错误：二元运算符的类型为'double'和'double'的操作数无效 [英] error: invalid operands of type 'double' and 'double' to binary operator

问题描述

我试图做以下问题，但我不断得到double和double类型的无效操作数的错误到二元运算符。我已经用我超级有限的能力切换了我能想到的一切，但仍然无法修复错误。谁能帮我？ （我在底部发布了我正在编写代码的问题）。非常感谢任何可以提供帮助的人



*编辑*我注意到当我用int切换双重代码时，我需要能够使用小数点当我使用浮动时，同样的错误发生在我身上

I was trying to do the following problem but I keep getting the error of invalid operands of type double and double to binary operator. I've switched everything I can think of with my super-limited abilities and still can't fix the error. Can anyone help me? (I posted the question at the very bottom that I am doing the code for). Thank you very much to anyone who can help

*edit* I noticed when i switch the double with int the code works but i need to be able to use decimal points.the same error happens to me when i use float

#include <cstdlib>
#include <iostream>
using namespace std;

/*
 * 
 */
int main(int argc, char** argv) {
    double principal1;
    double rate;
    double T;
    double amount;
    cout << "Enter principal1: ";
    cin >> principal1;
    cout << "Enter rate: ";
    cin >> rate;
    cout << "Enter times compounded annually: ";
    cin >> T;
    amount=(principal1)*((1+(rate/T))^T);
    cout << "The total amount is : " << amount << endl;
    
    return 0;
}

假设除原始投资外没有其他存款，余额为一年后的储蓄账户可以计算为

金额=委托人*（1 +利率/ T）^ T

本金是储蓄账户中的余额，比率是利率，T是利息在一年内复利的次数（如果利息按季度复利，则T为4）。

写一个要求本金，利率的程序，以及利息复合的次数。它应该显示类似的报告

编程挑战145
$ b $b兴趣费率：4.25

时间复合：4

本金：1000

利息：43.34

储蓄金额：1043.34

Assuming there are no deposits other than the original investment, the balance in a savings account after one year may be calculated as
Amount = Principal*(1 + Rate/T)^ T
Principal is the balance in the savings account, Rate is the interest rate, and T is the number of times the interest is compounded during a year (T is 4 if the interest is compounded quarterly).
Write a program that asks for the principal, the interest rate, and the number of times the interest is compounded. It should display a report similar to
Programming Challenges 145
Interest Rate:4.25
Times Compounded: 4
Principal:1000
Interest:43.34
Amount in Savings:1043.34

推荐答案

你没有显示编译错误所在的行。行 amount =（principal1）*（（1+（rate / T））^ T）; 这里有一个错误你不能在这里使用XOR ^运算符。如果您打算使用电源功能，请使用： http://www.cplusplus.com/reference/cmath/pow [ ^ ]。



另请参阅：

https：// msdn.microsoft.com/en-us/library/3akey979.aspx [ ^ ]。



编写代码而不阅读所用语言的基础知识会产生相反的效果。是不是很容易查阅一些参考网页（在这种情况下是MSDN）或书？我的建议是认真对待编程。

-SA

You did not show the line where the compilation error is. There is an error in the line amount=(principal1)*((1+(rate/T))^T); you cannot use XOR ^ operator here. If you meant to use power function, please use: http://www.cplusplus.com/reference/cmath/pow[^].

See also:
https://msdn.microsoft.com/en-us/library/3akey979.aspx[^].

It's very counter-productive to write code without reading on the basics of the language you use. Isn't it so easy to consult some reference Web page (MSDN, in this case) or a book? My advice is to get to programming seriously.

—SA

谢谢，我切换了我的代码如下，并使其工作。我不熟悉指数，现在我对他们很好





#include< cstdlib>

#include< iostream>

#include< cmath>

使用命名空间std;



/ *

*

* /

int main（）{

float principal1; //首席执行官

浮动利率; //利率

浮动T; //每年复合次数

浮动金额; //账户最终金额

浮动助手;

cout<< 输入principal1：;

cin>> principal1;

cout<< 输入率：;

cin>>率;

cout<< 输入时间每年复合：;

cin>> T;

helper =（1+（rate / T））; / *我不知道如何正确使用指数，所以我做了一个辅助变量的东西* /

amount =（principal1）* pow（helper，T）;

cout<< 一年后的平衡是：<<数量<<结束;



返回0;

}

Thank you , i switched my code to the following and made it work. I was unfamiliar with exponents and now I'm good with them


#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;

/*
*
*/
int main() {
float principal1; //starting principal
float rate; //interest rate
float T; //times compounded annually
float amount; //final amount in account
float helper;
cout << "Enter principal1: ";
cin >> principal1;
cout << "Enter rate: ";
cin >> rate;
cout << "Enter times compounded annually: ";
cin >> T;
helper=(1+(rate/T)); /*I didn't know how to properly use exponents with this so i made a helper variable thing */
amount=(principal1)* pow(helper,T);
cout << "The balance after a year is : " << amount << endl;

return 0;
}

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