Tre*_*135 -10 c++ calculator pow
我正在制作一个简单的计算器,你可以选择一个功能,然后2个输入来得到你的答案.这是一个简洁的小程序,除了权力之外,一切都运行顺畅.每个号码都正常工作.
但根据这个:5^2=24, 5^3=624.我在用pow(number1,number2).
#include <iostream>
#include<math.h>
using namespace std;
int main()
{
for(;;){
int func;
int number1;
int number2;
cout << "Input your function (+,-,x,/,Square,Power)(1,2,3,4,5,6) ";
cin >> func;
cout << "input #1: ";
cin >> number1;
cout << "input #2: ";
cin >> number2;
if (func==1){
int answer;
answer = number1 + number2;
cout << number1 << " + " << number2 << " = " << answer << endl;
}
else {
if (func==2){
int answer;
answer = number1 - number2;
cout << number1 << " - " << number2 << " = " << answer << endl;
}
else {
if (func==3){
int answer;
answer = number1 * number2;
cout << number1 << " x " << number2 << " = " << answer << endl;
}
else {
if (func==4){
int answer;
answer = number1 / number2;
int R = number1 % number2;
cout << number1 << " / " << number2<< " = " << answer << " R "<< R << endl;
}
else {
if (func==5){
int answer;
answer = pow(number1,0.5);
cout << "?" << number1 << "=" << answer << endl;
}
else {
if (func==6){
int answer;
answer = pow(number1,number2);
cout << "?" << number1 << "^" << number2 << "=" << answer << endl;
}
}
}
}
}
}
}
}
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pow()将数字作为浮点返回.真正回归的是24.99997或类似的东西.但是你通过将它分配给一个整数来截断它并且它出现为24.在这种情况下可以分配给一个整数,因为你知道答案将永远是一个整数所以你需要对它进行舍入 - 而不是截断它.舍入为整数的常规方法是添加0.5,如下所示:
int answer;
answer = pow(number1,number2)+0.5;
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另一方面,平方根可能不应该分配给一个整数 - 几乎每个平方根都有小数位.
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