两个宏的 C/C++ 幂

Question

我的二能力不如我应该的那么好，所以我想也许我可以#define 一些东西。

不幸的是，我对预处理器指令非常缺乏经验，而且我无法弄清楚如何执行诸如 for 循环之类的事情。我看了看：

• http://www.cplusplus.com/doc/tutorial/preprocessor
• https://msdn.microsoft.com/en-us/library/teas0593.aspx

但是他们都没有 for 循环的例子。我想要的只是能够写出pwrtwo(5) 之类的东西，而不是使用计算器算出 2⁵ 是 32。

Answer 1

为什么不正确地使用函数呢？这甚至允许我们在编译时使用强大的 constexpr 生成结果！

template <class T>
constexpr T pwrtwo(T exponent) {
    return (T(1) << exponent);
}

Answer 2

一个答案（不正确）表明答案应该是#define pwrtwo(x) (1 << ((x) - 1)。然而：

1 = 0000 0001 二进制模式

现在，当它要求 2 次方 5 时，然后

1 ;其中 x = 5

1 应该移动 4 次然后结果将是

0001 0000 ==> 2 次方 4

但这是错误的，这就是为什么实际答案应该是：

#define pwrtwo(x) (1 << (x))

Answer 3

2^x 是1 << x，所以：

#define pwrtwo(x) (1 << (x))

Answer 4

这个宏可以完成这项工作：

#ifndef PWRTWO
#define PWRTWO(EXP) (1 << (EXP))
#endif

但是，建议不要使用它，除非您限制或限制 EXP 的值范围！

尝试在int 的大小为 4 个字节的机器上将这些值输入到此宏中，尝试编译并运行，然后告诉我每个值会发生什么：

std::cout << PWRTWO(30) << std::endl; // Okay
std::cout << PWRTWO(31) << std::endl; // - value

// In the following cout statement within the macro, do not confuse the 
// (x >= 32) as what is actually being passed into the macro as an argument. 
// This is just a short hand notation to represent all values >= 32 where the actual
// numerical value would be entered into this macro statement as there is
// no defined variable x here in this context nor any comparison expression being passed to it.

// Compiler Warning C4293 '<<' shift count negative or two big, undefined behavoir
std::cout << PWRTWO(X >= 32) << std::endl; // In most cases on my machine it prints out 0. 
//However, since this is undefined behavior, there is no telling what it could or may do on another machine.

编辑

// Try this for loop to see the actual values being printed out as long as 
// the integer value on your machine is 32bit or 4 bytes in size.
std::cout << "Excepted values for 32bit integers" << std::endl;
std::cout << "----------------------------------\n";
for ( int i = 0; i < 31; i++ ) {
    std::cout << PWRTWO( i ) << std::endl;
}
std::cout << std::endl;

// Then print out the next one
std::cout << "First value to produce a negative result with int being 32bit." << std::endl;
std::cout << "------------------------------------------\n";
std::cout << PWRTWO( 31 ) << std::endl << std::endl;

// Then print out these as well : compiler warnings
std::cout << "Value Range that generates a compiler error." << std::endl;
std::cout << "-------------------------------------------\n";
for ( int i = 32; i <= 100; i++ ) {
    std::cout << PWRTWO( i ) << std::endl;
}