Reputation: 4662
I am playing around with some recursion, and trying to count the number of additions in a recursive fib function (code below). The problem is, g_add
never gets printed as anything but zero (0). In the debugger it is set correctly, but it wont print correctly. In fact, the code below is modified a bit, to sanity check that everything else is OK. In real life, g_add is set to zero initially, not to the 10 below, but look at the output is printing...
volatile int g_add = 10;
int rfib(int n)
{
if(n == 0) return 0;
else if(n == 1) return 1;
else {
++g_add;
return rfib(n-1) + rfib(n-2);
}
}
int _tmain(int argc, _TCHAR* argv[])
{
printf("Fib: %d\n", g_add);
for(int n =0; n<6;n++,g_add = 0)
{
printf("Fib %d is: %d - additions: %d\n", n, rfib(n), g_add);
}
}
And the output:
Fib: 10
Fib 0 is: 0 - additions: 10
Fib 1 is: 1 - additions: 0
Fib 2 is: 1 - additions: 0
Fib 3 is: 2 - additions: 0 (note, the debugger says it is 1)
Fib 4 is: 3 - additions: 0
Fib 5 is: 5 - additions: 0
Any thoughts on why g_add is not being printed correctly? And what I could do about it? I have tried with and without the volatile keyword. I think this might be related to the VS environment, rather than C++ so, I tagged both.
Upvotes: 0
Views: 283
Reputation: 264571
Order of evaluation of parameters to a function is not specified in the standard.
This compilers are allowed to do appropraite optimisations.
See: What are all the common undefined behaviours that a C++ programmer should know about?
In your situation.
Upvotes: 1
Reputation: 1142
You're assuming the order in which the parameters are evaluated.
Use this instead:
int fib = rfib(n);
printf("Fib %d is: %d - additions: %d\n", n, fib, g_add);
Upvotes: 6