Question

如果字符串是＆＃34; 8123＆＃34;，我首先将字符串编号转换为整数形式8123，然后将此编号发送到将其转换为二进制的函数。我得到的数字和unsigned long long一样大，但是一旦它通过了，输出就错了。有没有办法通过查看每个数字转换为二进制。即查看3,2,1和8以转换为二进制。

因此，而不是采取字符串＆＃34; 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999＆＃34;并将其转换为数字，有没有办法查看此字符串中的每个字符并将其转换为二进制文件？

非常感谢任何建议

Answer 1

伪代码：

string binary_string = ""
#an example
number = 81

while (number != 0)
    #append the string casted value of the remainder of (number / 2) 
    #to the front of binary_string
    binary_string = str(number % 2) + binary_string
    number = number / 2

e.g。 81：

binary_string = str（81％2）+ binary_string = str（1）+“”=“1”

number = 81/2 = 40

binary_string = str（40％2）+ binary_string = str（0）+“1”=“01”

number = 40/2 = 20

binary_string = str（20％2）+ binary_string = str（0）+“01”=“001”

number = 20/2 = 10

binary_string = str（10％2）+ binary_string = str（0）+“001”=“0001”

number = 10/2 = 5

binary_string = str（5％2）+ binary_string = str（1）+“0001”=“10001”

number = 5/2 = 2

binary_string = str（2％2）+ binary_string = str（0）+“10001”=“010001”

number = 2/2 = 1

binary_string = str（1％2）+ binary_string = str（1）+“010001”=“1010001”

81 - ＆gt; “1010001”

Answer 2

string dec2bin(string in) {
    for(size_t i = 0; i < in.length(); i++)
        in[i] -= '0';
    string out;
    while(in.length()) {
        out.insert(0, 1, '0' + (in[in.length()-1]&1));
        char overflow = 0;
        if(in[0]<=1) {
            overflow = 10;
            in.erase(0);
        }
        for(size_t i = 0; i<in.length(); i++) {
            in[i] += overflow;
            overflow = 10 * (in[i]&1);
            in[i] /= 2;
        }
    }
    return out;
}

将一串数字转换为二进制数

2 个答案: