问题描述
考虑以下代码:
#include <iostream>
#include <cmath>
int main() {
int i = 23;
int j = 1;
int base = 10;
int k = 2;
i += j * pow(base, k);
std::cout << i << std::endl;
}
它输出122"而不是123".它是 g++ 4.7.2 (MinGW, Windows XP) 中的错误吗?
It outputs "122" instead of "123". Is it a bug in g++ 4.7.2 (MinGW, Windows XP)?
推荐答案
std::pow() 适用于浮点数，它没有无限精度，并且您使用的标准库的实现可能实现 pow() 以一种(较差的)方式使这种缺乏无限精度变得相关.
std::pow() works with floating point numbers, which do not have infinite precision, and probably the implementation of the Standard Library you are using implements pow() in a (poor) way that makes this lack of infinite precision become relevant.
但是，您可以轻松定义自己的整数版本.在 C++11 中，您甚至可以将其设为 constexpr(以便尽可能在编译时计算结果):
However, you could easily define your own version that works with integers. In C++11, you can even make it constexpr (so that the result could be computed at compile-time when possible):
constexpr int int_pow(int b, int e)
{
return (e == 0) ? 1 : b * int_pow(b, e - 1);
}
这是一个现场示例.
尾递归形式(归功于 Dan Nissenbaum):
Tail-recursive form (credits to Dan Nissenbaum):
constexpr int int_pow(int b, int e, int res = 1)
{
return (e == 0) ? res : int_pow(b, e - 1, b * res);
}
这篇关于为什么我的 std::pow 整数数学给出了错误的答案?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！