Fake infinite loop [closed]

Question

The challenge is to write a fake infinite loop in any imperative programming language.

At first sight, the code should seem not to halt at all.

As a naive example:

int f(int x)
{
    return f(4.0);
}

int f(double x)
{
    return 0;
}

But remember
You're not allowed to use overflows:

int f(int x)
{
    if(x < 0) return -1;
    return f(x*x);
}

is illegal. When the number exceeds the integer limit, the function will stop. It's a good trick. But not good enough :)

The answer with the highest votes wins! Duration is 1 week.

Answer 1

Bash

trap 'exec "$0"' exit

It re-execs itself when it should exit. So it will never really exit.

It doesn't have the shebang. So you must run it like bash a.sh. And $0 will probably not be something executable.