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?? mathevrc.c

?? 完整的EVRC壓縮解壓縮算法源碼,附帶一個簡單的例子程序。
?? C
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	if (dSum != L_Sum)
	{
		/* overflow occurred */
		L_Sum = L_saturate(dSum);
        OP_COUNT(-4);        /* Complexity Count -- LT 6/96 */
	}

	return (L_Sum);
}

/***************************************************************************
 *
 *   FUNCTION NAME:mac_r
 *
 *   PURPOSE:
 *
 *     Multiply accumulate and round.  Fractionally multiply two 16
 *     bit numbers together with saturation.  Add that result to
 *     the 32 bit input with saturation.  Finally round the result
 *     into a 16 bit number.
 *
 *
 *   INPUTS:
 *
 *     var1
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var1 <= 0x0000 7fff.
 *     var2
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var2 <= 0x0000 7fff.
 *     L_var3
 *                     32 bit long signed integer (Longword) whose value
 *                     falls in the range
 *                     0x8000 0000 <= L_var2 <= 0x7fff ffff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0xffff 8000 <= swOut <= 0x0000 7fff.
 *
 *   IMPLEMENTATION:
 *
 *     Fractionally multiply two 16 bit numbers together with
 *     saturation.  The only numbers which will cause saturation on
 *     the multiply are 0x8000 * 0x8000.
 *
 *     Add that result to the 32 bit input with saturation.
 *     Round the 32 bit result by adding 0x0000 8000 to the input.
 *     The result may overflow due to the add.  If so, the result
 *     is saturated.  The 32 bit rounded number is then shifted
 *     down 16 bits and returned as a Shortword.
 *
 *     Please note that this is not a true multiply accumulate as
 *     most processors would implement it.  The 0x8000*0x8000
 *     causes and overflow for this instruction.  On most
 *     processors this would cause an overflow only if the 32 bit
 *     input added to it were positive or zero.
 *
 *   KEYWORDS: mac, multiply accumulate, macr
 *
 *************************************************************************/

Shortword mac_r(Longword L_var3, Shortword var1, Shortword var2)
{
    return (round32(L_mac(L_var3, var1, var2)));
}

/***************************************************************************
 *
 *   FUNCTION NAME:  msu_r
 *
 *   PURPOSE:
 *
 *     Multiply subtract and round.  Fractionally multiply two 16
 *     bit numbers together with saturation.  Subtract that result from
 *     the 32 bit input with saturation.  Finally round the result
 *     into a 16 bit number.
 *
 *
 *   INPUTS:
 *
 *     var1
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var1 <= 0x0000 7fff.
 *     var2
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var2 <= 0x0000 7fff.
 *     L_var3
 *                     32 bit long signed integer (Longword) whose value
 *                     falls in the range
 *                     0x8000 0000 <= L_var2 <= 0x7fff ffff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0xffff 8000 <= swOut <= 0x0000 7fff.
 *
 *   IMPLEMENTATION:
 *
 *     Fractionally multiply two 16 bit numbers together with
 *     saturation.  The only numbers which will cause saturation on
 *     the multiply are 0x8000 * 0x8000.
 *
 *     Subtract that result from the 32 bit input with saturation.
 *     Round the 32 bit result by adding 0x0000 8000 to the input.
 *     The result may overflow due to the add.  If so, the result
 *     is saturated.  The 32 bit rounded number is then shifted
 *     down 16 bits and returned as a Shortword.
 *
 *     Please note that this is not a true multiply accumulate as
 *     most processors would implement it.  The 0x8000*0x8000
 *     causes and overflow for this instruction.  On most
 *     processors this would cause an overflow only if the 32 bit
 *     input added to it were positive or zero.
 *
 *   KEYWORDS: mac, multiply accumulate, macr
 *
 *************************************************************************/

Shortword msu_r(Longword L_var3, Shortword var1, Shortword var2)
{
    return (round32(L_msu(L_var3, var1, var2)));
}

/***************************************************************************
 *
 *   FUNCTION NAME: mult
 *
 *   PURPOSE:
 *
 *     Perform a fractional multipy of the two 16 bit input numbers
 *     with saturation and truncation.
 *
 *   INPUTS:
 *
 *     var1
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var1 <= 0x0000 7fff.
 *     var2
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var2 <= 0x0000 7fff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0xffff 8000 <= swOut <= 0x0000 7fff.
 *
 *   IMPLEMENTATION:
 *
 *     Perform a fractional multipy of the two 16 bit input
 *     numbers.  If var1 == var2 == -0x8000, output 0x7fff.
 *     Otherwise output var1*var2 >> 15.  The output is a
 *     16 bit number.
 *
 *   KEYWORDS: mult, mulitply, mpy
 *
 *************************************************************************/

Shortword mult(Shortword var1, Shortword var2)
{
	Longword L_product;
	Shortword swOut;

    OP_COUNT(1);        /* Complexity Count -- LT 6/96 */

    L_product = L_mult(var1, var2); OP_COUNT(-1);
    swOut = extract_h(L_product);   OP_COUNT(-1);
	return (swOut);
}

/***************************************************************************
 *
 *   FUNCTION NAME: mult_r
 *
 *   PURPOSE:
 *
 *     Perform a fractional multipy and round of the two 16 bit
 *     input numbers with saturation.
 *
 *   INPUTS:
 *
 *     var1
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var1 <= 0x0000 7fff.
 *     var2
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var2 <= 0x0000 7fff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0xffff 8000 <= swOut <= 0x0000 7fff.
 *
 *   IMPLEMENTATION:
 *
 *     This routine is defined as the concatenation of the multiply
 *     operation and the round operation.
 *
 *     The fractional multiply (L_mult) produces a saturated 32 bit
 *     output.  This is followed by a an add of 0x0000 8000 to the
 *     32 bit result.  The result may overflow due to the add.  If
 *     so, the result is saturated.  The 32 bit rounded number is
 *     then shifted down 16 bits and returned as a Shortword.
 *
 *
 *   KEYWORDS: multiply and round, round, mult_r, mpyr
 *
 *************************************************************************/

Shortword mult_r(Shortword var1, Shortword var2)
{
	Shortword swOut;

    OP_COUNT(2);        /* Complexity Count -- LT 6/96 */

    swOut = round32(L_mult(var1, var2));
    OP_COUNT(-2);        /* Complexity Count -- LT 6/96 */
    return (swOut);
}

/***************************************************************************
 *
 *   FUNCTION NAME: negate
 *
 *   PURPOSE:
 *
 *     Negate the 16 bit input. 0x8000's negated value is 0x7fff.
 *
 *   INPUTS:
 *
 *     var1
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var1 <= 0x0000 7fff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0xffff 8001 <= swOut <= 0x0000 7fff.
 *
 *   KEYWORDS: negate, negative, invert
 *
 *************************************************************************/

Shortword negate(Shortword var1)
{
	Shortword swOut;

    OP_COUNT(1);        /* Complexity Count -- LT 6/96 */

    if (var1 == SW_MIN)
	{
		swOut = SW_MAX;
		giOverflow = 1;
	}
	else
		swOut = -var1;
	return (swOut);
}

/***************************************************************************
 *
 *   FUNCTION NAME: norm_l
 *
 *   PURPOSE:
 *
 *     Get normalize shift count:
 *
 *     A 32 bit number is input (possiblly unnormalized).  Output
 *     the positive (or zero) shift count required to normalize the
 *     input.
 *
 *   INPUTS:
 *
 *     L_var1
 *                     32 bit long signed integer (Longword) whose value
 *                     falls in the range
 *                     0x8000 0000 <= L_var1 <= 0x7fff ffff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0 <= swOut <= 31
 *
 *
 *
 *   IMPLEMENTATION:
 *
 *     Get normalize shift count:
 *
 *     A 32 bit number is input (possiblly unnormalized).  Output
 *     the positive (or zero) shift count required to normalize the
 *     input.
 *
 *     If zero in input, return 0 as the shift count.
 *
 *     For non-zero numbers, count the number of left shift
 *     required to get the number to fall into the range:
 *
 *     0x4000 0000 >= normlzd number >= 0x7fff ffff (positive number)
 *     or
 *     0x8000 0000 <= normlzd number < 0xc000 0000 (negative number)
 *
 *     Return the number of shifts.
 *
 *     This instruction corresponds exactly to the Full-Rate "norm"
 *     instruction.
 *
 *   KEYWORDS: norm, normalization
 *
 *************************************************************************/

Shortword norm_l(Longword L_var1)
{
	Shortword swShiftCnt;

    OP_COUNT(30);        /* Complexity Count -- LT 6/96 */

	if (L_var1 != 0)
	{
		if (!(L_var1 & LW_SIGN))
		{
			/* positive input */
			for (swShiftCnt = 0; !(L_var1 <= LW_MAX && L_var1 >= 0x40000000L);
				 swShiftCnt++)
			{
				L_var1 = L_var1 << 1;
			}
		}
		else
		{
			/* negative input */
			for (swShiftCnt = 0;
				 !(L_var1 >= LW_MIN && L_var1 < (Longword) 0xc0000000L);
				 swShiftCnt++)
			{
				L_var1 = L_var1 << 1;
			}
		}
	}
	else
	{
		swShiftCnt = 0;
	}
	return (swShiftCnt);
}

/***************************************************************************
 *
 *   FUNCTION NAME: norm_s
 *
 *   PURPOSE:
 *
 *     Get normalize shift count:
 *
 *     A 16 bit number is input (possiblly unnormalized).  Output
 *     the positive (or zero) shift count required to normalize the
 *     input.
 *
 *   INPUTS:
 *
 *     var1
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range 0xffff 8000 <= var1 <= 0x0000 7fff.
 *
 *   OUTPUTS:
 *
 *     none
 *
 *   RETURN VALUE:
 *     swOut
 *                     16 bit short signed integer (Shortword) whose value
 *                     falls in the range
 *                     0 <= swOut <= 15
 *
 *
 *
 *   IMPLEMENTATION:
 *
 *     Get normalize shift count:
 *
 *     A 16 bit number is input (possiblly unnormalized).  Output
 *     the positive (or zero) shift count required to normalize the
 *     input.
 *
 *     If zero in input, return 0 as the shift count.
 *
 *     For non-zero numbers, count the number of left shift
 *     required to get the number to fall into the range:
 *
 *     0x4000 >= normlzd number >= 0x7fff (positive number)
 *     or
 *     0x8000 <= normlzd number <  0xc000 (negative number)
 *
 *     Return the number of shifts.
 *
 *     This instruction corresponds exactly to the Full-Rate "norm"
 *     instruction.
 *
 *   KEYWORDS: norm, normalization
 *
 *************************************************************************/

Shortword norm_s(Shortword var1)
{
	short swShiftCnt;
	Longword L_var1;

    OP_COUNT(15);        /* Complexity Count -- LT 6/96 */

    L_var1 = L_deposit_h(var1);     OP_COUNT(-1);
    swShiftCnt = norm_l(L_var1);    OP_COUNT(-30);
	return (swShiftCnt);
}

/****************************************************************************
 *
 *     FUNCTION NAME: popOverflow
 *
 *     PURPOSE:
 *
 *        Pull the old overflow state from the "stack".  Replace the current
 *        overflow status with its predecessor.

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