debugging C++正在舍入函数的结果

klr1opcd  于 2022-12-13  发布在  其他
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I created a program with a function to convert numbers from a string to a numeric variable (the program will be copied below). I started off using a double; however, for some reason, whenever I built it, it kept rounding the last number up or down if I used decimals. I don't understand why so I would be grateful for any insight you guys could share with me.

#include <iostream>
using namespace std;

//variables initialised in a set of curly brackets are called local variables
//variables initialised outside a set of curly brackets are called global variables
// local variables can only be used in that set of curly brackets
// global variables can be used anywhere after it has been declared
// global variables are automatically set to 0 when it has not been assigned a value
//a void function is a function that returns nothing

//StringName.length() returns the length of a string

/*----------------------------------------FUNCTION-------------------------------------*/
int pow(int a, int b){
int c = 1;
for(int i=0;i<b;i++){
    c*=a;
}
return c;
}

float StringNoToNo(string a){
float b=0.0;
int y = 1;
int s = 0;
//cout<<"a.length() is: "<<a.length()<<endl;
for(int i = (a.length()); i>0;i--){
  //  cout<<"the loop: "<<(a.length()-i)<<endl;
    int z = a[i-1];
    //cout<<"z = "<<z<<endl;
    switch(z){
     case 48 ... 57:
      //    cout<<"(a[i]>=48)&&(a[i]<=57) is true"<<endl;
//cout<<"pow(10.0,(a.length()-i)) = "<<pow(10.0,(a.length()-i))<<endl;
//cout<<"a.length() - i = "<<(a.length()- i)<<endl;
    b += ((a[i-1]-48)* pow(10.0,(a.length()-i-s)));
  //  cout<<"b= "<<b<<endl;
    break;
     case 46:
        y=pow(10,(a.length()-i));
    //    cout<<"y = "<<y<<endl;
        if(s==0){
        s++;}else{
        goto v;
        }
        break;
    default:
        v:
      //  cout<<"(a[i]<48)||(a[i]>57) is true"<<endl;
        cout<< "the number was not written properly"<<endl;
        return 0;
        break;
    }
}
//cout<<" b = "<<b<<endl;
//cout<<"b/y = "<<b/y<<endl;
return (b/y);
}
/*----------------------------------------FUNCTION-------------------------------------*/
// overloading functions - you can create multiple functions with the same name so as long
// as they have different parameters.
// as long as the function is declared at the beginning, even if the function meant to
// overload it is written at the end of the code, it can still be used.

main() {
  string no;
  cout << "write a number: ";
  cin >> no;
  cout << "the number is: "<<StringNoToNo(no)<<endl;
}

At first, I used the double variable for the function and then changed it to float. After that, I used cout at various parts of the loop to see what was going on at every step of the way but still couldn't understand the problem.
The input and output I get are:

write a number: 1234.6789
the number is: 1234.68

Process returned 0 (0x0) execution time : 4.713 s Press any key to continue.
Of course, if I remove the // from the cout functions:

write a number: 1234.6789
the number is: a.length() is: 9
the loop: 0
z = 57
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 1
a.length() - i = 0
b= 9
the loop: 1
z = 56
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 10
a.length() - i = 1
b= 89
the loop: 2
z = 55
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 100
a.length() - i = 2
b= 789
the loop: 3
z = 54
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 1000
a.length() - i = 3
b= 6789
the loop: 4
z = 46
the loop: 5
z = 52
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 100000
a.length() - i = 5
b= 46789
the loop: 6
z = 51
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 1000000
a.length() - i = 6
b= 346789
the loop: 7
z = 50
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 10000000
a.length() - i = 7
b= 2.34679e+06
the loop: 8
z = 49
(a[i]>=48)&&(a[i]<=57) is true
pow(10.0,(a.length()-i)) = 100000000
a.length() - i = 8
b= 1.23468e+07
1234.68

Process returned 0 (0x0)   execution time : 24.055 s
Press any key to continue.
6xfqseft

6xfqseft1#

b= 1.23468e+07证明了你的函数没有舍入,它返回了正确的值。
实际上是cout <<在进行舍入,使用cout << precision(8)使其显示8位精度,而不是默认的6位。

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