assembly - STP 中的寄存器存储顺序

Question

在 AArch64 程序集中，以下行

stp x25, x30, [sp,#48]

在 sp+48 处存储 x25，在 sp+56 处存储 x30，对吗？

Answer 1

是的。来自 this manual ，第 C6-1237 页及以下:

Signed offset

[...]

64-bit variant

Applies when opc == 10.

STP <Xt1>, <Xt2>, [<Xn|SP>{, #<imm>}]

Decode for all variants of this encoding

boolean wback = FALSE;boolean postindex = FALSE;

[...]

Shared decode for all encodings

[...]

integer n = UInt(Rn);integer t = UInt(Rt);integer t2 = UInt(Rt2);[...]integer scale = 2 + UInt(opc<1>);integer datasize = 8 << scale;bits(64) offset = LSL(SignExtend(imm7, 64), scale);

[...]

Operation for all encodings

constant integer dbytes = datasize DIV 8;[...]if n == 31 then    CheckSPAlignment();    address = SP[];else    address = X[n];if !postindex then    address = address + offset;[...]data1 = X[t];[...]data2 = X[t2];Mem[address, dbytes, AccType_NORMAL] = data1;Mem[address+dbytes, dbytes, AccType_NORMAL] = data2;

让我们从头到尾了解一下。您的 stp x25, x30, [sp,#48] 是一个 64 位有符号偏移量 stp，它解码为:

n = 31
t = 25
t2 = 30
scale = 3 // since opc = 0b10
datasize = 64
offset = 48

将其插入到操作伪代码中，用变量替换它们的值，您就可以有效地得到:

CheckSPAlignment();
Mem[SP[] + 48, 8, AccType_NORMAL] = X[25];
Mem[SP[] + 56, 8, AccType_NORMAL] = X[30];