爲什麼計算不正確？

Question

這裏是我的代碼： -

#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; 

} 

爲什麼計算一些奇怪的長的答案？此外，編譯器顯示警告op和imain()函數未初始化（雖然它們在上面初始化）。我是C++新手。幫助將不勝感激。

Answer 1

我已經修改了你的代碼。實際的問題是你沒有分配返回值。我在你的InpuFunction和我的主要方法是不同的。除非您將InputFunction（稱爲方法）的值返回給main函數（調用方法），否則您的值不會更改。

#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 2

對於編譯器警告，你已經宣佈op和i，而不是用值初始化它們（如您a & b一樣）。

至於輸出，你實際得到了什麼輸出？

Answer 3

對於警告，當您聲明局部變量時，其值不會被設置。這意味着它的價值是不確定的，使用它會導致未定義的行爲。

在使用它們之前，您需要初始化局部變量。順便提一句，這可能是你的問題的根源。

您可能想要做

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

甚至更​​好，你並不真的需要在main功能i變量（或op），所以

m = InputFunction (&a, &b); 

應該足夠（更改後的InputFunction不走作爲參數的op）。

Answer 4

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

您從不將函數的返回值賦給主循環中的任何東西。

即，

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

您與您的所有功能（ADD，SUB，MUL等）相同的問題。返回值沒有分配給任何東西。