Lines Matching refs:Y
5 | input values X and Y. The entry point sREM computes the floating
6 | point (IEEE) REM of the input values X and Y.
10 | Double-extended value Y is pointed to by address in register
12 | of X and Y are both nonzero and finite; although either or both
18 | FREM(X,Y) or FMOD(X,Y), depending on entry point.
23 | Step 1. Save and strip signs of X and Y: signX := sign(X),
24 | signY := sign(Y), X := |X|, Y := |Y|,
28 | Step 2. Set L := expo(X)-expo(Y), k := 0, Q := 0.
35 | Step 3. Perform MOD(X,Y)
36 | 3.1 If R = Y, go to Step 9.
37 | 3.2 If R > Y, then { R := R - Y, Q := Q + 1}
42 | Step 4. At this point, R = X - QY = MOD(X,Y). Set
46 | Step 5. R = MOD(X,Y), but REM(X,Y) is requested.
47 | 5.1 If R < Y/2, then R = MOD(X,Y) = REM(X,Y). Go to
49 | 5.2 If R > Y/2, then { set Last_Subtract := true,
50 | Q := Q + 1, Y := signY*Y }. Go to Step 6.
51 | 5.3 This is the tricky case of R = Y/2. If Q is odd,
56 | Step 7. If Last_Subtract = true, R := R - Y.
60 | Step 9. At this point, R = 2^(-j)*X - Q Y = Y. Thus,
61 | X = 2^(j)*(Q+1)Y. set Q := 2^(j)*(Q+1),
85 .set Y,FP_SCR1 define
86 .set Y_Hi,Y+4
87 .set Y_Lo,Y+8
110 |..Save sign of X and Y
114 andil #0x00007FFF,%d3 | ...Y := |Y|
118 movel 8(%a0),%d5 | ...(D3,D4,D5) is |Y|
203 movel %d3,L_SCR1(%a6) | ...save biased expo(Y)
205 subl %d3,%d0 | ...L := expo(X)-expo(Y)
215 |..expo(X) < expo(Y). Thus X = mod(X,Y)
227 |..At this point carry = 0, R = (D1,D2), Y = (D4,D5)
228 cmpl %d4,%d1 | ...compare hi(R) and hi(Y)
230 cmpl %d5,%d2 | ...compare lo(R) and lo(Y)
233 |..At this point, R = Y
238 bcss R_LT_Y | ...borrow is set iff R < Y
241 |..If Carry is set, then Y < (Carry,D1,D2) < 2Y. Otherwise, Carry = 0
242 |..and Y < (D1,D2) < 2Y. Either way, perform R - Y
243 subl %d5,%d2 | ...lo(R) - lo(Y)
244 subxl %d4,%d1 | ...hi(R) - hi(Y)
249 |..At this point, Carry=0, R < Y. R = 2^(k-L)X - QY; k+j = L; j >= 0.
259 |..At this point, R=(Carry,D1,D2) = 2^(k-L)X - QY, j+k=L, j >= 0, R < 2Y.
264 |..k = L, j = 0, Carry = 0, R = (D1,D2) = X - QY, R < Y.
305 movew %d6,Y(%a6)
306 clrw Y+2(%a6)
322 movew %d6,Y(%a6)
334 movel L_SCR1(%a6),%d6 | ...new biased expo(Y)
335 subql #1,%d6 | ...biased expo(Y/2)
351 fsubx Y(%a6),%fp0 | ...no exceptions
392 |..R = 2^(-j)X - Q Y = Y, thus R = 0 and quotient = 2^j (Q+1)