〜运算符用C [英] ~ operator in C

问题描述

该程序的输出为-13。我从来没有完全理解〜运营商C.为什么它给-13作为输出？如何限制〜多项操作者只需4位？

 ＃包括LT＆;＆stdio.h中GT;
    ＃包括LT＆;＆CONIO.H GT;
    诠释的main（）
    {
        int类型的= 12;
        A =〜一;
        的printf（％d个，一）;
        残培（）;
        返回;
    }
 

解决方案

要限制效果位指定数目的，只要使用按位掩码，例如：

 的#include＆LT;＆stdio.h中GT;诠释主要（无效）{
    int类型的= 16; / * 10000 *二进制/
    INT B =〜A; / *威尔间preT b以-17的补* /
    INT C =（A和〜0xF的）| （〜A和0xF的）; / *将限制运营商最右边的4位，
                                        所以00000 01111变得和c将成为
                                        11111，不是11 ... 101111，所以C将是31 * /    的printf（一个是％D，B的％D，C为％d \\ n，A，B，C）;
    返回0;
}
 

输出：

 保罗@地方：〜/ src目录/ C /划伤$ ./comp
一个是16，​​b为-17，c为31
保罗@地方：〜/ src目录/ C / $从零开始
 

The output of this program is -13. I have never fully understood ~ operator in C. Why does it give -13 as output? How to limit ~ operator to just 4 bits of a number?

    #include<stdio.h>
    #include<conio.h>
    int main()
    {
        int a = 12;
        a = ~a;
        printf("%d",a);
        getch();
        return;
    } 

解决方案

To limit the effect to a specified number of bits, just use bitwise masks, e.g.:

#include <stdio.h>

int main(void) {
    int a = 16;             /* 10000 in binary */
    int b = ~a;             /* Will interpret b as -17 in two's complement */
    int c = (a & ~0xF) | (~a & 0xF); /* Will limit operator to rightmost 4 bits,
                                        so 00000 becomes 01111, and c will become
                                        11111, not 11...101111, so c will be 31     */

    printf("a is %d, b is %d, c is %d\n", a, b, c);
    return 0;
}

Outputs:

paul@local:~/src/c/scratch$ ./comp
a is 16, b is -17, c is 31
paul@local:~/src/c/scratch$

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