c++ - 将一串数字转换为二进制

Question

如果字符串是“8123”，我首先将字符串数字转换为整数形式 8123，然后将该数字发送到将其转换为二进制的函数。我得到了与 unsigned long long 一样大的数字来工作，但是一旦通过了，输出就会错误。有没有办法通过查看每个数字来转换为二进制。 即查找 3、2、1 和 8 来转换为二进制。

所以不要采用字符串“9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999” 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999”并将其转换为数字，有没有办法查看该字符串中的每个字符并将其转换为二进制？

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

例如81:

binary_string = str(81 % 2) + binary_string = str(1) + ""= "1"

数量 = 81/2 = 40

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

数量 = 40/2 = 20

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

数量 = 20/2 = 10

二进制字符串 = str(10 % 2) + 二进制字符串 = str(0) + "001"= "0001"

数字 = 10/2 = 5

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

数字 = 5/2 = 2

binary_string = str(2 % 2) + binary_string = str(0) + "10001"= "010001"

数字 = 2/2 = 1

二进制字符串 = str(1 % 2) + 二进制字符串 = str(1) + "010001"= "1010001"

81 ->“1010001”