本文整理汇总了C#中BigInteger.unCache方法的典型用法代码示例。如果您正苦于以下问题：C# BigInteger.unCache方法的具体用法？C# BigInteger.unCache怎么用？C# BigInteger.unCache使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类BigInteger的用法示例。
在下文中一共展示了BigInteger.unCache方法的6个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: inplaceAdd
/**
* Performs: {@code op1 += addend}. The number must to have place to hold a
* possible carry.
*/
internal static void inplaceAdd(BigInteger op1, int addend)
{
int carry = inplaceAdd(op1.digits, op1.numberLength, addend);
if (carry == 1) {
op1.digits[op1.numberLength] = 1;
op1.numberLength++;
}
op1.unCache();
}示例2: inplaceSubtract
/**
* Performs {@code op1 -= op2}. {@code op1} must have enough place to store
* the result (i.e. {@code op1.bitLength() >= op2.bitLength()}). Both
* should be positive (what implies that {@code op1 >= op2}).
*
* @param op1
* the input minuend, and the output result.
* @param op2
* the subtrahend
*/
internal static void inplaceSubtract(BigInteger op1, BigInteger op2)
{
// PRE: op1 >= op2 > 0
subtract (op1.digits, op1.digits, op1.numberLength, op2.digits,
op2.numberLength);
op1.cutOffLeadingZeroes ();
op1.unCache();
}示例3: completeInPlaceSubtract
/**
* Same as
*
* @link #inplaceSubtract(BigInteger, BigInteger), but without the
* restriction of non-positive values
* @param op1
* should have enough space to save the result
* @param op2
*/
internal static void completeInPlaceSubtract(BigInteger op1, BigInteger op2)
{
int resultSign = op1.compareTo (op2);
if (op1.sign == 0) {
java.lang.SystemJ.arraycopy (op2.digits, 0, op1.digits, 0, op2.numberLength);
op1.sign = -op2.sign;
} else if (op1.sign != op2.sign) {
add (op1.digits, op1.digits, op1.numberLength, op2.digits,
op2.numberLength);
op1.sign = resultSign;
} else {
int sign = unsignedArraysCompare (op1.digits,
op2.digits, op1.numberLength, op2.numberLength);
if (sign > 0) {
subtract (op1.digits, op1.digits, op1.numberLength, op2.digits,
op2.numberLength); // op1 = op1 - op2
// op1.sign remains equal
} else {
inverseSubtract (op1.digits, op1.digits, op1.numberLength,
op2.digits, op2.numberLength); // op1 = op2 - op1
op1.sign = -op1.sign;
}
}
op1.numberLength = java.lang.Math.max (op1.numberLength, op2.numberLength) + 1;
op1.cutOffLeadingZeroes ();
op1.unCache();
}示例4: completeInPlaceAdd
/**
* Same as @link #inplaceAdd(BigInteger, BigInteger), but without the restriction of
* non-positive values
* @param op1 any number
* @param op2 any number
*/
internal static void completeInPlaceAdd(BigInteger op1, BigInteger op2)
{
if (op1.sign == 0)
java.lang.SystemJ.arraycopy (op2.digits, 0, op1.digits, 0, op2.numberLength);
else if (op2.sign == 0)
return;
else if (op1.sign == op2.sign)
add (op1.digits, op1.digits, op1.numberLength, op2.digits,
op2.numberLength);
else {
int sign = unsignedArraysCompare(op1.digits,
op2.digits, op1.numberLength, op2.numberLength);
if (sign > 0)
subtract (op1.digits, op1.digits, op1.numberLength, op2.digits,
op2.numberLength);
else {
inverseSubtract (op1.digits, op1.digits, op1.numberLength,
op2.digits, op2.numberLength);
op1.sign = -op1.sign;
}
}
op1.numberLength = java.lang.Math.max (op1.numberLength, op2.numberLength) + 1;
op1.cutOffLeadingZeroes ();
op1.unCache();
}示例5: inplaceShiftRight
/**
* Performs {@code val >>= count} where {@code val} is a positive number.
*/
internal static void inplaceShiftRight(BigInteger val, int count)
{
int sign = val.signum();
if (count == 0 || val.signum() == 0)
return;
int intCount = count >> 5; // count of integers
val.numberLength -= intCount;
if (!shiftRight(val.digits, val.numberLength, val.digits, intCount,
count & 31)
&& sign < 0) {
// remainder not zero: add one to the result
int i;
for (i = 0; ( i < val.numberLength ) && ( val.digits[i] == -1 ); i++) {
val.digits[i] = 0;
}
if (i == val.numberLength) {
val.numberLength++;
}
val.digits[i]++;
}
val.cutOffLeadingZeroes();
val.unCache();
}示例6: inplaceShiftLeft
/**
* Performs {@code val <<= count}.
*/
// val should have enough place (and one digit more)
internal static void inplaceShiftLeft(BigInteger val, int count)
{
int intCount = count >> 5; // count of integers
val.numberLength += intCount
+ ( java.lang.Integer
.numberOfLeadingZeros(val.digits[val.numberLength - 1])
- ( count & 31 ) >= 0 ? 0 : 1 );
shiftLeft(val.digits, val.digits, intCount, count & 31);
val.cutOffLeadingZeroes();
val.unCache();
}