请问这个方法一共有6个参数
fun1(int a1, int a2, int a3, int a4, int a5, int a6)
下面是方法体里面的具体arm指令:
.text:00001450 var_5C = -0x5C
.text:00001450 var_58 = -0x58
.text:00001450 var_54 = -0x54
.text:00001450 var_50 = -0x50
.text:00001450 anonymous_1 = -0x4C
.text:00001450 anonymous_3 = -0x48
.text:00001450 anonymous_5 = -0x44
.text:00001450 var_40 = -0x40
.text:00001450 var_30 = -0x30
.text:00001450 anonymous_0 = -0x2C
.text:00001450 anonymous_2 = -0x28
.text:00001450 anonymous_4 = -0x24
.text:00001450 var_20 = -0x20
.text:00001450 arg_0 = 8
.text:00001450 arg_4 = 0xC
.text:00001450
.text:00001450 ; __unwind {
.text:00001450 PUSH {R4-R7,LR}
.text:00001452 ADD R7, SP, #0xC
.text:00001454 PUSH.W {R8-R11}
.text:00001458 SUB SP, SP, #0x44
.text:0000145A MOV R11, R0
.text:0000145C LDR R0, =(__stack_chk_guard_ptr - 0x1466)
.text:0000145E MOV R10, R3
.text:00001460 MOV R9, R2
.text:00001462 ADD R0, PC ; __stack_chk_guard_ptr
.text:00001464 MOV R6, R1
.text:00001466 MOV.W R8, #0
.text:0000146A LDR R5, [R0] ; __stack_chk_guard
.text:0000146C LDR R0, [R5]
.text:0000146E STR R0, [SP,#0x60+var_20]
.text:00001470 LSLS R0, R1, #0x1C
.text:00001472 BNE loc_14F2
.text:00001474 ADD R4, SP, #0x60+var_50
.text:00001476 LDR R1, [R7,#arg_4]
.text:00001478 MOVS R2, #0x10
.text:0000147A MOV R0, R4
我知道参数多于4个会放在栈中传递,
请问参数a6 是在哪里取出来的啊?
