巧用数组减少if语句

假设我们要写一个判断用户输入整数奇偶性的程式，能够用以下代码实现：

long num;

cin >> num;
if ( num % 2L ) {
cout << "Odd!\n";
} else {
cout << "Even!\n";
}

但是这个 if 判断是没必要的，我们能够利用数组来避免：

const char *msg[] = { "Even", "Odd" };
long num;

cin >> num;
cout << msg[num & 1L] << '\n'; // can also be `num % 2'

这种方法不但对这个程式有效，不少别的应用也能够使用这个方法来减少 if 语句。这个方法也能够用于 C 程式。以下是奇偶判断程式完整代码：

/*
* FileName: odd_or_even.cpp
* Author: Antigloss at http://stdcpp.cn
* LastModifiedDate: 2005-7-22 22:30
* Purpose: Tell if a given number is odd or even
*/

#include <cstdlib> // for EXIT_SUCCESS
#include <iostream>
#include <limits> // for numeric_limits

// flush the input buffer
inline void flush_stdin()
{
std::cin.clear(); // clear error state of the stream
// clear data left at the input buffer
std::cin.ignore( std::numeric_limits< std::streamsize >::max(), '\n' );
} // end of flush_stdin

int main()
{
long num;
const char *msg[] = { "Even", "Odd" };

for (;;) {
std::cout << "Please input an integer(q to end): ";

if ( std::cin >> num ) {
std::cout << msg[num & 1L] << '\n'; // we can also use `num % 2L'
} else {
std::cin.clear(); // clear error state before reading from the input stream
if ( std::cin.get() == 'q' ) {
flush_stdin();
break;
}
std::cerr << "You should input an INTEGER!\n";
}
flush_stdin();
}

std::cout << "Thanks for using our product!\nPress ENTER to quit...";
std::cin.get();
return EXIT_SUCCESS;
}