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C++17 Folding expressions

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Folding expressions
Variadic Template(C++ 11)
재귀 함수 호출을 종료시켜주는 코드가
필요하다
Sum(T t) 함수가 가독성도 떨어지고 생
산성도 떨어지기 때문에 C++ 17에
Folding expressions 이 나옴
https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4191.html
https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4295.html
재귀 함수 호출을 종료시켜주는 코드가
없어도 됨
+ - * / % ^ & | = < > << >> += -= *=
/= %= ^= &= |= <<= >>= == != <= >= &&
|| , .* ->*
연산자 사용가능
https://en.cppreference.com/w/cpp/language/fold
unary right fold(단항 오른쪽 접기)
(E op ...) -> (E1 op (... op (EN-1 op EN)))
입력 : fun(1, 2, 3, 4, 5), 표현식 : t + …
(1 + (2 + (3 + (4 + 5)))
unary left fold(단항 왼쪽 접기)
(... op E) -> (((E1 op E2) op ...) op EN)
입력 : fun(1, 2, 3, 4, 5), 표현식 : … + t
((((1 + 2) + 3) +4) + 5)
unary fold-expression
print("단");print("짠");
과 같다
unary fold-expression
N이 0일 경우 기본값은 아래와 같이 들어가며 저기에 들어갈 수 없는 타입일 경우
에러가 난다
binary right fold(이항 오른쪽 접기)
(E op ... op I) -> (E1 op (... op (EN−1 op (EN op I))))
func1(3, 5, 7, 9)
->(3 + (5 + (7 + (9 + 1/*init*/))))
binary left fold(이항 왼쪽 접기)
(I op ... op E) -> ((((I op E1) op E2) op ...) op EN)
func2(3, 5, 7, 9)
->((((1/*init*/ + 3) + 5) + 7) + 9)
binary fold
binary fold
cast 보다 우선순위가 낮은 연산자가 있으면 안됨
()로 묶으면 해결된다
binary fold

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C++17 Folding expressions

  • 2. Variadic Template(C++ 11) 재귀 함수 호출을 종료시켜주는 코드가 필요하다 Sum(T t) 함수가 가독성도 떨어지고 생 산성도 떨어지기 때문에 C++ 17에 Folding expressions 이 나옴 https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4191.html https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n4295.html
  • 3. 재귀 함수 호출을 종료시켜주는 코드가 없어도 됨 + - * / % ^ & | = < > << >> += -= *= /= %= ^= &= |= <<= >>= == != <= >= && || , .* ->* 연산자 사용가능 https://en.cppreference.com/w/cpp/language/fold
  • 4. unary right fold(단항 오른쪽 접기) (E op ...) -> (E1 op (... op (EN-1 op EN))) 입력 : fun(1, 2, 3, 4, 5), 표현식 : t + … (1 + (2 + (3 + (4 + 5))) unary left fold(단항 왼쪽 접기) (... op E) -> (((E1 op E2) op ...) op EN) 입력 : fun(1, 2, 3, 4, 5), 표현식 : … + t ((((1 + 2) + 3) +4) + 5)
  • 6. unary fold-expression N이 0일 경우 기본값은 아래와 같이 들어가며 저기에 들어갈 수 없는 타입일 경우 에러가 난다
  • 7. binary right fold(이항 오른쪽 접기) (E op ... op I) -> (E1 op (... op (EN−1 op (EN op I)))) func1(3, 5, 7, 9) ->(3 + (5 + (7 + (9 + 1/*init*/)))) binary left fold(이항 왼쪽 접기) (I op ... op E) -> ((((I op E1) op E2) op ...) op EN) func2(3, 5, 7, 9) ->((((1/*init*/ + 3) + 5) + 7) + 9)
  • 9. binary fold cast 보다 우선순위가 낮은 연산자가 있으면 안됨 ()로 묶으면 해결된다