我试图编写一个程序,使用用户输入的二进制数将二进制转换为base10
这是原始的低效率的'代码:
void BASE10() {
int col_0, col_1, col_2, col_3, col_4, col_5, col_6, col_7;
cout << "Please enter the Binary number you would like to convert ONE digit at a time" << endl;
cin >> col_0;
cin >> col_1;
cin >> col_2;
cin >> col_3;
cin >> col_4;
cin >> col_5;
cin >> col_6;
cin >> col_7;
int num = 0;
if (col_0 == 1) {
num = num +128;
}
if (col_1 == 1) {
num = num +64;
}
if (col_2 == 1) {
num = num +32;
}
if (col_3 == 1) {
num = num +16;
}
if (col_4 == 1) {
num = num +8;
}
if (col_5 == 1) {
num = num +4;
}
if (col_6 == 1) {
num = num +2;
}
if (col_7 == 1) {
num = num +1;
}
cout << num << endl;
Restart();
我希望使用for
循环来传递单个字符串,而不是这个,用户输入一个整数数组,然后可以在计算中使用。
我该怎么做?
答案 0 :(得分:3)
#include <iostream>
#include <string>
#include <bitset>
int main()
{
const std::string str = "00100101";
std::cout << std::bitset<8>(str).to_ulong() << '\n';
}
答案 1 :(得分:0)
#include <iostream>
#include <string>
using namespace std;
int main(int argc, char const *argv[])
{
int val = 0;
// Save the integer value of ascii 0s for convenience in converting
// input characters to integer representations.
static int ASCII_OFFSET_0 = static_cast<int>('0');
// Read input data from user.
string input;
cin >> input;
// now that we have some input, presumably all 0s and 1s,
// convert the characters of the strings to integers, and compute
// the total value in base 10 of the binary string represented by input.
// NB: It may be desirable to validate the input as only being '0' or '1'
// characters. But, other characters won't 'break' this.
for (int i=0; i < input.length(); i++)
{
// Current is the integer representation of character input[i]
// with the characters ascii offset removed. i.e '0' == (int)48
int current = static_cast<int>(input.at(i)) - ASCII_OFFSET_0;
// Utilize arithmetic shift operator to calculate the value associated
// with the position it falls at in input.
// NB: first character of input is considered highest order bit.
int shift_value = (input.length() - (i + 1));
// Bitwise or is used to continually update the true value we have read
// operates on the notion that 10 or 1 (binary) == 11 (binary)
val |= current << shift_value;
}
// Val now contains an integer representation of the string given to cin
cout << val << endl;
return 0;
}
此方法允许您接受任何长度的二进制表示。
编辑:添加注释以解释该方法以及如何将二进制字符串转换为整数。