sticking together 4 hex bytes question

Hello I have four DECIMAL numer for example 31(1F) 47(2F) 63(3F) FF(255)
how using those decimal numbers i get a HEX number of 1F2F3FFF?
Thanks.

Réponses (2)

Ameer Hamza
Ameer Hamza le 1 Nov 2020
Modifié(e) : Ameer Hamza le 1 Nov 2020
Try this code
x = [31 47 63 255];
y = reshape(dec2hex(x).', 1, [])
Result
>> y
y =
'1F2F3FFF'

6 commentaires

fima v
fima v le 1 Nov 2020
Hello Ameer, you are transposing the vector to be vertical from horisontal.
what is the meaning of reshaping 1,[] ?
Thanks.
Just to make it a row vector. The output without it is
>> x = [31 47 63 255];
>> dec2hex(x).'
ans =
2×4 char array
'123F'
'FFFF'
fima v
fima v le 1 Nov 2020
Modifié(e) : fima v le 1 Nov 2020
Hello Ameer,Yes that i understood, after you created a row vector.
we have another line of reshaping 1,[]
what is the meaning of it?
Ameer Hamza
Ameer Hamza le 1 Nov 2020
reshape(dec2hex(x).', 1, []) means that take the output of dec2hex(x).' and convert it into a vector with one row.
fima v
fima v le 1 Nov 2020
Modifié(e) : fima v le 1 Nov 2020
but X was a row vector in the beginning :-)
How its different now?
maybe it means make row vector with one column? that will explain wjy we get one number 1X1 vector.
Thanks.
I think it is better you run the following lines one by one. Then you will get an idea of what is happening
x = [31 47 63 255];
y = dec2hex(x)
y = y.'
y = reshape(y, 1, [])

Connectez-vous pour commenter.

Walter Roberson
Walter Roberson le 1 Nov 2020
x = [31 47 63 255];
swap(typecast(uint8(x), 'uint64'))

3 commentaires

I think that the second line should be
dec2hex(swapbytes(typecast(uint8(x), 'uint32')))
or also this
dec2hex(typecast(uint8(fliplr(x)), 'uint32'))
Walter Roberson
Walter Roberson le 1 Nov 2020
ah yes you are right, swapbytes and uint32
fima v
fima v le 1 Nov 2020
Modifié(e) : fima v le 1 Nov 2020
Hello Walter, can i please have intuition for the line Ameer posted?
first you take my array and convert each member from uint8 to uint32 are you dound zero padding?
then in each member you swap its numbers ,Why?
Then you convert each member from decimal to hex ,but our number is not decimal its uint_32
and we still have an array,how it came to be only one number?
Thanks.

Connectez-vous pour commenter.

Question posée :

le 1 Nov 2020

Commenté :

le 1 Nov 2020

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by