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POOnly 7 bits for Java char?
 

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  1. POOnly 7 bits for Java char?
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    <p>I'm trying to make a UUencode algorithm and part of my code contains this:</p> <pre><code>for(int x = 0; x &lt; my_chars.length; x++) { if((x+1) % 3 == 0) { char first = my_chars[x-2]; char second = my_chars[x-1]; char third = my_chars[x]; int first_binary = Integer.parseInt(Integer.toBinaryString(first)); int second_binary = Integer.parseInt(Integer.toBinaryString(second)); int third_binary = Integer.parseInt(Integer.toBinaryString(third)); int n = (((first &lt;&lt; 8) | second) &lt;&lt; 8) | third; System.out.print(my_chars[x-2] + "" + my_chars[x-1] + my_chars[x] + Integer.toBinaryString(n)); } } System.out.println(); System.out.println(Integer.toBinaryString('s')); </code></pre> <p>What I'm trying to achieve is to combine those 8 bits from the chars that I get into a big 24 bits int. The problem I'm facing is that the result is a 23 bit int. Say my first 3 chars were:</p> <ul> <li>'T' with a binary representation of <code>01010100</code> </li> <li>'u' with a binary representation of <code>01110101</code> </li> <li>'r' with a binary representation of <code>01110010</code> </li> </ul> <p>The result that I get from my program is a int formed from these bits: <code>10101000111010101110010</code> Which is missing the 0 at the beginning from the representation of 'T'.</p> <p>Also I have included the last 2 lines of code because the binary string that I get from 's' is: <code>1110011</code> which is missing the 0 at the beginning.</p> <p>I have checked if I scrolled by mistake to the right but it does not seem that I have done so. </p>
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    1. USDuncan Jones
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    1. COASCII letters are 7 bit. Everything makes sense. If you're one bit short, you can just think of a leading zero, `n` has 32 bits anyway.
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      1. USjlordo
 

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