val a1 = client.newCall(newRequest("http://a/1"))
    val a2 = client.newCall(newRequest("http://a/2"))
    val a3 = client.newCall(newRequest("http://a/3"))
    val a4 = client.newCall(newRequest("http://a/4"))
    val a5 = client.newCall(newRequest("http://a/5"))
    a1.enqueue(callback)
    a2.enqueue(callback)
    a3.enqueue(callback)
    a4.enqueue(callback)
    a5.enqueue(callback)

    //
    // 1. Subscribers to events posted on different threads can be interleaved with each other
    //    freely. (A event on one thread, B event on another could yield any of
    //    [a1, a2, a3, b1, b2], [a1, b2, a2, a3, b2], [a1, b2, b3, a2, a3], etc.)
    // 2. It's possible for subscribers to actually be dispatched to in a different order than they
    //    were added to the queue. It's easily possible for one thread to take the head of the

        UniAddress a3 = mock(UniAddress.class);
        when(nsc.getAllByName("server", true)).thenReturn(new Address[] { a2, a3 });
        SmbResourceLocatorImpl l3 = locator("smb://server/");
        assertSame(a2, l3.getAddress());
        verify(nsc, times(1)).getAllByName("server", true);
        assertTrue(l3.hasNextAddress());
        assertSame(a3, l3.getNextAddress());

    val a3 = "2001:db8::1:0:0:1"
    assertThat(parse("http://[2001:db8:0:0:1:0:0:1]").host).isEqualTo(a3)
    assertThat(parse("http://[2001:0db8:0:0:1:0:0:1]").host).isEqualTo(a3)
    assertThat(parse("http://[2001:db8::1:0:0:1]").host).isEqualTo(a3)
    assertThat(parse("http://[2001:db8::0:1:0:0:1]").host).isEqualTo(a3)
    assertThat(parse("http://[2001:0db8::1:0:0:1]").host).isEqualTo(a3)

// See issue 43661.

TEXT ·a1(SB), 0, $0-0
	CMPL runtime·writeBarrier(SB), $0
	MOVL $0, R15
	RET
TEXT ·a2(SB), 0, $0-0
	CMPL runtime·writeBarrier(SB), $0
	MOVQ $0, R15
	RET
TEXT ·a3(SB), 0, $0-0
	CMPL runtime·writeBarrier(SB), $0
	XORL R15, R15
	RET
TEXT ·a4(SB), 0, $0-0
	CMPL runtime·writeBarrier(SB), $0
	XORQ R15, R15
	RET
TEXT ·a5(SB), 0, $0-0
	CMPL runtime·writeBarrier(SB), $0

00002f0 89 17 85 f8 16 94 f2 8a b2 a3 f5 b6 6d df 75 cd
0000300 90 dd 64 bd 5d 55 4e f2 55 19 1b b7 cc ef 1b ea
0000310 2e 05 9c f4 aa 1e a8 cd a6 82 c7 59 0f 5e 9d e0
0000320 bb fc 6c d6 99 23 eb 36 ad c6 c5 e1 d8 e1 e2 3e
0000330 d9 90 5a f7 91 5d 6f bc 33 6d 98 47 d2 7c 2e 2f
0000340 99 a4 25 72 85 49 2c be 0b 5b af 8f e5 6e 81 a6
0000350 a3 5a 6f 39 53 3a ab 7a 8b 1e 26 f7 46 6c 7d 26

	register["T2"] = riscv.REG_T2
	register["S0"] = riscv.REG_S0
	register["S1"] = riscv.REG_S1
	register["A0"] = riscv.REG_A0
	register["A1"] = riscv.REG_A1
	register["A2"] = riscv.REG_A2
	register["A3"] = riscv.REG_A3
	register["A4"] = riscv.REG_A4
	register["A5"] = riscv.REG_A5
	register["A6"] = riscv.REG_A6
	register["A7"] = riscv.REG_A7
	register["S2"] = riscv.REG_S2
	register["S3"] = riscv.REG_S3

   *     Iterators.peekingIterator(Iterators.forArray("a", "b"));
   * String a1 = peekingIterator.peek(); // returns "a"
   * String a2 = peekingIterator.peek(); // also returns "a"
   * String a3 = peekingIterator.next(); // also returns "a"
   * }
   *
   * <p>Any structural changes to the underlying iteration (aside from those performed by the

	</li>
</ul>

<p>
Given the declarations
</p>

<pre>
type (
	A0 = []string
	A1 = A0
	A2 = struct{ a, b int }
	A3 = int
	A4 = func(A3, float64) *A0
	A5 = func(x int, _ float64) *[]string

	B0 A0
	B1 []string
	B2 struct{ a, b int }
	B3 struct{ a, c int }
	B4 func(int, float64) *B0
	B5 func(x int, y float64) *A1

the columns into wide, overlapping limbs. // r0 = a0×b0 + 19×(a1×b4 + a2×b3 + a3×b2 + a4×b1) r0 := mul(a0, b0) r0 = addMul19(r0, a1, b4) r0 = addMul19(r0, a2, b3) r0 = addMul19(r0, a3, b2) r0 = addMul19(r0, a4, b1) // r1 = a0×b1 + a1×b0 + 19×(a2×b4 + a3×b3 + a4×b2) r1 := mul(a0, b1) r1 = addMul(r1, a1, b0) r1 = addMul19(r1, a2, b4) r1 = addMul19(r1, a3, b3) r1 = addMul19(r1, a4, b2) // r2 = a0×b2 + a1×b1 + a2×b0 + 19×(a3×b4 + a4×b3) r2 := mul(a0, b2) r2 = addMul(r2, a1, b1) r2 = addMul(r2, a2, b0)...