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C++ bitset any()用法及代码示例


bitset::any()是C++ STL中的内置函数,如果数字中至少设置了一位,则返回True。如果未设置所有位或数字为零,则返回False。

用法:

bool count() 

参数:该函数不接受任何参数。


返回值:该函数返回一个布尔值。如果设置了布尔值的任何一位,则返回的布尔值为True。如果未设置任何位,则为False。

下面的程序演示了bitset::any()函数。

示例1:

// CPP program to illustrate the 
// bitset::any() function 
#include <bits/stdc++.h> 
using namespace std; 
  
int main() 
{ 
    // initialization 
    bitset<4> b1(string("1100")); 
    bitset<6> b2(string("000000")); 
  
    // function to check if any of 
    // its bits are set or not 
    bool result1 = b1.any(); 
    if (result1) 
        cout << b1 << " has a minimum of one-bit set"
             << endl; 
    else
        cout << b1 << " does not have any bits set"
             << endl; 
  
    // function to check if any of 
    // its bits are set or not 
    bool result2 = b2.any(); 
    if (result2) 
        cout << b2 << "  has a minimum of one-bit set"
             << endl; 
    else
        cout << b2 << " does not have any bits set"
             << endl; 
  
    return 0; 
}
输出:
1100 has a minimum of one-bit set
000000 does not have any bits set

示例2:

// CPP program to illustrate the 
// bitset::any() function 
// when the input is as a number 
  
#include <bits/stdc++.h> 
using namespace std; 
  
int main() 
{ 
    // initialization 
    bitset<2> b1(3); 
    bitset<3> b2(0); 
  
    // function to check if any of 
    // its bits are set or not 
    bool result1 = b1.any(); 
    if (result1) 
        cout << b1 << " has a minimum of one-bit set"
             << endl; 
    else
        cout << b1 << " does not have any bit set"
             << endl; 
  
    // function to check if any of 
    // its bits are set or not 
    bool result2 = b2.all(); 
    if (result2) 
        cout << b2 << " has a minimum of one-bit set"
             << endl; 
    else
        cout << b2 << " does not have any bit set"
             << endl; 
  
    return 0; 
}
输出:
11 has a minimum of one-bit set
000 does not have any bit set


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注:本文由纯净天空筛选整理自gopaldave大神的英文原创作品 bitset any() in C++ STL。非经特殊声明,原始代码版权归原作者所有,本译文未经允许或授权,请勿转载或复制。