Tags: X86 ARM MIPS ARM64 ASM L1 .
What does this code do?
$SG3103 DB '%d', 0aH, 00H
_main PROC
push 0
call DWORD PTR __imp___time64
push edx
push eax
push OFFSET $SG3103 ; '%d'
call DWORD PTR __imp__printf
add esp, 16
xor eax, eax
ret 0
_main ENDP
Additional question: why MSVC replaced time() with time64()? Is it correct? Dangerous? What printf() will print after year 2038?
main PROC
PUSH {r4,lr}
MOV r0,#0
BL time
MOV r1,r0
ADR r0,|L0.32|
BL __2printf
MOV r0,#0
POP {r4,pc}
ENDP
|L0.32|
DCB "%d\n",0
main PROC
PUSH {r4,lr}
MOVS r0,#0
BL time
MOVS r1,r0
ADR r0,|L0.20|
BL __2printf
MOVS r0,#0
POP {r4,pc}
ENDP
|L0.20|
DCB "%d\n",0
main:
stp x29, x30, [sp, -16]!
mov x0, 0
add x29, sp, 0
bl time
mov x1, x0
ldp x29, x30, [sp], 16
adrp x0, .LC0
add x0, x0, :lo12:.LC0
b printf
.LC0:
.string "%d\n"
main:
var_10 = -0x10
var_4 = -4
lui $gp, (__gnu_local_gp >> 16)
addiu $sp, -0x20
la $gp, (__gnu_local_gp & 0xFFFF)
sw $ra, 0x20+var_4($sp)
sw $gp, 0x20+var_10($sp)
lw $t9, (time & 0xFFFF)($gp)
or $at, $zero
jalr $t9
move $a0, $zero
lw $gp, 0x20+var_10($sp)
lui $a0, ($LC0 >> 16) # "%d\n"
lw $t9, (printf & 0xFFFF)($gp)
lw $ra, 0x20+var_4($sp)
la $a0, ($LC0 & 0xFFFF) # "%d\n"
move $a1, $v0
jr $t9
addiu $sp, 0x20
$LC0: .ascii "%d\n"<0> # DATA XREF: main+28
More challenges: challenges.re; about solutions: challenges.re/#Solutions.