Question

使用较长版本(something).operator[]()而不仅仅是(something)[]有什么好处？

例如：

std::array<int, 10> arr1;
std::array<int, 10> arr2;

for(int i = 0; i < arr1.size(); i++) 
    std::cout << arr1[i] << ' '; 
    std::cout << std::endl;

for(int i = 0; i < arr2.size(); i++) 
    std::cout << arr2.operator[](i) << ' '; 
    std::cout << std::endl;

Answer 1

没有。 []只是用户定义类型operator[]的语法糖。您自己定义这些函数时只需要operator语法。这适用于所有运营商，例如operator()，operator[]，operator new，operator=，...

Answer 2

语法糖

使用g ++ -g -std = gnu ++ 0x ...

编译

0000000000400554 <main>:
#include <array>

int main() {
  400554:       55                      push   %rbp
  400555:       48 89 e5                mov    %rsp,%rbp
  400558:       48 83 ec 60             sub    $0x60,%rsp

    std::array<int, 10> arr1;
    std::array<int, 10> arr2;

    arr1[6];
  40055c:       48 8d 45 d0             lea    -0x30(%rbp),%rax
  400560:       be 06 00 00 00          mov    $0x6,%esi
  400565:       48 89 c7                mov    %rax,%rdi
  400568:       e8 19 00 00 00          callq  400586 <std::array<int, 10ul>::operator[](unsigned long)>

    arr2.operator[](6);
  40056d:       48 8d 45 a0             lea    -0x60(%rbp),%rax
  400571:       be 06 00 00 00          mov    $0x6,%esi
  400576:       48 89 c7                mov    %rax,%rdi
  400579:       e8 08 00 00 00          callq  400586 <std::array<int, 10ul>::operator[](unsigned long)>
  40057e:       b8 00 00 00 00          mov    $0x0,%eax
}
  400583:       c9                      leaveq
  400584:       c3                      retq
  400585:       90                      nop

运营商[]长短版本

2 个答案: