c ++中的整数的幂 [英] power of an integer in c++

问题描述

我需要从 pow（a，b）获得结果作为整数（a和b都是整数）。当前包含（int）pow（（double）a，（double）b）的计算是错误的。也许有人可以帮助一个函数，用pow（a，b）与整数，并返回一个整数吗？

但这里是奇怪的部分：在Linux中使用Geany（和g ++ / gcc编译器）并且只有 pow（a，b）脚本编译和工作正常。但在大学，我有Dev-C ++（和MS Windows）。在Dev-C ++中，脚本没有编译时出现错误 [Warning]从 double'转换为 int' / p>

我需要使这个scrpit在Windows下（和Mingw编译器）工作。

/ p>

-skazhy

解决方案

  int myPow（int x，int p）{
 if（p == 0）return 1; 
 if（p == 1）return x; 
 return x * myPow（x，p-1）; 
} 
  

I need to get the result from pow(a,b) as an integer (both a and b are integers too). currently the calculations where (int) pow( (double)a, (double)b) is included are wrong. Maybe someone can help with a function that does the pow(a,b) with integers and returns an integer too?

But here is the odd part: I made my script in Linux with Geany (and g++/gcc compiler) and had just pow(a,b) the script compiled and worked fine. But in university I have Dev-C++ (and MS Windows). In Dev-C++ the script didn't compile with an error [Warning] converting to int' from double'

I need to make this scrpit work under Windows (and Mingw compiler) too.

Thanks in advance,

-skazhy

解决方案

A nice recursive approach you can show off:

int myPow(int x, int p) {
  if (p == 0) return 1;
  if (p == 1) return x;
  return x * myPow(x, p-1);
}

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