How to convert a string to a 8 bit Ascii one row binary vector?

Hello everyone. I have for example the string:"abcd" i want to convert it to "01100001011000100110001101100100" (see the apendded zeros). as a logical vector.

thanks!

Answer 1

[Azzi Abdelmalek]

s='abcd'
p=dec2bin(double('abcd'),8)'
out=p(:)'

[Jack Day]

thank you Azzi. But unfortunately, dec2bin doesnt convert well the full ascii table. 'abcd' is converted well, but ╞ , ╨ etc. does not.

i need a convertion of 8 bit ascii of all the characters. thanks

[Stephen23]

@Jack Day: ╞ , ╨ are not in the ASCII character set. ASCII only uses seven bits.

Those characters are in the Unicode "Box Drawing" set, which starts at 0X2500 (9472 decimal). There is no way that these can fit into seven (or even eight) bits!

To convert into a logical vector with eight bits per character, as the question requires:

 reshape(dec2bin(+'abcd',8)',1,[])=='1'