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Question

I have the following piece of code :

int f(int &x, int c){
    c = c - 1;
    if (c == 0) return 1;
    x = x + 1;
    return f(x, c)*x;
}

Now, suppose I call the above function like this :

int p = 5;
std::cout << f(p, p) << std::endl;

The output is 9^4 , since x is passed by reference, hence the final value of x should be 9, but when the return statement of the above function is changed to :

return x*f(x, c);

the output is 3024 (6*7*8*9) . Why is there a difference in output ? Has it anything to do with the order of evaluation of Operator* ? If we are asked to predict the output of the above piece of code, is it fixed, compiler-dependent or unspecified ?

Answer 1

When you write:

f(x,c)*x

the compiler may choose to retrieve the stored value in x (for the second operand) either before or after calling f . So there are many possible ways that execution could proceed. The compiler does not have to use any consistency in this choice.

To avoid the problem you could write:

auto x_temp = x;
return f(x, c) * x_temp;

Note: It is unspecified behaviour; not undefined behaviour because there is a sequence point before and after any function call (or in C++11 terminology, statements within a function are indeterminately-sequenced with respect to the calling code, not unsequenced).

Answer 2

The cause is that f() function has side effect on its x parameter. The variable passed to this parameter is incremented by the value of the second parameter c when the function returns.

Therefore when you swap the order of the operand, you get different results as x contains different values before and after the function is called.

However, note that behaviour of the code written in such way is undefined as compiler is free to swap evaluation of operand in any order. So it can behave differently on different platforms, compilers or even with different optimization settings. Because of that it's generally necessary to avoid such side effects. For details see http://en.cppreference.com/w/c/language/eval_order