Need help on below code and queries.
My understanding of the code below suggests it will get stuck in an infinite loop.
What is the purpose of DemoFunction2?
//
0:000> uf 010024d0
asmdemo2!DemoFunction2:
010024d0 55 push ebp
010024d1 8bec mov ebp,esp
010024d3 8b5508 mov edx,dword ptr [ebp+8]
010024d6 33c0 xor eax,eax
010024d8 b920000000 mov ecx,20h
010024dd d1ea shr edx,1
010024df 7301 jnc asmdemo2!DemoFunction2+0x12 (010024e2)
010024e1 40 inc eax
010024e2 e2f9 loop asmdemo2!DemoFunction2+0xd (010024dd)
010024e4 5d pop ebp
010024e5 c3 ret
0:000> r
eax=80002418 ebx=7ffd7000 ecx=00682295 edx=00000000 esi=80002418
edi=00000002
eip=010024d0 esp=0006fe98 ebp=0006fea8 iopl=0 nv up ei pl zr na
pe nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000
efl=00000246
asmdemo2!DemoFunction2:
010024d0 55 push ebp
0:000> dps esp
0006fe98 0100251c asmdemo2!main+0x20
0006fe9c 80002418
0006fea0 00000002
0006fea4 00000000
0006fea8 0006ff88
0006feac 01002969 asmdemo2!_mainCRTStartup+0x12c
0006feb0 00000002
0006feb4 00682270
0006feb8 006822b8
0006febc f395c17d
0006fec0 00000000
0006fec4 00000000
0006fec8 7ffd7000
0006fecc 00000000
0006fed0 00000000
0006fed4 00000000
0006fed8 00000094
0006fedc 00000006
0006fee0 00000000
0006fee4 00001771
0006fee8 00000002
0006feec 76726553
0006fef0 20656369
0006fef4 6b636150
0006fef8 00003120
0006fefc 00000000
0006ff00 00000000
0006ff04 00000000
0006ff08 00000000
0006ff0c 00000000
0006ff10 00000000
0006ff14 00000000
No it is not an infinite loop.
At the beginning ecx is set to 32.
Every time the loop instruction executes ecx is decremented by 1.
When ecx reaches 0, the loop instruction will fall through (not jump) and the instructions thereafter will be executed (pop+ret).
Obviously the code is inefficient.
A simple
popcnt eax,[esp+4]
ret
Would do the same function much faster, but it works.
It counts the number of set bits in the first argument, using the cdecl calling convention.