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?
最佳答案
是的。来自 this manual ,第 C6-1237 页及以下:
Signed offset
[...]
64-bit variant
Applies whenopc == 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];
关于assembly - STP 中的寄存器存储顺序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57153954/