C++ 二进制转十进制？

Question

我做了一个将二进制数转换为十进制数的函数，但是如果我对二进制数进行高位，它只会返回 256 ？？？什么可能导致这种情况？我正在使用 int 变量。任何帮助将非常感激

    #include <iostream>

using namespace std;

int FromBin (int n)
{
    int increment;
    int Result;
    increment = 1;
    Result = 0;
    while(n != 0)
    {
        if (n % 10 == 1){
            Result = Result+increment;
            n = n-1;
        }
        n = n/10;
        increment = increment*2;
    }
    cout<<Result;
}

void ToBin(int n)
{
    if (n / 2 != 0) {
        ToBin(n / 2);
    }
    cout<<n % 2;
}

int main()
{
    int choice;
    int n;
    cout<<"Choose a function: press 0 for decimals to binary, press 1 for binary to decimal\n";
    cin>>choice;
    if (choice == 0){
        cout<<"Enter a number: \n";
        cin>>n;
        ToBin(n);
    }
    else if (choice == 1){
        cout<<"Enter a number: \n";
        cin>>n;
        FromBin(n);
    }
    else{
        cout<<"Invalid input";
    }
}

我是 C++ 新手，所以我不明白这一点......：/

Answer 1

这是你在这里进行的一个很酷的程序......这是我为你的问题找到的可能解决方案......

 /* C++ program to convert binary number into decimal */
 #include <iostream>
     using namespace std;
 int main()
 {
     long bin, dec = 0, rem, num, base = 1;
     cout << "Enter the binary number(1s and 0s) : ";
     cin >> num;
     bin = num;
     while (num > 0)
     {
         rem = num % 10;
         dec = dec + rem * base;
         base = base * 2;
         num = num / 10;
     }
     cout << "The decimal equivalent of " << bin << " : " << dec << endl;
     return 0;
 }

Answer 2

这就是我认为你正在拍摄的。您可以通过从 切换到int来处理更大的数字long。

long fromBin(long n)
{
    long factor = 1;
    long total = 0;

    while (n != 0)
    {
        total += (n%10) * factor;
        n /= 10;
        factor *= 2;
    }

    return total;
}

现场演示