I am trying to translate assembly code back into C code but I noticed this one operation called sarq. I think q is for what size the address is but I do not know what the sarq does to the address. I commented on what I believe the code does.
.LC0 .string "ans %d\n"
main:
.LFB0: val = -8(%rbp), result = -12(%rbp)
pushq %rbp
movq %rsp, %rbp
subq $16, %rsp
movabsq $53162464113523643, %rax
movq %rax, -8(%rbp) //val(variable) address -8,inputs value in %rax
movl $0, -12(%rbp) //result(variable) address -12, inputs 0
jmp .L2 //starts loop
.L3:
movq -8(%rbp), %rax //moves value in val into rax
andl $1, %eax //dunno what eax is but adds 1 into it
xorl %eax, -12(%rbp) //compares the value of eax and result to see if they are not equal. so compares 1 to 0
sarq -8(%rbp) //does something to val?
.L2:
cmpq $0, -8(%rbp) //compares val to 0
jg .L3 //if greater, goes to L3
movl -12(%rbp), %eax //else, moves value from result into eax
movl %eax, %esi //moves eax into esi
movl $.LC0, %edi //Moves w/e $.LC0 is into edi. Based on the top, edi now holds that string?
movl $0, %eax //moves 0 into eax
call printf //print statement
leave
ret
sar is an arithmetic right shift. The single operand form shifts its operand right by one place, filling the topmost bit with the sign of the number. the suffix q indicates that the operand is a 64 bit operand (a quadword). Thus sarq -8(%rbp) shifts the quadword eight bytes below %rbp right by one place.