How does the C code that prints from 1 to 1000 without loops or conditional statements work?

问题

I've found C code that prints from 1 to 1000 without loops or conditionals : But I don't understand how it works. Can anyone go through the code and explain each line?

#include <stdio.h>
#include <stdlib.h>

void main(int j) {
  printf("%d\n", j);
  (&main + (&exit - &main)*(j/1000))(j+1);
}

回答1:

Don't ever write code like that.

For j<1000, j/1000 is zero (integer division). So:

(&main + (&exit - &main)*(j/1000))(j+1);

is equivalent to:

(&main + (&exit - &main)*0)(j+1);

Which is:

(&main)(j+1);

Which calls main with j+1.

If j == 1000, then the same lines comes out as:

(&main + (&exit - &main)*1)(j+1);

Which boils down to

(&exit)(j+1);

Which is exit(j+1) and leaves the program.

(&exit)(j+1) and exit(j+1) are essentially the same thing - quoting C99 §6.3.2.1/4:

A function designator is an expression that has function type. Except when it is the operand of the sizeof operator or the unary & operator, a function designator with type "function returning type" is converted to an expression that has type "pointer to function returning type".

exit is a function designator. Even without the unary & address-of operator, it is treated as a pointer to function. (The & just makes it explicit.)

And function calls are described in §6.5.2.2/1 and following:

The expression that denotes the called function shall have type pointer to function returning void or returning an object type other than an array type.

So exit(j+1) works because of the automatic conversion of the function type to a pointer-to-function type, and (&exit)(j+1) works as well with an explicit conversion to a pointer-to-function type.

That being said, the above code is not conforming (main takes either two arguments or none at all), and &exit - &main is, I believe, undefined according to §6.5.6/9:

When two pointers are subtracted, both shall point to elements of the same array object, or one past the last element of the array object; ...

The addition (&main + ...) would be valid in itself, and could be used, if the quantity added was zero, since §6.5.6/7 says:

For the purposes of these operators, a pointer to an object that is not an element of an array behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type.

So adding zero to &main would be ok (but not much use).

回答2:

It uses recursion, pointer arithmetic, and exploits the rounding behavior of integer division.

The j/1000 term rounds down to 0 for all j < 1000; once j reaches 1000, it evaluates to 1.

Now if you have a + (b - a) * n, where n is either 0 or 1, you end up with a if n == 0, and b if n == 1. Using &main (the address of main()) and &exit for a and b, the term (&main + (&exit - &main) * (j/1000)) returns &main when j is below 1000, &exit otherwise. The resulting function pointer is then fed the argument j+1.

This whole construct results in recursive behavior: while j is below 1000, main calls itself recursively; when j reaches 1000, it calls exit instead, making the program exit with exit code 1001 (which is kind of dirty, but works).

来源：https://stackoverflow.com/questions/7937789/how-does-the-c-code-that-prints-from-1-to-1000-without-loops-or-conditional-stat

标签

function-pointers