Can anyone explain this output to me?

So I was playing around with some thought experiments where I imagined what would happen when two functions became mutually recursive. One such one was what if both functions could potentially fall into an infinite loop.

To that end, I thought up of this simple example:

#include <iostream>
#include <cstdlib>

int foo(int x);
int bar(int x);


int foo(int x)
{
    return bar(x + 1);
}

int bar(int x)
{
    return foo(x - 1);
}

int main(int argc, char **argv)
{
    if (argc > 1)
        std::cout << "The value is: " << foo(atoi(argv[1])) << std::endl;

    return 0;
}

Interestingly, this will actually print out absolutely nothing if you compile it with g++. Compile it with any -O switch and it falls into an infinite loop.

Thoughts? Is this potentially a compiler bug, or is this to be expected? 开发者_高级运维I'd think in the case of -O optimizations, it would realize that foo(x) and bar(x) return just x.

I didn't expect it to actually "optimize away" the call and completely ignore printing out "The value is" to standard input.

Edit: I compiled this as just g++ source.cpp (-O1/2/3), under Cygwin using GCC 4.5.0. The -OX versions actually loop infinitely without the stack overflowing.

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Since you didn't specify which C++ you're on about I'll give the C++0x answer:

This is in fact undefined behaviour in C++0x.

I can't fully explain the behaviour if you're using C++03, but since your stack will eventually overflow, you're still really within the realm of implementation-dependent behaviour. Expecting anything sane to happen here is a fool's errand.

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From a mathematical point of view, f(x) = b(x + 1) and b(x) = f(x - 1) are not sufficient to define f and b. One possible solution is f(x) = x and b(x) = x - 1, while another one is f(x) = 0 and g(x) = 0 (and there are infinitely many more ways). So mathematically, the question of what the function should return doesn't have an answer. From a computational point of view, the situation is even worse, because after all, you are not defining mathematical functions, you are specifying how a computation is to be performed. foo(x - 1) means "perform the computations specified in the function foo() when we pass the value of x - 1 as a parameter", and similarly for bar(x + 1). So when each function says, unconditionally, that the other function is to be called, you have specified that you wish for an infinite recursion to take place. As others have mentioned, some languages (or language versions) treat this as a well-defined situation (where the desired outcome is that the program hangs), while others treat it as an undefined situation (i.e., the compiler may do anything it likes).

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As your recursion has no stopping condition it is only natural that you get into an infinite loop. Your assumption, that these functions return "just x" is false. They are bound to never return.

The compiler cannot just optimize away recursion.