Question

这是我的代码： -

#include <iostream>
using namespace std;


int Add (int *x , int *y)
{
    int a=*x;
    int b=*y;
    int c=a+b;

    return (c);
}

int Sub (int *x , int *y)
{
    int a=*x;
    int b=*y;
    int c=a-b;

    return (c);
}

int Mul (int *x , int *y)
{
    int a=*x;
    int b=*y;
    int c=a*b;

    return (c);
}

int Div (int *x , int *y)
{
    int a=*x;
    int b=*y;
    int c=a/b;

    return (c);
}

int Mod (int *x , int *y)
{
    int a=*x;
    int b=*y;
    int c=a%b;

    return (c);
}



int InputFunction (int *a , int *b , char op)
{
    int x=*a;
    int y=*b;
    int c=0;

    cout<<"Please enter first number : ";
    cin>>x;

    cout<<"Please enter second number : ";
    cin>>y;

    cout<<endl;

    cout<<"Please choose an operator to perform the operation :- "<<endl<<endl;
    cout<<" \t \t \t + for addition"<<endl;
    cout<<" \t \t \t - for sunbtraction"<<endl;
    cout<<" \t \t \t x for mutiplication"<<endl;
    cout<<" \t \t \t / for division"<<endl;
    cout<<" \t \t \t % for modulus"<<endl<<endl<<endl;
    cout<<" \t \t    Your choice : ";
    cin>>op;

    switch (op)
    {
        case '+':
                 Add (&x , &y);
                 break;

        case '-':
                 Sub (&x , &y);
                 break;

        case 'x':
                 Mul (&x , &y);
                 break;

        case '/':
                 Div (&x , &y);
                 break;

        case '%':
                 Mod (&x , &y);
                 break;

        default:
                 cout<<"Your symbol is not recognized!";
                 break;
    }

    int i=c;


    return (i);
}

int main()
{
    int a=0;
    int b=0;
    char op;
    char ch;
    int i;

    do
    {
    InputFunction (&a , &b , op);

    int m=i;

    cout<<" \t \t Your answer : "<<m<<endl<<endl;
    cout<<"Do you want to repeat the program ? (Y/N) ";
    cin>>ch;

    }while (ch == 'Y' || ch == 'y');

    cout<<"Good- Bye"<<endl;

    return 0;

}

为什么要计算一些奇怪的长答案？此外，编译器还显示警告op函数中的i和main()未初始化（尽管它们已在上面初始化）。我是C ++的新手。帮助将不胜感激。

Answer 1

int InputFunction (int *a , int *b , char op)

您永远不会将函数的返回值赋给主循环中的任何内容。

即，

int i = InputFuction(...);
int a = Add(1, 2);
int b = Sub(1, 2);
// etc.

您的所有功能（Add，Sub，Mul等）都存在同样的问题。返回值未分配给任何内容。

Answer 2

我已经为你修改了代码。实际问题是您没有分配返回值。我在你的InpuFunction中，我的主要方法是不同的。除非将InputFunction（称为方法）的值返回到main函数（调用方法），否则您的值不会更改。

#include <iostream>
using namespace std;


int Add (int *x , int *y)
{

  int a=*x;
  int b=*y;
  int c=a+b;

  return (c);
}

int Sub (int *x , int *y)
{
  int a=*x;
  int b=*y;
  int c=a-b;

  return (c);
}

int Mul (int *x , int *y)
{
  int a=*x;
  int b=*y;
  int c=a*b;

  return (c);
}

int Div (int *x , int *y)
{
  int a=*x;
  int b=*y;
  int c=a/b;

  return (c);
}

int Mod (int *x , int *y)
{
  int a=*x;
  int b=*y;
  int c=a%b;

  return (c);
}



int InputFunction (char op)
{
  int x;
  int y;
  int c=0;

  cout<<"Please enter first number : ";
  cin>>x;
  cout<<"Please enter second number : ";
  cin>>y;
  cout << " x = " << x << " y = " << y << endl; 

  cout<<endl;

  cout<<"Please choose an operator to perform the operation :- "<<endl<<endl;
  cout<<" \t \t \t + for addition"<<endl;
  cout<<" \t \t \t - for sunbtraction"<<endl;
  cout<<" \t \t \t x for mutiplication"<<endl;
  cout<<" \t \t \t / for division"<<endl;
  cout<<" \t \t \t % for modulus"<<endl<<endl<<endl;
  cout<<" \t \t    Your choice : ";
  cin>>op;


  switch (op)
    {
    case '+':
      c = Add (&x , &y);
      break;

    case '-':
      c = Sub (&x , &y);
      break;

    case 'x':
      c = Mul (&x , &y);
      break;

    case '/':
      c = Div (&x , &y);
      break;

    case '%':
      c = Mod (&x , &y);
      break;

    default:
      cout<<"Your symbol is not recognized!";
      break;
    }
  cout << "c is " << c << endl;
  int i=c;


  return (i);
}

int main()
{

  char op;
  char ch;
  int i;

    do
      {
    int m = InputFunction (op);

    cout<<" \t \t Your answer : "<<m<<endl<<endl;
    cout<<"Do you want to repeat the program ? (Y/N) ";
    cin>>ch;

      }while (ch == 'Y' || ch == 'y');

    cout<<"Good- Bye"<<endl;

    return 0;

}

Answer 3

对于编译器警告，您已声明op和i，但未使用值初始化它们（与a＆amp; b一样）。

至于输出，你实际得到了什么输出？

Answer 4

对于警告，当您声明局部变量时，不会设置其值。这意味着它的价值是不确定的，使用它将导致不确定的行为。

您需要在使用之前初始化局部变量。顺便说一下，它可能是您问题的根本原因。

您可能想要这样做。

i = InputFunction (&a, &b, op);

或者甚至更好，您在i函数中确实不需要op变量（或main），所以

m = InputFunction (&a, &b);

应该足够了（在您将InputFunction更改为不将op作为参数后）。

为什么计算不对？

4 个答案: