十六进制到十进制转换器不起作用（C++）

Question

So, as part of an assignment, I wrote a program to convert hexadecimal to decimal. But I was not able to get the desired result. Can someone please pin point the error in this code?

#include<bits/stdc++.h>
#include<math.h>
using namespace std;

int hexaToDecimal(string n){
    int ans = 0;
    int power =1;
    int s = n.size();

    for(int i=s-1; i>=0; i--){
        if(n[i] >= '0' && n[i] <= '9'){
            ans = ans + power*(n[i]);
        }
        else if(n[i] >= 'A' && n[i] <= 'F'){
            ans = ans + power*(n[i]-'A' + 10);
        }
        power = power * 16;
    }
    return ans;
}

        


int main(){
    string n;
    cin>>n;
    cout<<hexaToDecimal(n)<<endl;
    return 0;
}

Answer 1

Simpler way to go about it:

unsigned fromHex(const string &s) { 
    unsigned result = 0;
    for (char c : s) 
        result = result << 4 | hexDigit(c);
    return result;
}

unsigned hexDigit(char c) { 
    return c > ‘9’ ? c - ‘A’ + 10: c - ‘0’;
}

Answer 2

You may add - '0' like - 'A' . There is the code:

#include<bits/stdc++.h>
#include<math.h>
using namespace std;
int hexaToDecimal(string n){
    int ans = 0;
    int power =1;
    int s = n.size();

    for(int i=s-1; i>=0; i--){
        if(n[i] >= '0' && n[i] <= '9'){
            ans = ans + power*(n[i] - '0'); //THERE.
        }
        else if(n[i] >= 'A' && n[i] <= 'F'){
            ans = ans + power*(n[i]-'A' + 10);
        }
        power = power * 16;
    }
    return ans;
}

        


int main(){
    string n;
    cin>>n;
    cout<<hexaToDecimal(n)<<endl;
    return 0;
}

Changes only near //THERE.

Answer 3

Here is a different approach on how to solve your problem. You could use std::hex which is defined in the headerfile #include<iostream> Which is a simpler way to do it.

For Example:

#include <iostream>


int main()
{
    int HexNum;
    std::cin >> std::hex >> HexNum;
    std::cout << HexNum << std::endl;

    return 0;
}

Output:

10F
271