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Spell IEEE-754 consistently.
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@ -107,11 +107,11 @@ normalized values by adding 1 to the significand.
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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 $-((1 + \frac{1}{4} + \frac{1}{16}) \times 2^{128-127}) = -(1 \frac{5}{16} \times 2^{1}) = -(1.3125 \times 2^{1}) = -2.625$
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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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\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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\item IEEE754 formats:
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\item IEEE-754 formats:
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\begin{tabular}{|l|l|l|}
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\begin{tabular}{|l|l|l|}
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\hline
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\hline
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& IEEE754 32-bit & IEEE754 64-bit \\
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& IEEE-754 32-bit & IEEE-754 64-bit \\
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\hline
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\hline
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sign & 1 bit & 1 bit \\
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sign & 1 bit & 1 bit \\
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exponent & 8 bits (excess-127) & 11 bits (excess-1023) \\
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exponent & 8 bits (excess-127) & 11 bits (excess-1023) \\
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@ -127,7 +127,7 @@ the sign is zero, the number represents positive infinity.
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\item When the exponent is all ones, the mantissa is all zeros, and
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\item When the exponent is all ones, the mantissa is all zeros, and
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the sign is one, the number represents negative infinity.
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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 IEEE754 number in memory
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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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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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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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IEEE number grows upward and downward in the same fashion as signed integers.
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