How to Concatenate Hex Numbers?

Question

I've been trying to concatenate 4 hex numbers and can't seem to do it.

Example:

int a = 0x01;
int b = 0x00;
int c = 0x20;
int d = 0xF1;
//Result should be 0x010020F1

The results that I am getting using sprintf() and bitwise operations always have cut off zeros, giving me answers like 1020F1, which is much different than what I want. Anybody have a better method?

If you want to use `snprintf()`, then: `snprintf(buffer, sizeof(buffer), "0x%.2X%.2X%.2X%.2X", a, b, c, d);` should do the trick unless `buffer` is shorter than 11 bytes long. — Jonathan Leffler, Apr 25 '13 at 07:18
If you want only to print that, use `printf("0x%02x%02x%02x%02x\n", a,b,c,d);` — Eddy_Em, Apr 25 '13 at 07:18
@Eddy_Em: being ultra persnickety, use `X` to get the `F1` part of the required result. — Jonathan Leffler, Apr 25 '13 at 07:20

Alex · Accepted Answer · 2019-04-28T15:44:51.150

23

Supposing unsigned int a,b,c,d;

unsigned int result = (a<<24) | (b<<16)| (c<<8) | d;

But this is essentially implementation dependent since C++ standard only specifies minimal sizes of integers.

So for uint32_t a, b, c, d:

uint32_t result = (a<<24) | (b<<16)| (c<<8) | d;

edited Apr 28 '19 at 15:44

answered Apr 25 '13 at 07:15

Alex

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11
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This answer correctly uses `unsigned int`. We should emphasize the **unsigned**. If the identifiers are `int` as in the question, `a<<24` may be undefined. – Eric Postpischil Apr 25 '13 at 13:40
In the general case, and for maximum portability (that is, to machines with 16-bit `int`) some casts are necessary: `uint32_t result = ((uint32_t )a<<24) | ((uint32_t )b<<16)| (c<<8) | d;` – Steve Summit Aug 11 '21 at 17:30

How to Concatenate Hex Numbers?

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