转换基地10到基地2个，C编程 [英] Convert base 10 to base 2, C programming

问题描述

我还是新编程，我想请问一下 C 语言大约具体问题。我使用$ C $个cblocks编译器。

I'm still new with programming and I'd like to ask about one specific question about the Clanguage. I'm using codeblocks compiler.

下面是一块code的：

Here is a piece of code:

int n, c, k;

scanf("%d",&n);

for(c = 31;c >= 0;c--)
{
    k = n >> c;

    if(k & 1)
        printf("1");
    else
        printf("0");
}

return 0;

这是我从一个网站了，它的基地，以二进一转换。

that I got from one website and it converts the base to the binary one.

我的问题是，我在那里有一个，如果和放大器; else语句，为什么会出现，如果（K＆安培; 1）的printf（1）？我以为使得k既可以是1和0，如果我使用（K＆安培; 1）有两个选项，如（0安培; 1）= 0（1和1）= 1。可以请任何人给我解释一下这个？
 非常感谢你。

My question is, where I have an if&else statement, why is there if(k & 1) printf("1") ?? I thought that k can be both 1 and 0 and if I use (k & 1) there are two options such as (0 & 1) = 0 and (1 & 1) = 1. Can please anybody explain me this? Thank you very much.

推荐答案

的内部的前pression如果语句总是判断为真或假。在整数而言，0表示假的，和任何非零意味着真。当你有 K＆安培; 1 ，这意味着，如果 K 的最低显著位为1，那么前pression计算结果为1，因此被认为真正。如果 K 的最低显著位为0，则前pression计算结果为0，因此被认为是假的。

The expression inside of an if statement always evaluates to true or false. In terms of integers, 0 means false, and anything non-zero means true. When you have k & 1, that means that if the least-significant bit of k is 1, then that expression evaluates to 1 and is therefore considered true. If the least-significant bit of k is 0, then the expression evaluates to 0 and is therefore considered to be false.

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