global variable is not incrementing properly - c

Question

int count;
int main(){
count=0;
printf("%6d\t %6d\t %6d\t ",i+1,multiFib(i),count);
}
int multiFib(int x){
if (x<2){
return 1;
count++;
}}
In this code, count always gives 0 result. I made it a global variable, not to define it in the function again. How can I increase that variable?
It outputs in order; that is not problem. It gives the result:
1 1 0
-
2 1 0
-
3 2 0
So, count is not increasing — but why?

Answer 1

Although order of evaluation is not guaranteed (see multiple other answers), there is another, more fundamental problem (actually, two of them):
int multiFib(int x)
{
if (x<2){
return 1;
count++;
}
}
You attempt to increment count after you've returned from your function. Your compiler should be warning about unreachable code.
You don't return a value from multiFib() when the input is bigger than 2.
The fragment of main() that you show references an undefined variable i. I assume that is because you deleted a for (i = 0; i < 40; i++) loop from before the printf() statement. To fix the 'order of evaluation' problem, you might want to use:
int count;
int main(void)
{
for (int i = 0; i < 40; i++)
{
count = 0;
int fib = multiFib(i);
printf("%3d %10d %10d\n", i+1, fib, count);
}
return(0);
}
This evaluates the number of calls to multiFib() for each top-level call to the function.
For the multiFib() function itself, you probably need:
int multiFib(int x)
{
count++;
if (x < 2)
return x;
else
return multiFib(x-1) + multiFib(x-2);
}
Though this is an appallingly slow implementation of Fibonacci numbers so you shouldn't use this as an answer to your homework (but it does make a 'working' function).
Example output:
1 0 1
2 1 1
3 1 3
4 2 5
5 3 9
6 5 15
7 8 25
8 13 41
9 21 67
10 34 109
11 55 177
12 89 287
13 144 465
14 233 753
15 377 1219
16 610 1973
17 987 3193
18 1597 5167
19 2584 8361
20 4181 13529
21 6765 21891
22 10946 35421
23 17711 57313
24 28657 92735
25 46368 150049
26 75025 242785
27 121393 392835
28 196418 635621
29 317811 1028457
30 514229 1664079
31 832040 2692537
32 1346269 4356617
33 2178309 7049155
34 3524578 11405773
35 5702887 18454929
36 9227465 29860703
37 14930352 48315633
38 24157817 78176337
39 39088169 126491971
40 63245986 204668309
Note that if you go much beyond 40, you start getting arithmetic overflow. It also gets to be very (very, very) slow.

Answer 2

This:
if (x<2){
return 1;
count++;
}
You return from the function before you increment count. So it's never incremented.
Return doesn't just set the returned value. It actually exits the function at that point.
Just swap the order of those two lines. Also, you should keep an eye on your compiler warnings. Any decent compiler will warn you that the count++; line is unreachable.

Answer 3

printf("%6d\t %6d\t %6d\t ",i+1,multiFib(i),count);
The order of function argument evaluation is unspecified in C. multiFib can be evaluated before or after count in the printf function call.

Answer 4

The order of evaluation of function arguments is not specified.
In your code,
printf("%6d\t %6d\t %6d\t ",i+1,multiFib(i),count);
The argument to printf function multiFib(i) might be getting called after printing count.
Try this,
printf("%6d\t",multiFib(i));
printf("%6d\t %6d\t ",i+1,count);
to check the result.

Answer 5

As it was said in another answers, problem is that order of evaluation of function arguments is not specified, so in your example:
printf("%6d\t %6d\t %6d\t ",i+1,multiFib(i),count);
You can not be sure if count will be evaluated before or after passing multiFib(i) to the printf() function.
Please look here for more detailed information about evaluation order in C++, or here for evaluation order in C.
You might also find useful to force evaluation order by separating your printf into two calls:
printf("%6d\t",multiFib(i));
printf("%6d\t %6d\t ",i+1,count);
like Nishith Jain M R suggested in his answer.