如何在数组中获取特定的用户输入？

Question

我正在编写一个程序，通过加倍 (link to wikihow article) 将二进制数转换为十进制数。

如果用户输入不是 1 或 0，那么它不是二进制数，在这种情况下，我希望循环“中断”并说如下内容：

“糟糕！二进制数只有 1 或 0”。

如果不是“则”循环应该继续。

那是我想编写类似的代码

for(int digits = 0; digits != digitsINbinNum; ++digits){
      if(a condition that checks if user input is anything else than 1 or 0){
         coût << ""Oops! Binary numbers have only 1 or 0" << endl; 
         break;
          }else{
        cin >> binArray[digits];/*<-----------Here's the part where I am trying to do that*/
      }
    }

有关详细信息，请参阅下面给出的代码：

#include <iostream>
#include <iterator>

using namespace std;

int main(){
    int digitsINbinNum;
    cout << "If you don't mind. Please enter the number of digits in your binary number: ";
    cin >> digitsINbinNum;
    int binArray[digitsINbinNum];

    cout << "Enter the binary number: ";
    for(int digits = 0; digits != digitsINbinNum; ++digits){
        cin >> binArray[digits];/*<-----------Here's the part where I am trying to do that*/
    }

/*using the doubling method as found in wikihow.com*/
    int total = 0;
    for(int posiOFdigit = 0; posiOFdigit != sizeof(binNum[noOFdigits]); posiOFdigit++){
        total = total * 2 + binNum[posiOFdigit];
    }

    /*Printing the number*/
    cout << "Decimal form of ";
    for(int n = 0; n != noOFdigits; n++){
        cout << binNum[n];
    }
    cout << " is " << total;
    return 0;
}

Answer 1

二进制数转十进制数的逻辑可以参考题中给定的link。

修改给定的代码，使其尽可能接近问题的参考代码。

注意：由于 ISO C++ 禁止变长数组，我正在更改 int binArray[digits] 到 int *binArray = (int *)malloc(sizeof(int) * digitsINbinNum);。 这种修改使它成为一个整数指针，并在运行时获得所需大小的内存。

#include <iostream>

using namespace std;

int main(){
    int digitsINbinNum,
        /* variable to keep decimal conversion of given binary */
        decimal_val = 0;
    bool is_binary = true;
    cout << "If you don't mind. Please enter the number of digits in your binary number: ";
    cin >> digitsINbinNum;
    /*
     ISO C++ forbids variable length array,
     making it int pointer and allocating dynamic memory
    */
    int *binArray = (int *)malloc(sizeof(int) * digitsINbinNum);
    if (binArray == NULL)
    {
        cout << "Memory allocation failure" << endl;
        exit -1;
    }
    cout << "Enter the binary number: ";
    for(int digits = 0; digits != digitsINbinNum; ++digits){
        cin >> binArray[digits];
        /*<-----------Here's the part where I am trying to do that*/
        /* doubling method logic for conversion of given binary to decimal */
        if ((binArray[digits] == 0) ||
            (binArray[digits] == 1))
        {
            decimal_val = (decimal_val * 2) + binArray[digits];
        }
        else /* not a binary number */
        {
            is_binary = false;
            cout << "Oops! Binary numbers have only 1 or 0" << endl;
            break;
        }
    }

    /* if conversion is successful: print result */
    if (is_binary)
    {   
        cout << "Decimal Value for given binary is: " << decimal_val << endl;
    }
    if (binArray)
    {
        free(binArray);
    }
    return 0;
}

Answer 2

你不需要一个数组。这是一个简单的解决方案：

#include <iostream>

int main(){
    int digitsINbinNum;
    std::cout << "If you don't mind. Please enter the number of digits in your binary number: ";
    std::cin >> digitsINbinNum;

    std::cout << "Enter the binary number: ";
    int ret = 0;
    for(int digits = 0; digits != digitsINbinNum; ++digits) {
        int bin;
        std::cin >> bin;
        if (bin == 1 || bin == 0) {
            ret = 2 * ret + bin;
        } else {
            std::cout << "Oops! Binary numbers have only 1 or 0" << std::endl;
            return -1;
        }
    }
    std::cout << ret << std::endl;

    return 0;
}