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Clarify magnitude comparison logic for IEEE-754
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John Winans
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@ -43,66 +43,71 @@ is the portion of the {\em mantissa} that is to the right of the binary-point.
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point numbers will start that way. Thus we can save memory when storing
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normalized values by inserting a `1.' to the left of significand.
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{
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\small
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\setlength{\unitlength}{.15in}
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\begin{picture}(32,4)(0,0)
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\put(0,1){\line(1,0){32}} % bottom line
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\put(0,2){\line(1,0){32}} % top line
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\put(0,1){\line(0,1){2}} % left vertical
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\put(0,2){\makebox(1,1){\tiny 31}} % left end bit number marker
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\put(32,1){\line(0,1){2}} % vertical right end
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\put(31,2){\makebox(1,1){\tiny 0}} % right end bit number marker
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\DrawBitBoxIEEEFloat{11000000001010000000000000000000}
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\put(0,0){\makebox(1,1){\small sign}}
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\put(1,0){\makebox(8,1){\small exponent}}
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\put(9,0){\makebox(23,1){\small significand}}
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\put(0,1){\makebox(1,1){1}} % sign
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\put(1,1){\line(0,1){2}} % seperator
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\put(1,2){\makebox(1,1){\tiny 30}} % bit marker
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\put(1,1){\makebox(1,1){1}} % exponent
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\put(2,1){\makebox(1,1){0}}
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\put(3,1){\makebox(1,1){0}}
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\put(4,1){\makebox(1,1){0}}
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\put(5,1){\makebox(1,1){0}}
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\put(6,1){\makebox(1,1){0}}
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\put(7,1){\makebox(1,1){0}}
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\put(8,1){\makebox(1,1){0}}
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\put(8,2){\makebox(1,1){\tiny 23}} % bit marker
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\put(9,1){\line(0,1){2}} % seperator
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\put(9,2){\makebox(1,1){\tiny 22}} % bit marker
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\put(9,1){\makebox(1,1){0}}
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\put(10,1){\makebox(1,1){1}}
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\put(11,1){\makebox(1,1){0}}
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\put(12,1){\makebox(1,1){1}}
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\put(13,1){\makebox(1,1){0}}
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\put(14,1){\makebox(1,1){0}}
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\put(15,1){\makebox(1,1){0}}
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\put(16,1){\makebox(1,1){0}}
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\put(17,1){\makebox(1,1){0}}
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\put(18,1){\makebox(1,1){0}}
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\put(19,1){\makebox(1,1){0}}
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\put(20,1){\makebox(1,1){0}}
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\put(21,1){\makebox(1,1){0}}
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\put(22,1){\makebox(1,1){0}}
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\put(23,1){\makebox(1,1){0}}
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\put(24,1){\makebox(1,1){0}}
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\put(25,1){\makebox(1,1){0}}
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\put(26,1){\makebox(1,1){0}}
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\put(27,1){\makebox(1,1){0}}
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\put(28,1){\makebox(1,1){0}}
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\put(29,1){\makebox(1,1){0}}
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\put(30,1){\makebox(1,1){0}}
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\put(31,1){\makebox(1,1){0}}
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\end{picture}
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}
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%{
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%\small
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%\setlength{\unitlength}{.15in}
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%\begin{picture}(32,4)(0,0)
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% \put(0,1){\line(1,0){32}} % bottom line
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% \put(0,2){\line(1,0){32}} % top line
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%
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% \put(0,1){\line(0,1){2}} % left vertical
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% \put(0,2){\makebox(1,1){\tiny 31}} % left end bit number marker
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%
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% \put(32,1){\line(0,1){2}} % vertical right end
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% \put(31,2){\makebox(1,1){\tiny 0}} % right end bit number marker
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%
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% \put(0,0){\makebox(1,1){\small sign}}
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% \put(1,0){\makebox(8,1){\small exponent}}
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% \put(9,0){\makebox(23,1){\small significand}}
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%
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% \put(0,1){\makebox(1,1){1}} % sign
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%
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% \put(1,1){\line(0,1){2}} % seperator
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% \put(1,2){\makebox(1,1){\tiny 30}} % bit marker
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%
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% \put(1,1){\makebox(1,1){1}} % exponent
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% \put(2,1){\makebox(1,1){0}}
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% \put(3,1){\makebox(1,1){0}}
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% \put(4,1){\makebox(1,1){0}}
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% \put(5,1){\makebox(1,1){0}}
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% \put(6,1){\makebox(1,1){0}}
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% \put(7,1){\makebox(1,1){0}}
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% \put(8,1){\makebox(1,1){0}}
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%
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% \put(8,2){\makebox(1,1){\tiny 23}} % bit marker
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% \put(9,1){\line(0,1){2}} % seperator
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% \put(9,2){\makebox(1,1){\tiny 22}} % bit marker
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%
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% \put(9,1){\makebox(1,1){0}}
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% \put(10,1){\makebox(1,1){1}}
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% \put(11,1){\makebox(1,1){0}}
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% \put(12,1){\makebox(1,1){1}}
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% \put(13,1){\makebox(1,1){0}}
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% \put(14,1){\makebox(1,1){0}}
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% \put(15,1){\makebox(1,1){0}}
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% \put(16,1){\makebox(1,1){0}}
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% \put(17,1){\makebox(1,1){0}}
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% \put(18,1){\makebox(1,1){0}}
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% \put(19,1){\makebox(1,1){0}}
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% \put(20,1){\makebox(1,1){0}}
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% \put(21,1){\makebox(1,1){0}}
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% \put(22,1){\makebox(1,1){0}}
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% \put(23,1){\makebox(1,1){0}}
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% \put(24,1){\makebox(1,1){0}}
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% \put(25,1){\makebox(1,1){0}}
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% \put(26,1){\makebox(1,1){0}}
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% \put(27,1){\makebox(1,1){0}}
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% \put(28,1){\makebox(1,1){0}}
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% \put(29,1){\makebox(1,1){0}}
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% \put(30,1){\makebox(1,1){0}}
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% \put(31,1){\makebox(1,1){0}}
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%\end{picture}
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%}
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%\item $-((1 + \frac{1}{4} + \frac{1}{16}) \times 2^{128-127}) = -(1 \frac{5}{16} \times 2^{1}) = -(1.3125 \times 2^{1}) = -2.625$
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\item $-((1 + \frac{1}{4} + \frac{1}{16}) \times 2^{128-127}) = -((1 + \frac{1}{4} + \frac{1}{16}) \times 2^1) = -(2 + \frac{1}{2} + \frac{1}{8}) = -(2 + .5 + .125) = -2.625$
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@ -127,12 +132,25 @@ the sign is zero, the number represents positive infinity.
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\item When the exponent is all ones, the significand is all zeros, and
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the sign is one, the number represents negative infinity.
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\item Note that the binary representation of an IEEE-754 number in memory
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can be compared for magnitude with another one using the same logic as for
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comparing two's complement signed integers because the magnitude of an
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IEEE number grows upward and downward in the same fashion as signed integers.
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This is why we use excess notation and locate the significand's sign bit on
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the left of the exponent.
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\item Observe that the binary representation of a pair of IEEE-754 numbers
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(when one or both are positive) can be compared for magnitude
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by treating them as if they are two's complement signed integers.
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This is because an IEEE number is stored in {\em signed magnitude} format and
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therefore positive floating point values will grow upward and downward in the
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same fashion as for unsigned integers and that since negative floating point
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values will have its MSB set, they will `appear` to be less than a positive
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floating point value.
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When comparing two negative IEEE float values by treating them both as two's
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complement signed integers, the order will be reversed because IEEE float values
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with larger (that is, increasingly negative) magnitudes will appear to decrease
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in value when interpreted as signed integers.
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This works this way because excess notation is used in the format of the
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exponent and why the significand's sign bit is located on the left of
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the exponent.\footnote{I know this is true and was done on purpose because
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Bill Cody, chairman of IEEE committee P754 that designed the IEEE-754 standard,
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told me so personally circa 1991.}
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\item Note that zero is a special case number. Recall that a normalized
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number has an implied 1-bit to the left of the significand\ldots\ which
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@ -1626,3 +1626,69 @@
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\newcommand\rvddt{{\tt rvddt}}
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\newcommand\hex[1]{{\tt 0x#1}}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% IEEE-754 Floating Point Number
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%
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% #1 sign
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% #2 exponent
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% #3 significand
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\newcommand\DrawIEEEFloat[1]{
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\StrLen{#1}[\numchars]
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\begin{scope}[shift={(0,.75)}]
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\DrawInsnBitstring{\numchars}{#1}{}
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\DrawInsnBoxSeg{\numchars}{31}{31}{sign}
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\DrawInsnBoxSeg{\numchars}{30}{23}{exponent}
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\DrawInsnBoxSeg{\numchars}{22}{0}{significand}
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\draw {[rounded corners=\SignBoxCornerRadius] (1.35, -.6) -- (1.35, .6) -- (.65, .6) -- (.65, -.6) -- cycle}; % sign bit
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\end{scope}
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\DrawHexMarkersRel{\numchars}
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% \DrawInsnBoxRel{31}{31}{}
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% \DrawInsnBoxRel{30}{23}{exponent}
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% \DrawInsnBoxRel{22}{0}{significand}
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% \draw(\InsnBoxTypePosY,.75) node[right]{\hyperref[insnformat:rtype]{R-type}};
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% \draw(\InsnBoxMnemonicPosY,.75) node[right]{\tt #4};
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% \draw(\InsnBoxMnemonicArgPosY,.75) node[right]{\tt #5};
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% \begin{scope}[shift={(31-31,0)}]\DrawBitstringX{#1}\end{scope}
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% \begin{scope}[shift={(31-30,0)}]\DrawBitstringX{#2}\end{scope}
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% \begin{scope}[shift={(31-22,0)}]\DrawBitstringX{#3}\end{scope}
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}
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\newcommand\DrawBitBoxIEEEFloat[1]{
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\BeginTikzPicture
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\begin{scope}[shift={(0,0)}]\DrawIEEEFloat{#1}\end{scope}
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\EndTikzPicture
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}
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%
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%\newcommand\DrawInsnTypeB[1]{
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% \StrLen{#1}[\numchars]
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% \begin{scope}[shift={(0,.75)}]
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% \DrawInsnBitstring{\numchars}{#1}{\hyperref[insnformat:btype]{B-type}}
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% \DrawInsnBoxSeg{\numchars}{31}{25}{imm[12\textbar10:5]}
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% \DrawInsnBoxSeg{\numchars}{24}{20}{rs2}
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% \DrawInsnBoxSeg{\numchars}{19}{15}{rs1}
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% \DrawInsnBoxSeg{\numchars}{14}{12}{funct3}
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% \DrawInsnBoxSeg{\numchars}{11}{7}{imm[4:1\textbar11]}
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% \DrawInsnBoxSeg{\numchars}{6}{0}{opcode}
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%
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% % add some hint bits in for imm fields
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% \draw {[rounded corners=\SignBoxCornerRadius] (1.35, -.6) -- (1.35, .6) -- (.65, .6) -- (.65, -.6) -- cycle}; % sign bit
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% \draw (32-7-.5, -.75) -- (32-7-.5, .1); % imm[11]
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% \draw (32-30-.5, -.75) -- (32-30.5, .1); % imm[12]
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%
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% \end{scope}
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%
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% \DrawHexMarkersRel{\numchars}
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%}
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