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8247782: typos in java.math
Reviewed-by: rriggs, lancea, darcy
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34c79640e7
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4f2edacaaf
2 changed files with 4 additions and 4 deletions
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@ -934,7 +934,7 @@ public class BigDecimal extends Number implements Comparable<BigDecimal> {
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// At this point, val == sign * significand * 2**exponent.
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/*
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* Special case zero to supress nonterminating normalization and bogus
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* Special case zero to suppress nonterminating normalization and bogus
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* scale calculation.
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*/
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if (significand == 0) {
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@ -4052,7 +4052,7 @@ public class BigDecimal extends Number implements Comparable<BigDecimal> {
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pows[i] = pows[i - 1].multiply(BigInteger.TEN);
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}
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// Based on the following facts:
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// 1. pows is a private local varible;
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// 1. pows is a private local variable;
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// 2. the following store is a volatile store.
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// the newly created array elements can be safely published.
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BIG_TEN_POWERS_TABLE = pows;
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@ -2751,7 +2751,7 @@ public class BigInteger extends Number implements Comparable<BigInteger> {
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BigInteger base2 = (this.signum < 0 || this.compareTo(m1) >= 0
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? this.mod(m1) : this);
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// Caculate (base ** exponent) mod m1.
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// Calculate (base ** exponent) mod m1.
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BigInteger a1 = (m1.equals(ONE) ? ZERO :
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base2.oddModPow(exponent, m1));
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@ -2905,7 +2905,7 @@ public class BigInteger extends Number implements Comparable<BigInteger> {
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* This means that if you have a k-bit window, to compute n^z,
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* where z is the high k bits of the exponent, 1/2 of the time
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* it requires no squarings. 1/4 of the time, it requires 1
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* squaring, ... 1/2^(k-1) of the time, it reqires k-2 squarings.
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* squaring, ... 1/2^(k-1) of the time, it requires k-2 squarings.
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* And the remaining 1/2^(k-1) of the time, the top k bits are a
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* 1 followed by k-1 0 bits, so it again only requires k-2
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* squarings, not k-1. The average of these is 1. Add that
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