What is the correct way to obtain (-1)^n?

Question

Many algorithms require to compute (-1)^n (both integer), usually as a factor in a series. That is, a factor that is -1 for odd n and 1 for even n. In a C++ environment, one often sees:

#include<iostream> #include<cmath> int main(){    int n = 13;    std::cout << std::pow(-1, n) << std::endl; } 

What is better or the usual convention? (or something else),

std::pow(-1, n) std::pow(-1, n%2) (n%2?-1:1) (1-2*(n%2))  // (gives incorrect value for negative n) 

EDIT:

In addition, user @SeverinPappadeux proposed another alternative based on (a global?) array lookups. My version of it is:

const int res[] {-1, 1, -1}; // three elements are needed for negative modulo results const int* const m1pow = res + 1;  ... m1pow[n%2] 

---

This is not probably not going to settle the question but, by using the emitted code we can discard some options.

First without optimization, the final contenders are:

   1 - ((n & 1) << 1); 

(7 operation, no memory access)

  mov eax, DWORD PTR [rbp-20]   add eax, eax   and eax, 2   mov edx, 1   sub edx, eax   mov eax, edx   mov DWORD PTR [rbp-16], eax 

and

   retvals[n&1]; 

(5 operations, memory --registers?-- access)

  mov eax, DWORD PTR [rbp-20]   and eax, 1   cdqe   mov eax, DWORD PTR main::retvals[0+rax*4]   mov DWORD PTR [rbp-8], eax 

Now with optimization (-O3)

   1 - ((n & 1) << 1); 

(4 operation, no memory access)

  add edx, edx   mov ebp, 1   and edx, 2   sub ebp, edx 

.

  retvals[n&1]; 

(4 operations, memory --registers?-- access)

  mov eax, edx   and eax, 1   movsx rcx, eax   mov r12d, DWORD PTR main::retvals[0+rcx*4] 

.

   n%2?-1:1; 

(4 operations, no memory access)

  cmp eax, 1   sbb ebx, ebx   and ebx, 2   sub ebx, 1 

The test are here. I had to some some acrobatics to have meaningful code that doesn't elide operations all together.

Conclusion (for now)

So at the end it depends on the level optimization and expressiveness:

• 1 - ((n & 1) << 1); is always good but not very expressive.
• retvals[n&1]; pays a price for memory access.
• n%2?-1:1; is expressive and good but only with optimization.

Answer 1

You can use (n & 1) instead of n % 2 and << 1 instead of * 2 if you want to be super-pedantic, er I mean optimized.
So the fastest way to compute in an 8086 processor is:

1 - ((n & 1) << 1)

I just want to clarify where this answer is coming from. The original poster alfC did an excellent job of posting a lot of different ways to compute (-1)^n some being faster than others.
Nowadays with processors being as fast as they are and optimizing compilers being as good as they are we usually value readability over the slight (even negligible) improvements from shaving a few CPU cycles from an operation.
There was a time when one pass compilers ruled the earth and MUL operations were new and decadent; in those days a power of 2 operation was an invitation for gratuitous optimization.