Problem statement: Write a C program to count number of bits set to 1 in an Integer.

問題陳述：編寫一個C程序來計算Integer中設置為1的位數 。

Solution: We can use bitwise operator here to solve the problem.

解決方案：我們可以在這里使用按位運算符來解決問題。

Pre-requisite: Input number n

前提條件 ：輸入數字n

Algorithm:

算法：

1) Set count=12) Do bit wise AND with n and 1. n & 1let n be a₇a₆a₅a₄a₃a₂a₁a₀1->00000001So doing bitwise AND (refer to published article on bitwise operators) will result in all bits 0 except the LSB which will be a0. If a0 is 0 then the all bits are not setThus,IFn& 1 == 1count++;END IFRight shift n by 1n=n>>13) IF n==0Print countELSEREPEAT step 1, 2

Example with Explanation:

解釋示例：

Checking for 7 7->00000111Initially, count=0So first iteration: n=7 //00000111 n & 1 =1 so , count=1 n=n>>1 (n=3) //00000011So second iteration: n=3 //00000011 n & 1 =1 count=2 n=n>>1 (n=1) //00000001So third iteration: n=1 //00000001 n & 1 =1 count=3 n=n>>1 (n=0) //00000000 So, Print count, 3 .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

C implementation

C實現

#include <stdio.h>int main() {unsigned int n;printf("enter the integer\n");scanf("%d",&n);int count=0;while(n!=0){if(n & 1 == 1){ //if current bit 1count++;//increase count}n=n>>1;//right shift}printf("no of bits those are 1 ");printf("in its binary representation: %d\n",count);return 0; }

Output

輸出量

First run: enter the integer 7 no of bits those are 1 in its binary representation: 3Second run: enter the integer 12 no of bits those are 1 in its binary representation: 2

翻譯自: https://www.includehelp.com/c-programs/count-number-of-bits-set-to-1-in-an-integer.aspx

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