Note on Padding
RFC 70
Document  Type 
RFC  Unknown
(October 1970; No errata)
Updated by RFC 228



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Last updated  20130302  
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Network Working Group S. Crocker Request for Comments #70 UCLA 15 October 70 A Note on Padding The padding on a message is a string of the form 10*. For Hosts with word lengths 16, 32, 48, etc., bits long, this string is necessarily in the last word received from the Imp. For Hosts with word lengths which are not a multiple of 16 (but which are at least 16 bits long), the 1 bit will be in either the last word or the next to last word. Of course if the 1 bit is in the next to last word, the last word is all zero. An unpleasant coding task is discovering the bit position of the 1 bit within its word. One obvious technique is to repeatedly test the loworder bit, shifting the word right one bit position if the loworder bit is zero. The following techniques are more pleasant. Isolating the LowOrder Bit Let W be a nonzero word, where the word length is n. Then W is of the form x....x10....0 \__ __/\__ __/ V V nk1 k where 0<=k<n and the x's are arbitrary bits. Assuming two's complement arithmetic, W1 = x....x01....1 _ _ W = x....x10....0 _ _ _ W = x....x01....1 By using AND, OR and exclusive OR with various pairs of these quantities, useful new forms are obtained. For example, [Page 1] Network Working Group A Note on Padding RFC 70 W AND W1 xx...x00....0 \__ __/\__ __/ V V nk1 k thus removing the loworder 1 bit; also W AND W = 0....010....0 __ __/__ __/ V V nk1 k thus isolating the loworder bit. Below, we will focus solely on this last result; however, in a particular application it may be advantageous to use a variation. Determining the Position of an Isolated Bit The two obvious techniques for finding the bit position of an isolated bit are to shift repetitively with tests, as above, and to use floating normalization hardware. On the PDP10, in particular, the JFFO instruction is made to order*. On machines with hexadecimal normalization, e.g. IBM 360's and XDS Sigma 7's, the normalization hardware may not be very convenient. A different approach uses division and table lookup. k A word with a single bit on has an unsigned integer value of 2 for k 0<=k<n. If we choose a p such that mod(2 ,p) is distinct for each 0<=k<n, we can make a table of length p which gives the correspondence k between mod(2 ,p) and k. The remainder of this paper is concerned with the selection of an appropriate divisor p for each word length n. *Some of the CDC machines have a "population count" instruction which k gives the number of bits in a word. Note the 2 1 has exactly k bits on. [Page 2] Network Working Group A Note on Padding RFC 70 Example Let n = 8 and p = 11 Then 0 mod(2, 11) = 1 1 mod(2, 11) = 2 2 mod(2, 11) = 4 3 mod(2, 11) = 8 4 mod(2, 11) = 5 5 mod(2, 11) = 10 6 mod(2, 11) = 9 7 mod(2, 11) = 7 This yields a table of the form remainder bit position 0  1 0 2 1 3  4 2 5 4 6  7 7 8 3 9 6 10 5 [Page 3] Network Working Group A Note on Padding RFC 70 Good Divisors The divisor p should be as small as possible in order to minimize the length of the table. Since the divisor must generate n distinct remainders, the divisor will certainly need to be at least n. A remainder of zero, however, can occur only if the divisor is a power of jShow full document text