#include <iostream>
using namespace std;
int recur(int x) {
1 and recur(--x);
cout << x;
return x;
}
int main() {
recur(10);
return 0;
}
views:
171answers:
4
+5
A:
Thats an infinite recursion. So it will seg fault when it runs out of stack space.
Naveen
2010-05-11 08:18:32
+2
A:
It doesn't have a termination condition for the recursion, and so will recurse until you run out of stack space.
Andreas Brinck
2010-05-11 08:19:03
+4
A:
1 and recur(--x);
is equivalent to
recur(--x);
Clearly you are making infinite recursive calls which leads to stack overflow followed by segmentation fault.
I guess you meant
x and recur(--x);
which makes the recursive call only when x is non-zero.
codaddict
2010-05-11 08:20:32
...and in a beautifully non-obvious way. Prefer the `if (x <= 0) return 0;` that tzaman suggested because it is more readable and is safer in the event that later code changes cause `x` to skip 0 on its way to negative infinity.
msw
2010-05-11 08:58:39
+2
A:
recur is an infinite loop; you need to put a base condition on there so it stops calling itself.
E.g. (at the top of the function) if (x <= 0) return 0;
Also, what's the point of the 1 and ? It's a no-op... maybe you meant x and, which would stop the recursion when x reached 0, provided you only ever called recur with a positive number (negative values would still cause the infinite loop).
tzaman
2010-05-11 08:20:43
Negative values won't cause an infinite loop, but you will likely still blow your stack, so its basically the same thing.
Dennis Zickefoose
2010-05-11 08:35:08
A recursive function isn't __looping,__ it's __recursing.__ Therefor it isn't an __endless loop,__ but an __endless recursion.__
sbi
2010-05-11 08:46:52