How to convert char to hex stored in uint8_t form?

问题

Suppose I have these variables,

const uint8_t ndef_default_msg[33] = {
    0xd1, 0x02, 0x1c, 0x53, 0x70, 0x91, 0x01, 0x09,
    0x54, 0x02, 0x65, 0x6e, 0x4c, 0x69, 0x62, 0x6e,
    0x66, 0x63, 0x51, 0x01, 0x0b, 0x55, 0x03, 0x6c,
    0x69, 0x62, 0x6e, 0x66, 0x63, 0x2e, 0x6f, 0x72,
    0x67
};
uint8_t *ndef_msg;
char *ndef_input = NULL;

How can I convert ndef_input (which is just a plain text, like "hello") to hex and save into ndef_msg? As you can see ndef_default_msg is in hex form. Data inside ndef_msg should be something like that as well.

A bit of background, in the original program (source code), the program will open a file, get the data and put it inside ndef_msg, which then will be written into a card. But I don't understand how it can take the data and convert to hex.

I want to simplify the program so it will directly ask user for text (instead of asking for a file).

回答1:

Why not read it into ndef_msg directly, (minus the \0 if it suppose to be a pure array). The hex are just for presentation, you could have just picked decimal or octal with no consequence for the content.

void print_hex(uint8_t *s, size_t len) {
    for(int i = 0; i < len; i++) {
        printf("0x%02x, ", s[i]);
    }
    printf("\n");
}

int main()
{
    uint8_t ndef_msg[34] = {0};

    scanf("%33s", ndef_msg);
    print_hex(ndef_msg, strlen((char*)ndef_msg));


    return 0;
}

You probably need to handle the reading of the string differently to allow for whitespace and perhaps ignore \0, this is just to illustrate my point.

回答2:

Maybe not very elegant but simple: define a look up table that maps a character code (0 to 255) into a desired value.

// something like this:
for( i = 0; i < INPUT_LEN; ++i ) {
    value_in = input[i];
    value_out = lut[value_in];
    array_out[i] = value_out;
}

I used such not-elegant solutions a couple of times (e.g. for color mapping) and it worked equally good as other fancy solutions

回答3:

If I understand right, you read data of hex format stored in ndef_input, parse it and save the value in ndef_msg.

you may use

// parse the hex string and store it in an int variable
int temp_int;
sscanf(ndef_input, "%x", &temp_int);
// covert it to uint8_t type
ndef_msg = malloc(sizeof(uint8_t));
*ndef_msg = (uint8_t)temp_int;

回答4:

char *ndef_input="Z";

uint8_t b=90; //assume this is your character Z in decimal ascii code 90 and HEX = 5A
uint8_t LSB = b & 0x0F; // this is LSB 10 decimal = A
uint8_t MSB = (b & 0xF0)>>4; // this is MSB 5 in decimal = 5 in Hex
cout << "MSB" << MSB << "LSB" << LSB;

回答5:

I hope that can help you

/*
 *  DESCRIPTION
 *    Converts a block in ASCII representation to binary
 *  PARAMETERS
 *    char * inMessage      : message in ASCII format, '\0' terminated
 *  OUTPUTS
 *    uint8 * outMessage    : output message in binary format
 *                    Format: outMessage[i], where i is byte number
 *  RETURN VALUE
 *    uint32                : number of converted bytes
 */
uint32 ascii2hex_block( uint8 * outMessage, int32 out_len, const char * inMessage )
{
    #define SET_BIT(U,N)   ((U) |=   (0x1 << (N)))

    int32   i = 0;
    int32   k = 0;
    int32   blockLen = 0;
    char    inChar;
    uint8   hexVal;
    uint32  retVal = 0;

    while ( inMessage[blockLen]!='\0' )     blockLen++;
    blockLen = blockLen >> 1;

    if (blockLen <= out_len)                        // not enough space in output
    {
        retVal = blockLen;
        for (i = 0; i < blockLen; i++)
        {
            outMessage[i] = 0;
            inChar = inMessage[k];
            hexVal = ascii2hex( inChar );
            if (hexVal == 0xff) retVal = 0;     // found an invalid character
            if ( (hexVal & (0x1 << 0) ) != 0 )  SET_BIT( outMessage[i], 4 );
            if ( (hexVal & (0x1 << 1) ) != 0 )  SET_BIT( outMessage[i], 5 );
            if ( (hexVal & (0x1 << 2) ) != 0 )  SET_BIT( outMessage[i], 6 );
            if ( (hexVal & (0x1 << 3) ) != 0 )  HELPER_SET_BIT( outMessage[i], 7 );
            k++;
            inChar = inMessage[k];
            hexVal = ascii2hex( inChar );
            if ( (hexVal & (0x1 << 0) ) != 0 )  SET_BIT( outMessage[i], 0 );
            if ( (hexVal & (0x1 << 1) ) != 0 )  SET_BIT( outMessage[i], 1 );
            if ( (hexVal & (0x1 << 2) ) != 0 )  SET_BIT( outMessage[i], 2 );
            if ( (hexVal & (0x1 << 3) ) != 0 )  SET_BIT( outMessage[i], 3 );
            k++;
        }
    }   

    return retVal;
}

And ascii2hex is defined as follow:

/*
 *  DESCRIPTION
 *    Converts an ascii character ('0'..'f' or '0'..'F') to corresponding integer value.
 *    In case of invalid ascii character, return value is 0xff
 *  USAGE
 *    uint8 ascii2hex( char inASCII );
 *  PARAMETERS
 *    char inASCII  : ascii character to convert
 *  RETURN VALUE
 *    uint8         : value of inASCII, 0xff for invalid input
 */
uint8 ascii2hex( char inASCII )
{
    uint8 retHex=0xff;

    if( (inASCII>=48) && (inASCII<=57) )
        retHex = inASCII-48;
    else if( (inASCII>=65) && (inASCII<=70) )
        retHex = inASCII-55;
    else if( (inASCII>=97) && (inASCII<=102) )
        retHex = inASCII-87;

    return retHex;
}

来源：https://stackoverflow.com/questions/9396265/how-to-convert-char-to-hex-stored-in-uint8-t-form

标签

hex

uint