Converting a struct to a hex string in C

Question

I have the following struct

typedef struct __attribute__((packed)) word {
    uint16_t value;
    uint8_t flag
} Word;

I want to convert it to a hex string. For example if value = 0xabcd and flag = 0x01 I want to get the string 01abcd

if I do some pointer juggling

Word word;
word.value = 0xabcd;
wordk.flag = 0x01;

printf("word: %X\n", *(int *)&word);

I get the output that I want (word: 1ABCD) but this doesn't seem safe and when I tried to do this after looking at some of the answer here

char ptr[3];
memcpy(ptr, word, 3);
printf("word: %02X%02X%02X\n", ptr[2], ptr[1], ptr[0]);

I got word: 01FFFFFFABFFFFFFCD, for some reason the first two bytes are being extended to a full int

Adrian Mole · Accepted Answer · 2022-03-10 12:08:16Z

3

There's no real gain from messing around with pointers or type-punning, if all you want is to output the values of the two structure members. Just print them individually:

printf("word: %02x%04x\n", (unsigned int)word.flag, (unsigned int)word.value);

edited Mar 10, 2022 at 12:08

answered Mar 10, 2022 at 11:58

Adrian Mole

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4 Comments

Ingo Leonhardt Over a year ago

%x expects an unsigned int parameter so you should add casts ro avoid UB

Eric Postpischil Over a year ago

@AdrianMole: No, they will not. uint8_t will be promoted to int. uint16_t will be promoted to or already is unsigned int if that is 16 bits. Otherwise, it will be promoted to int.

Adrian Mole Over a year ago

@EricPostpischil Oops. excellent point ... edited. (Those integer promotion rules can still bite me!)

Lundin Over a year ago

The gain from "messing around with pointers" would be massively improved execution speed, much smaller binary size and no stdio.h format string/va arg/non-existent type safety hazards present in the program (as illustrated by the above comments). How exactly is faster, smaller and safer "no real gain"?

klutt · Accepted Answer · 2022-03-10 12:01:13Z

2

Use a simple sprintf to convert to a string:

int main(void)
{
    Word v = { 0xabcd, 0x01 };
    
    char s[10];
    
    sprintf(s, "%02x%04x", v.flag, v.value);
    
    puts(s);
}

answered Mar 10, 2022 at 12:01

klutt

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Comments

Lundin · Accepted Answer · 2022-03-10 15:41:40Z

I want to get the string 01abcd

So you want to print the binary representation of the struct on a little endian machine backwards. There's no obvious advantage of using sprintf for this apart from it being quick & dirty to implement.

If you want something with performance, then hacking this out manually isn't rocket science - simply iterate over the struct byte by byte and convert each nibble to the corresponding hex character:

void stringify (const uint8_t* data, size_t size, char* dst)
{
  for(size_t i=0; i<size; i++)
  {
    uint8_t byte = data[size-i-1]; // read starting from the end of the data
    dst[i*2]   = "0123456789ABCDEF"[ byte >> 4  ]; // ms nibble
    dst[i*2+1] = "0123456789ABCDEF"[ byte & 0xF ]; // ls nibble
  }
  dst[size*2] = '\0';
}

This will give "01ABCD" on a little endian machine and "01CDAB" on a big endian machine.

(Optionally add restrict to data and dst parameters but I doubt it will improve performance since the const correctness already blocks aliasing at some extent.)

chux · Accepted Answer · 2022-03-10 16:30:02Z

0

Could join them mathematically first.

printf("word: %06lx\n", wordk.flag * 0x10000L + word.value);

Using long in case we are on a 16-bit int machine.

answered Mar 10, 2022 at 16:30

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