Could someone explain how this code works?
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David
el 10 de Oct. de 2013
Comentada: Cedric
el 12 de Oct. de 2013
So, we were given this as an example of recursion. It's a program that computes the factorial of a given scalar.
function f = factorial(x)
temp = 0;
if (x == 1)
temp = 1;
else
temp = x*factorial(x-1);
end
I get that x is inputted and if it's value is 0, then it returns 1 right off the bat. However, if x is not equal to 1 then the operation temp = x * factorial(x - 1); is carried out. So, x - 1 is then passed to factorial(), this is where I'm lost. What happens now? How does the value of temp not end up as 0, seeing as it's reset to that at the beginning of the function? Also, how does the code know when to stop and return the result?
f = temp;
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Cedric
el 10 de Oct. de 2013
Editada: Cedric
el 11 de Oct. de 2013
The line
temp = 0 ;
is useless, and the temp which appear afterwards should be f, the output argument.
Hint: consider all "instances" of the factorial function as different, and draw a schematics, e.g.
x = factorial(3)
-> factorial, x=3 [1st instance]
| ..
| else
| f = x * factorial(3-1)
| -> factorial, x=2 [2nd instance]
| | ..
| | else
| | f = x * factorial(2-1)
| | -> factorial, x=1 [3rd in.]
| | | if x == 1
| | | f = 1 ;
| | <-
| | so f = 2 * 1
| <-
| so f = 3 * 2
so x = 6
EDIT: I just realized that there was a typo that you spotted!
This was not
factorial(3 - 2)
but
factorial(3 - 1)
Thank you, I made the correction.
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Más respuestas (1)
Youssef Khmou
el 10 de Oct. de 2013
David, The variable temp is local (inside the function), as long as the iterative variable x didnt arrive at 1 the process continues, "temp" is set only inside function , N=4 :
inside func insde func
N=4 -> (Temp=0,..,Temp=3)-->N=3 (Temp=0,Temp=2) .....N=1
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