First please view below codes...
I was so confuses with recursion function...I mean how could the output will be 2? I hope that someone could explain how recursion works in details to me please? Very thanks you.Code:#include <stdio.h> int strange (int n); main(void) { int n = 4, result; result = strange (n); printf("%d", result); } int strange (int n) { int ans; if (n ==1) ans = 0; else ans = 1 + strange (n/2); return (ans); }
Have you tried tracing the code yet? Note down the value of n and ans for the current call of strange.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
If you follow your code properly it will be 2. Starting off with n = 4, since n != 1 you send it back through and divide it by 2. So now n = 2, but since its != 1 you send it back through again and divide it by 2. Now n = 1 so you return 0. As you follow your code back up you return 1 + 0 = 1. Then going up another step you return 1 + 1 which = 2 thus your final answer is 2. What did you expect it to return?
Ya, of course I try to trace this program before but I still can't obtain the correct output...Now I understand some but why should I return
1 + 0 = 1 then going up to 1 + 1 again ?
Ok...now I fully understand how this program works....but I face another more difficult recursion
I try to understand this program clearly and trace out...but I still doesn't get why the output is 8 but not 9?Code:#include <stdio.h> int FUNCTION (int n); main(void) { int n = 6, valid; valid = FUNCTION (n); printf("%d", valid); } int FUNCTION (int n) { if ( n == 1) return 1; else if (n<= 0) return 0; else return ( FUNCTION ( n - 2) + FUNCTION (n - 1) ) ; }
Brush up again ur concepts....u are mugging up and not getting the concepts...if u FULLY UNDERSTAND THE PREVIOUS ONE...then this is also of the same type
=8Code:n=6 FUNCTION ( n - 2) + FUNCTION (n - 1) =FUNCTION ( 4) + FUNCTION (5) =2FUNCTION (4) + FUNCTION (3) =2{FUNCTION (2)+FUNCTION (3)} + FUNCTION (3) =2FUNCTION (0)+2FUNCTION (1)+FUNCTION (2)} + FUNCTION (2) + FUNCTION (1) =2FUNCTION (0)+2FUNCTION (1)+FUNCTION (2)} + FUNCTION (0) + FUNCTION (1) + FUNCTION (1) =2(0+2+1)+0+1+1 =6+1+1
" I failed in some subjects in exam , but my friend passed in all . Now he is an engineer in Microsoft and I am the owner of Microsoft !! "
- Bill Gates .
Ok...I try to understand all the solution you post to me but I seems can't understand start from line 4. I mean where the 2 come from? And why the 5 will become 3...? Sorry if it sounds like I didn't try my best to understand it but I really does try to understand it...
Maybe try an easier example:Code:int SOME_FUNC(int n) { if (n <= 1) return 1; else return n * SOME_FUNC(n-1); } int main(void) { int result = SOME_FUNC(5); printf("%d", result); }
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Advice: Take only as directed - If symptoms persist, please see your debugger
Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
u have to break each n every
FUNCTION(n)=FUNCTION(n-1)+FUNCTION(n-2)
so
FUNCTION(5)=FUNCTION(4)+FUNCTION(3)
" I failed in some subjects in exam , but my friend passed in all . Now he is an engineer in Microsoft and I am the owner of Microsoft !! "
- Bill Gates .
And for reference, here's a non-recursive version of SOME_FUNC:If you can understand this one, then you'll understand what the other one is doing.Code:int SOME_FUNC(int n) { int output = 1; while (n > 1) { output *= n; --n; } return output; }
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Advice: Take only as directed - If symptoms persist, please see your debugger
Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
