This program was compiled on MS Visual C++ 08 /*Fibonacci Numbers*/ #include<stdio.h> #include<limits.h> void fibonacci(int n) { unsigned long long fib0 = 0; /*First Fibonacci Number*/ unsigned long long fib1 = 1; /*Second Fibonacci Number*/ unsigned long long fibn = 1; /*Nth Fibonacci Number*/ int count = 3; /*Hold Count*/ printf(" 1 :%25llu \n 2 :%25lld \n",fib0,fib1); while(count <= n ) { fibn = fib0 + fib1 ; if((fibn < 0) || (fibn > ULLONG_MAX)){ puts("\nOverflow\n"); break; } printf("%3d :%25llu \n",count,fibn); fib0 = fib1; fib1 = fibn; count++; } return ; } int main(void) { unsigned long temp = 0; puts("Fibonacci Numbers"); fibonacci(100); /*Print the first 100 Fibonacci Numbers*/ return 0; } This is a part of the output : Fibonacci Numbers ...snip... 90 : 1779979416004714189 91 : 2880067194370816120 92 : 4660046610375530309 93 : 7540113804746346429 94 : 12200160415121876738 95 : 1293530146158671551 96 : 13493690561280548289 97 : 14787220707439219840 98 : 9834167195010216513 99 : 6174643828739884737 100 : 16008811023750101250 Why are the numbers after 95th Fibonacci numbers (including it) wrong?

Tarique wrote: ) <snip> ) unsigned long long fibn = 1; /*Nth Fibonacci Number*/ ) <snip> ) if((fibn < 0) || (fibn > ULLONG_MAX)){ This can never happen. Some compilers would even warn about an if condition never being true, or about unreachable code, or something similar. ) puts("\nOverflow\n"); ) <snip> SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT

Why do you treat fib1 as long long when it is declared as unsigned long long? How can 'fibn' be less than zero when it is an unsigned type? Also how can it be greater than ULLONG_MAX? One method would be: if (ULLONG_MAX - fib1 < fib0) { puts("Overflow."); break; } Why the precision specifiers? Are you sure about the output?

Here you need if(fibn < fib0 || fibn < fib1) unsigned numbers wrap silently. You need a huge integer library to calculate high Fibonacci numbers effectively.

Try this modification: #include<stdio.h> #include<limits.h> void fibonacci(int n) { unsigned long long fib0 = 0; /*First Fibonacci Number*/ unsigned long long fib1 = 1; /*Second Fibonacci Number*/ unsigned long long fibn = 1; /*Nth Fibonacci Number*/ int count = 3; /*Hold Count*/ printf(" 1 :%25llu \n 2 :%25llu \n",fib0,fib1); while(count <= n ) { fibn = fib0 + fib1 ; /* if((fibn < 0) || (fibn > ULLONG_MAX)){ puts("\nOverflow\n"); break; } */ if (ULLONG_MAX - fib1 < fib0) { puts("Overflow!"); break; } printf("%3d :%25llu \n",count,fibn); fib0 = fib1; fib1 = fibn; count++; } return ; } int main(void) { unsigned long temp = 0; puts("Fibonacci Numbers"); fibonacci(100); /*Print the first 100 Fibonacci Numbers*/ return 0; } Relavant output is: 88 : 679891637638612258 89 : 1100087778366101931 90 : 1779979416004714189 91 : 2880067194370816120 92 : 4660046610375530309 93 : 7540113804746346429 94 : 12200160415121876738 Overflow!

Overlooked that..changed it Initially i was using a long long integer,but then changed it to unsigned int. Did not remove the fibn < 0 check. Since i was getting -ve numbers as output(some of them..which was obviously due to overflow),it seemed to be at least a temporary fix! It's a little easier to actually add any two numbers in the output when they are right aligned! Well yes.I did check the numbers prior to 90,the smaller ones are easier to check...did some random checks with larger numbers. The 95th one is obviously wrong! It is smaller than the 94th one.

<snip> Unsigned numbers cannot overflow in C. As for signed values, you must check for possible overflow /before/ the suspect calculation. Once overflow has occured the behaviour of your program is undefined. Some compilers have an option to enable overflow detection. This might be easier than doing so manually before every calculation.

Umm..I am combining two questions together. These were the suggestions : 1. if(fibn < fib0 || fibn < fib1) from Mr.Malcolm McLean 2. if (ULLONG_MAX - fib1 < fib0) from Santosh Can you please explain the logic ?

fibn is set to fib0 + fib1. So if fibn is less than either, some overflow must have occurred. If greater than either, there cannot be overflow. This holds true for any two positive integers represented by a fixed number of bits. Santosh is saying effectively the same thing. The overflow occurs if fib1 + fib0 > ULLONG_MAX. However we cannot sum fib0 and fib1, because that in itself woyuld give overflow. So he rearranges the equation.

Tarique wrote: ) Umm..I am combining two questions together. ) ) These were the suggestions : ) 1. if(fibn < fib0 || fibn < fib1) from Mr.Malcolm McLean Actually, if(fibn < fib0) is enough. If overflow occurs, then fibn will be like this: fibn = (fib0 + fib1) - (ULLONG_MAX + 1) You can algebraically rewrite this to: fibn = fib0 - (ULLONG_MAX+1 - fib1) Knowing that fib1 is smaller than ULLONG_MAX+1, you can deduce that if overflow occurs, fibn < fib0. Same holds for fibn < fib1, symmetrically. ) 2. if (ULLONG_MAX - fib1 < fib0) from Santosh You really want to check: if ((fib0 + fib1) > ULLONG_MAX) But that will not work because of overflow. If you rewrite it algebraically, you get the above comparison. (Move fib1 to the right of the comparator.) SaSW, Willem -- Disclaimer: I am in no way responsible for any of the statements made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you ! #EOT

In addition to Malcolm's and Willem's explanations also note that method one is used after the concerned calculation while method 2 can be used before. But this doesn't matter for unsigned calculations.

Apparently the C standard says that unsigned arithmetic does not overflow, therefore the problem in your code is nothing to do with overflow. Even though the problem in your code clearly *is* to do with overflowing the range of your datatype. In this case, I think you can test for overflow by making the sure each successive fibonacci number is > the previous number. If you are particularly interesting in calculating big fibonaccis, try using double datatype. These will be approximate.

Bartc said: Right. It is, instead, to do with the OP's apparent belief that standard integer types are infinitely wide. No, unsigned integer arithmetic doesn't overflow, any more than a clock overflows at midnight. Or use, or even write and then use, a bignum library.

I would like to differ here. If i believed that standard integer types are infinitely wide,i would not have bothered to at least try to check if bounds were transgressed. The problem was that I was using (signed) long long integer instead,as I have mentioned earlier in the thread.When i changed that to unsigned type,i should have changed the *failure* condition. I definitely learned it today!

Tarique wrote: [...] [...] Because you test for overflow after the overflow has occured. Note that fibn is unsigned so (fibn < 0) can never be true, and fibn is an unsigned long long so (fibn > ULLONG_MAX) can never be true, so your test amounts to if (0) { /* dead code */ } Consider how your test differs from this while(count <= n ) { if(fib1 > ULLONG_MAX - fib0) { puts("\nOverflow\n"); break; } fibn = fib0 + fib1 ; where fib0 must be less than or equal to ULLONG_MAX (since it is an unsigned long long), so ULLONG_MAX - fib0 always has a defined value for which the test against fib1 makes sense. Remember that fib1 > ULLONG_MAX - fib0 (which can be true) and fib1 + fib0 > ULLONG_MAX (which can never be true) do not mean the same thing.

The overflow/wraparound thing has been discussed here before. Clearly I think that 'overflow' is a more appropriate term for this behaviour than 'wraparound', but the C standard dictates otherwise. But, when I add 2 unsigned ints on my machine, in assembler, sometimes the Carry flag gets set. OK, that's not in the province of C, but to me it says 'Overflow'. It's a piece of useful information ignored by C (possibly, why overflow itself is ignored). The clock would have the same problems, if you're trying to measure time and it doesn't have a day counter. There are many examples in everyday life of counters that 'wrap' yet it would be unwise to ignore the obvious overflow that has occurred: electric meters apparently showing a large negative kWh consumption, decrepit cars with just 35 miles/km on the odometer, etc.

Actually, if you want to check an addition c = a + b for wrapping, you only need to check _either_ (c < a) _or_ (c < b). You don't need to check both. If one is true then the other must be true.