C++ - 十进制到二进制的转换
我写了一个'简单的'(我花了30分钟)程序,将十进制数字转换为二进制。我确信还有更简单的方法,你能告诉我吗? 这里是代码。
#include <iostream>
#include <stdlib.h>
using namespace std;
int a1, a2, remainder;
int tab = 0;
int maxtab = 0;
int table[0];
int main()
{
system("clear");
cout << "Enter a decimal number: ";
cin >> a1;
a2 = a1; //we need our number for later on so we save it in another variable
while (a1!=0) //dividing by two until we hit 0
{
remainder = a1%2; //getting a remainder - decimal number(1 or 0)
a1 = a1/2; //dividing our number by two
maxtab++; //+1 to max elements of the table
}
maxtab--; //-1 to max elements of the table (when dividing finishes it adds 1 additional elemnt that we don't want and it's equal to 0)
a1 = a2; //we must do calculations one more time so we're gatting back our original number
table[0] = table[maxtab]; //we set the number of elements in our table to maxtab (we don't get 10's of 0's)
while (a1!=0) //same calculations 2nd time but adding every 1 or 0 (remainder) to separate element in table
{
remainder = a1%2; //getting a remainder
a1 = a1/2; //dividing by 2
table[tab] = remainder; //adding 0 or 1 to an element
tab++; //tab (element count) increases by 1 so next remainder is saved in another element
}
tab--; //same as with maxtab--
cout << "Your binary number: ";
while (tab>=0) //until we get to the 0 (1st) element of the table
{
cout << table[tab] << " "; //write the value of an element (0 or 1)
tab--; //decreasing by 1 so we show 0's and 1's FROM THE BACK (correct way)
}
cout << endl;
return 0;
}
顺便说一下,这很复杂,但我已经尽力了。
编辑 - 这是我最终使用的解决方案:
std::string toBinary(int n)
{
std::string r;
while(n!=0) {r=(n%2==0 ?"0":"1")+r; n/=2;}
return r;
}
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3
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std::bitset
][1] 有一个.to_string()
方法,返回一个std::string
,持有二进制的文本表示,有前导零的填充。根据你的数据需要选择bitset的宽度,例如,
std::bitset
可以从32位整数得到32个字符的字符串。一个 "int "变量不是十进制的,是二进制的。你要找的是数字的二进制字符串表示,你可以通过应用掩码过滤单个比特,然后打印出来。
你想做的事,比如。