What's the shortest code to cause a stack overflow? [closed]

Question

Read this line, and do what it says twice.

All these answers and no Befunge? I'd wager a fair amount it's shortest solution of them all:

1

Not kidding. Try it yourself: http://www.quirkster.com/iano/js/befunge.html

EDIT: I guess I need to explain this one. The 1 operand pushes a 1 onto Befunge's internal stack and the lack of anything else puts it in a loop under the rules of the language.

Using the interpreter provided, you will eventually--and I mean eventually--hit a point where the Javascript array that represents the Befunge stack becomes too large for the browser to reallocate. If you had a simple Befunge interpreter with a smaller and bounded stack--as is the case with most of the languages below--this program would cause a more noticeable overflow faster.

You could also try this in C#.net

throw new StackOverflowException();

Nemerle:

This crashes the compiler with a StackOverflowException:

def o(){[o()]}

My current best (in x86 assembly) is:

push eax
jmp short $-1

which results in 3 bytes of object code (50 EB FD). For 16-bit code, this is also possible:

call $

which also results in 3 bytes (E8 FD FF).

PIC18

The PIC18 answer given by TK results in the following instructions (binary):

overflow
   PUSH
   0000 0000 0000 0101
   CALL overflow
   1110 1100 0000 0000
   0000 0000 0000 0000

However, CALL alone will perform a stack overflow:

CALL $
1110 1100 0000 0000
0000 0000 0000 0000

Smaller, faster PIC18

But RCALL (relative call) is smaller still (not global memory, so no need for the extra 2 bytes):

RCALL $
1101 1000 0000 0000

So the smallest on the PIC18 is a single instruction, 16 bits (two bytes). This would take 2 instruction cycles per loop. At 4 clock cycles per instruction cycle you've got 8 clock cycles. The PIC18 has a 31 level stack, so after the 32nd loop it will overflow the stack, in 256 clock cycles. At 64MHz, you would overflow the stack in 4 micro seconds and 2 bytes.

PIC16F5x (even smaller and faster)

However, the PIC16F5x series uses 12 bit instructions:

CALL $
1001 0000 0000

Again, two instruction cycles per loop, 4 clocks per instruction so 8 clock cycles per loop.

However, the PIC16F5x has a two level stack, so on the third loop it would overflow, in 24 instructions. At 20MHz, it would overflow in 1.2 micro seconds and 1.5 bytes.

Intel 4004

The Intel 4004 has an 8 bit call subroutine instruction:

CALL $
0101 0000

For the curious that corresponds to an ascii 'P'. With a 3 level stack that takes 24 clock cycles for a total of 32.4 micro seconds and one byte. (Unless you overclock your 4004 - come on, you know you want to.)

Which is as small as the befunge answer, but much, much faster than the befunge code running in current interpreters.