So your program is now reduced to this code: $ hexdump sum.bin 0000000 55 48 89 e5 b8 ff aa 00 00 c9 c3 000000b

Test your 2 files md5 sum.bin sum2.bin MD5 (sum.bin) = a0ccc94bcdc860a81ff28252f56c2257 MD5 (sum2.bin) = a0ccc94bcdc860a81ff28252f56c2257

We could probe our code with a selfmade userland loader: $./uloader sum2.bin Display Opcodes to exec: 55 48 89 e5 b8 ff aa 00 00 c9 c3 End opcodes code to exec address: exec_code =0x100100080 new crafted Proc : address = 0x100100080 returned value ==>aaff

----BEGIN Code--- #include #include > #include #include

int main( int argc, char argv[] ){ unsigned int (proc)(); unsigned int fdprog=0; unsigned int exec_code=NULL; unsigned char ptr=NULL; unsigned int returned_value=0x0; exec_code=(int ) malloc( 100 ); ptr=( char )exec_code; fdprog=open(argv[1], S_IRUSR ); printf("Display Opcodes to exec:\n"); while( read(fdprog, ptr, sizeof(unsigned char)) ){ printf(" %02x ", ptr ); ptr++; }

        printf("\nEnd opcodes\n");                                                                                      
        printf("code to exec address: exec_code =%p \n",exec_code);                                                              
        proc=(unsigned int (*)() ) exec_code;                                                                          
        printf("new crafted Proc : address = %p \n",proc);                                                                       
        returned_value= (*proc)();    //here is the magic bro! :)
                                                                                                                        
        printf("returned value ==>%lx \n",returned_value);                                                                            return 0;                                                                                                               }                        

----END Code ---

Saludos! Jorge A. Garcia.