如何在字符串中存储ascii值为0而不终止它？

Question

将输入作为十六进制字符串,然后将其转换为C中的char字符串.十六进制字符串可以包含0x00,转换后转换为Ascii中的0.这会终止字符串.我必须将值存储在char字符串中,因为API使用它.
我的代码到目前为止:

int hex_to_int(unsigned char c) {
    int first =0;
    int second =0;
    int result=0;
    if(c>=97 && c<=102)
        c-=32;
    first=c / 16 - 3;
    second =c % 16;
    result = first*10 + second;
    if(result > 9) result--;
        return result;
}
unsigned char hex_to_ascii(unsigned char c, unsigned char d){
    unsigned char a='0';
    int high = hex_to_int(c) * 16;
    int low = hex_to_int(d);
    a= high+low;
    return a;
}
unsigned char* HextoString(unsigned char *st){
    int length = strlen((const char*)st);
    unsigned char* result=(unsigned char*)malloc(length/2+1);
    unsigned char arr[500];
    int i;
    unsigned char buf = 0;
    int j=0;
    for(i = 0; i < length; i++)
    {
        if(i % 2 != 0)
    {
        arr[j++]=(unsigned char)hex_to_ascii(buf, st[i]);
    }
    else
    {
         buf = st[i];
    }
    }
    arr[length/2+1]='\0';
    memcpy(result,arr,length/2+1);
    return result;
}

Answer 1

您可以将任何值存储在char数组中.但是如果要存储值0x00,则不能在此数组上使用字符串函数.因此,您必须使用整数变量来存储要存储的数据的长度.然后,您可以编写使用此整数的函数.