C++ 错误 2679 : binary'>>' : no operator found

Question

我在构建程序时收到以下错误消息:

errorC2657: binary '>>': no operator found which takes a right hand operand of type'const char[2]'(or there is no acceptable conversion)

这是我的程序的源代码:

#include <iostream>
#include <string> 

using namespace std;

int main()
{
int number1;
int number2;
int intSum;
int intdiff;
int intprod;
int intaver;
cout << "Enter the 1st and 2nd ODD integers (separate them by a space):     ";
cin >> number1 >>" " >> number2;
intSum = abs(number1 + number2);
intdiff = abs(number1 - number2);
intprod = abs(number1 * number2);
intaver = abs((number1 + number2) / 2);
if ( number1%2 != 0)
    cout << "You must enter ODD integers! See you next time." <<endl;
else if ( number2%2 != 0)

    cout << "You must enter ODD integers! See you next time." <<endl;
else
{
    cout << "The sum of the integers is"<<intSum<<endl;
    cout << "The difference of the integers is"<<intdiff<<endl;
    cout << "The product of the integers is"<<intprod<<endl;
    cout << "The average of the integers is"<<intaver<<endl;   
    if (number1 > number2)
        cout << "The first one "<< number1 <<" is bigger." <<endl;
    else if (number2 > number1)
        cout << "The second one "<< number2 <<" is bigger." <<endl;
    else
        cout <<"Both numbers are the same."<<endl;
    }
    return 0;
}

请帮我解决问题。我是初学者。

Answer 1

这个声明:

cin >> number1 >> " " >> number2;

尝试将您的输入写入到number1，然后是字符串文字(const char[2] )，然后是 number2。这不是你想要的，特别是因为流足够聪明，可以自己跳过空白:

cin >> number1 >> number2;

你可以用下面的代码看到这一点:

#include <iostream>
int main() {
  int i1, i2;
  std::cout << "Please enter two integers: ";
  std::cin >> i1 >> i2;
  std::cout << "You entered " << i1 << " and " << i2 << '\n';
  return 0;
}

它在运行时是这样工作的:

Please enter two integers:     4     77
You entered 4 and 77