C ++上的指数的LLDB错误 [英] LLDB Error with Exponents on C++

问题描述

我更新了计算器代码，并添加了指数函数。但是，当我试图得到的方程的答案，我得到这个错误：
（lldb）
任何帮助将非常感谢，因为这是我的第一天与C + +！
是的，这就是！
这是我的代码！

I have updated my calculator code, and am adding an exponent function. However, when I try to get the answer to the equation, i get this error: (lldb) Any help would be greatly appreciated, as This is my first day with C++! Yep, that's all! Here's my code!

#include <math.h>
#include <iostream>
int int1, int2, answer;
bool bValue(true);
std::string oper;
std::string cont;
using namespace std;
std::string typeOfMath;
int a;
int b;
int answerExponent;


int main(int argc, const char * argv[]){

// Taking user input, the first number of the calculator, the operator, and second number. Addition, Substraction, Multiplication, Division
    cout<<"______________________________________________\n";
    cout<<"|Welcome to The ExpCalc! Do you want to do   |\n";
    cout<<"|Exponent Math, or Basic Math(+, -, X, %)    |\n";
    cout<<"|Type in 'B' for basic Math, and'E' for      |\n";
    cout<<"|Exponential Math! Enjoy! (C) John L. Carveth|\n";
    cout<<"|____________________________________________|\n";
    cin>> typeOfMath;
    if(typeOfMath == "Basic" ||
       typeOfMath == "basic" ||
       typeOfMath == "b" ||
       typeOfMath =="B")
    {
        cout << "Hello! Please Type in your first integer!\n";
        cin>> int1;
        cout<<"Great! Now Enter your Operation: ex. *, /, +, -...\n";
        cin>> oper;
        cout<<"Now all we need is the last int!\n";
        cin>> int2;

        if (oper == "+") {
            answer = int1 + int2;
        }
        if (oper == "-") {
            answer = int1 - int2;

        }if (oper == "*") {
            answer = int1 * int2;
        }if (oper == "/") {
            answer = int1 / int2;
        }
        cout<<answer << "\n";
        cout<<"Thanks for Using The ExpCalc!\n";

    }else if(typeOfMath == "Exp" ||typeOfMath == "E" ||typeOfMath == "e" ||typeOfMath == "Exponent"){
        cout<<"Enter the desired Base. Example: 2^3, where 2 is the base.\n";
        cin>> a;
        cout<<"Now what is the desired exponent/power of the base? Ex. 2^3 where 3 is the exponent!\n";
        cin>>b;
        answerExponent = (pow(a,b));
        cout<< answerExponent;
    } else(cout<<"Wrong String!");
}

请帮助！我可能会问很多问题，所以请不要疯了！我也在Mac上使用Xcode 4！

Please help! I will probably ask a lot of questions aswell, so please dont get mad! I am also on a Mac, using Xcode 4!

推荐答案

将此行添加到您的包含：

Add this line to your includes:

#include< string>

获取代码编译并运行正确的输出为2 ^ 3，在Visual Studio和GCC 4.7.2使用ideone.com（点击这里查看输出）。但是，我的编译器仍会发出警告，因为从 double 到 int 。更改此：

With that done, I am able to get the code to compile and run with the correct output for 2^3, in both Visual Studio, and in GCC 4.7.2 using ideone.com (click here to see the output). However, my compiler still emits a warning because of a conversion from double to int which you should probably attend to by casting. Change this:

answerExponent =（pow（a，b））;

到此：

answerExponent = static_cast< int>（pow（a，b））;

有了这个说法，编译器会发出这个警告的原因，通过投射你基本上只是告诉编译器关闭， 。一个更好的方法是避免铸造的需要。不要执行上述更改，请更改此行：

With that said, the compiler is emitting that warning for a reason, and by casting you are basically just telling the compiler to "shut up and do it anyway". A better approach is to avoid the need for a cast. Instead of doing the above change, change this line:

int answerExponent;

到此：

double answerExponent;

这样更有意义，因为调用 pow 和 double s作为参数没有什么意义如果你打算丢弃数字的小数部分后。

This makes more sense, because there is little point in calling pow with doubles as arguments if you are going to throw away the fractional part of the number afterwards.

这篇关于C ++上的指数的LLDB错误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IT屋！