数据转换-整数字节数组

0. 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中完成整数与8位字节串的转换功能(10'):
int Int2ByteArr(unsigned int i, unsigned char * ba);
int ByteArr2Int(unsigned char * ba,unsigned int * i);

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 int i=123456789;
Int2ByteArr(i, ba);
//结果:ba  = {0x07,0x5B,0xCD,0x15};

unsigned char ba [] = {0x07,0x5B,0xCD,0x15};
ByteArr2Int(ba, &i);
//结果: i=123456789

3 提交代码(或代码链接)和运行结果
//main.c
#include<stdio.h>
#include"utils.h"
int main(){
 
  int a;
  char bytearr[100];
  printf("input an int:");
  scanf("%d",&a);
  INT2ByteArr(a,bytearr);
  printf("ByteArr is: %s\n",bytearr);
 
  char barr[100];
  int h;
  printf("input bytearr:");
  scanf("%s",barr);
  ByteArr2INT(barr,&h);
  printf("int is: %d\n",h);
   
}
//utils.c
#include<stdio.h>
#include"utils.h"
#include<string.h>
int Hex2Char(int fromi,char *toc)
{
    if(fromi>=0&&fromi<=9)
    {
        *toc= fromi+'0';
    }
    else if(fromi>=10&&fromi<=15)
    {
        *toc = fromi+'A'-10;
    }
    else
    {
        printf("error");
     }
    return 0;
}
int Char2Hex(char fromc,int *toi)
{
    if(fromc>='0'&& fromc<='9')
    {
             *toi= fromc-'0';
    }
    else if(fromc>='A'&& fromc<='F')
    {
             *toi= fromc-'A'+10;
   
        }
        else
        {
            printf("error");
        }
    return 0;
}

int INT2ByteArr(int i,char *ba){
   int j;
   int a;
   int sum=0;
   int k=i;
   for(j=0;;j++)
   {
     k=k/16;
     if(k!=0)
      sum++;
     else
       break;
   }
   for(j=sum;j>=0;j--){
     a = i%16;
     Hex2Char(a,&ba[j]);
     i=i/16;   

   }
   ba[sum+1]='\0';


}
int ByteArr2INT(char *ba,int *i)
{
   int len;
   int j;
   int n=0;
   *i=0;
   len = strlen(ba);
   for(j=0;j<len;j++)
   {
      Char2Hex(ba[j],&n);
      //printf("%d\n",n);
      *i=(*i)*16+n; 

   }
   
}

posted @ 2022-06-02 15:38  20191316王秋雨  阅读(45)  评论(0编辑  收藏  举报