C ++如何为std :: bitset参数分配输入值? [英] C++ How can I assign an input value to a std::bitset argument?

问题描述

我想编写一个简单的程序，该程序将从输入中获取位数，并在输出中显示在给定位数上写入的二进制数(示例:我键入3:它显示000、001、010、011 ，100、101、110、111). 我遇到的唯一问题是在第二个 for 循环中，当我尝试在 bitset< bits > 中分配变量时，它想要常量数字. 如果您能帮助我找到解决方案，我将非常感激. 这是代码:

I want to make a simple program that will take number of bits from the input and as an output show binary numbers, written on given bits (example: I type 3: it shows 000, 001, 010, 011, 100, 101, 110, 111). The only problem I get is in the second for-loop, when I try to assign variable in bitset<bits>, but it wants constant number. If you could help me find the solution I would be really greatful. Here's the code:

#include <iostream>
#include <bitset>
#include <cmath>

using namespace std;

int main() {
    int maximum_value = 0,x_temp=10;
    //cin >> x_temp;
    int const bits = x_temp;

    for (int i = 1; i <= bits; i++) {
        maximum_value += pow(2, bits - i);
    }
    for (int i = maximum_value; i >= 0; i--)
        cout << bitset<bits>(maximum_value - i) << endl;
    return 0;
}

推荐答案

数字(非类型"，如C ++所称)模板参数必须为编译时常量，因此您不能使用用户提供的数字.请改用较大的常数(例如64).您需要另一个将限制输出的整数:

A numeric ("non-type", as C++ calls it) template parameter must be a compile-time constant, so you cannot use a user-supplied number. Use a large constant number (e.g. 64) instead. You need another integer that will limit your output:

int x_temp = 10;
cin >> x_temp;
int const bits = 64;
...

这里64是您可以使用的某种最大值，因为bitset具有一个带有unsigned long long参数的构造函数，该参数具有64位(至少；可能更多).

Here 64 is some sort of a maximal value you can use, because bitset has a constructor with an unsigned long long argument, which has 64 bits (at least; may be more).

但是，如果在中间计算中使用int，则您的代码最多可靠地支持14位(无溢出).如果要支持14位以上的字符(例如64位)，请使用更大的类型，例如uint32_t或uint64_t.

However, if you use int for your intermediate calculations, your code supports a maximum of 14 bits reliably (without overflow). If you want to support more than 14 bits (e.g. 64), use a larger type, like uint32_t or uint64_t.

保留比所需的位数更多的一个问题是将显示其他位数.要删除它们，请使用substr:

A problem with holding more bits than needed is that the additional bits will be displayed. To cut them out, use substr:

cout << bitset<64>(...).to_string().substr(64 - x_temp);

此处to_string将其转换为具有64个字符的字符串，并且substr剪切最后一个字符，其编号为x_temp.

Here to_string converts it to string with 64 characters, and substr cuts the last characters, whose number is x_temp.

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