数据转换-位串字节数组

2022-06-02 14:35:36 阅读：231 来源： 互联网

标签：ba int bs ByteArr2Bitstr char 数组位串 include 字节

一、任务详情

在openEuler(推荐)或Ubuntu或Windows(不推荐)中完成下面任务

1 参考《GMT 0009-2012 SM2密码算法使用规范》第6节“数据转换” 在utils.h和utils.c中完成位串与８位字节串的转换功能(10')：
int Bitstr2ByteArr(unsigned char * bs, unsigned char * ba，int *lba);
int ByteArr2Bitstr(unsigned char * ba, unsigned char * bs, int *lbs);

2 并写出测试代码测试上述函数(不能与下面代码一样)，比如(10'):

unsigned char bs [] = "1010010100011100";
int len;
Bitstr2ByteArr(bs, char * ba, &len);
//结果：ba = {0x5, 0x1C}; len=2

char ba[] = {0x5, 0x1C}
ByteArr2Bitstr(char * ba, char * bs);
//结果：bs= "1010010100011100"
3 提交代码（或代码链接）和运行结果

主函数main1.c

#include <stdio.h>
#include <string.h>
#include "utils.h"
int main(){
        int bitList[8] = { 0 };
    int a;
    sscanf("32", "%x", &a);
    ByteArr2Bitstr(a, bitList);
    printf("0x32 be changed to\n");
    for (int i =0 ;i<8;++i)
        printf("%i",bitList[i]);//八位长
        printf("\n");
    int b,n;
    n=Bitstr2ByteArr(bitList,b);
    printf("the data will be changed to\n");
    printf("%x\n",n);
}
~

utils1.c

#include <stdio.h>
#include <string.h>
#include "utils.h"
//20191223
int ByteArr2Bitstr(int ba,int * bs){
    for (int i =0 ;i<8;++i)
        {
            int nTmp = (1 << i);
            bs[7-i] = (( ba & nTmp) == nTmp )? 1 : 0;
        }
}

int Bitstr2ByteArr(int * bs,int ba)
{
    int n = 0;
    for (int i = 0 ; i < 8;++i)
    {
        n += bs[7-i] * (1 << i);
    }
    *bs = n;
    return n;
}

头文件utils1.h

#ifndef  _UTILS_H_
#define  _UTILS_H_

int ByteArr2Bitstr(int ba,int * bs);
int Bitstr2ByteArr(int * bs,int ba);
#endif

编译指令

gcc test/main1.c src/utils1.c -o bin/20191223test2 -Iinclude

运行结果

标签：ba,int,bs,ByteArr2Bitstr,char,数组,位串,include,字节
来源： https://www.cnblogs.com/zzjjyy123/p/16337621.html