mov

Instruction 1: LEA DX, MESSAGE ; Move the address of MESSAGE in register DX Instruction 2: MOV DX, OFFSET MESSAGE ; Move the address of MESSAGE in register DX Questions: Are the above instructions equal? They seem to work similarly, but I have just started programming assembly so I can not say. If both of them are similar, then which one of above is the better way to do the above task? Note: I have already read this question On my 32-bit system, the instructions match opcodes like this: 8d 15 c8 90 04 08 lea 0x80490c8,%edx ba c8 90 04 08 mov $0x80490c8,%edx So you use a whole extra byte when

I want to play in .mov file in android. But videoview or mediaplayer doesn't support this meida format. How can i add the support to it? In general Android doesn't support any other media formats than the one listed here . That being said, there are quite a few 3rd party players that enable playback of more exotic formats, most of which are probably based around ffmpeg. You might want to take a look at the open source Dolpin Player (actual player also available in the Play store ) for Android for some more pointers - not sure if mov playback is supported by default though. However, since most

Concerning the following small code, which was illustrated in another post about the size of structure and all the possibilities to align data correctly : struct { char Data1; short Data2; int Data3; char Data4; } x; unsigned fun ( void ) { x.Data1=1; x.Data2=2; x.Data3=3; x.Data4=4; return(sizeof(x)); } I get the corresponding disassembly (with 64 bits) 0000000000000000 <fun>: 0: 55 push %rbp 1: 48 89 e5 mov %rsp,%rbp 4: c6 05 00 00 00 00 01 movb $0x1,0x0(%rip) # b <fun+0xb> b: 66 c7 05 00 00 00 00 movw $0x2,0x0(%rip) # 14 <fun+0x14> 12: 02 00 14: c7 05 00 00 00 00 03 movl $0x3,0x0(%rip) # 1e

From what I've read about mov , it copies the second argument into the first argument. Then, what does this do? movl 8(%ebp), %edx It copies whatever is in edx to the first parameter of the function (since an offset of +8 from ebp is a parameter)? I feel like what this really means is moving the first parameter into the edx register, but I read on Wikipedia that it is the other way around? movl 8(%ebp), %edx is in "AT&T Syntax"; in this syntax, the source comes first and the destination second. So yes, your belief is correct. Most documentation uses the "Intel Syntax", which has the reverse

69A8AB13 int 3 69A8AB14 int 3 69A8AB15 mov edi,edi 69A8AB17 push ebp 69A8AB18 mov ebp,esp mov edi,edi doesn't make sense for me,what's it for? It's a 2 byte NOP instruction. It gets included at the beginning of any function in an image compiled with the /hotpatch option: http://msdn.microsoft.com/en-us/library/ms173507.aspx -scott See also http://blogs.msdn.com/b/oldnewthing/archive/2011/09/21/10214405.aspx According to this page: StackExchange's Reverse Engineering In x86-64 mov edi,edi is not a NOP. In x86-64 it zeroes the top 32 bits of rdi. I though it was important enough to point it out,

I'm looking Intel datasheet: Intel® 64 and IA-32 Architectures Software Developer’s Manual and I can't find the difference between MOVDQA : Move Aligned Double Quadword MOVAPS : Move Aligned Packed Single-Precision In Intel datasheet I can find for both instructions: This instruction can be used to load an XMM register from a 128-bit memory location, to store the contents of an XMM register into a 128-bit memory location, or to move data between two XMM registers. The only difference is: To move a double quadword to or from unaligned memory locations, use the MOVDQU instruction. and To move

I am using ffmpeg to convert videos to mp4 format. Everything works fine except with MOV format. The command I use for everything is: ffmpeg -i input.mov -strict experimental -sameq -s vga -aspect 1.7777 -vcodec libx264 -preset fast -crf 22 -y output.mp4 but the output I keep getting is: ffmpeg version 0.9, Copyright (c) 2000-2011 the FFmpeg developers built on Mar 12 2012 11:01:05 with gcc 4.4.5 configuration: --enable-libx264 --enable-gpl --disable-yasm libavutil 51. 32. 0 / 51. 32. 0 libavcodec 53. 42. 0 / 53. 42. 0 libavformat 53. 24. 0 / 53. 24. 0 libavdevice 53. 4. 0 / 53. 4. 0

What does this instruction do? mov (%r11,%r12,1), %edx Look here . It says In the AT&T Syntax, memory is referenced in the following way, segment-override:signed-offset(base,index,scale) Down on the page there are some examples. I find this the best: GAS memory operand NASM memory operand ------------------ ------------------- (%ecx,%ebx,2) [ecx+ebx*2] mov source, destination in AT&T syntax copies the value from source to destination. Also consider the size of edx. How many bytes (4) do you think mov will copy ? mov (%r11,%r12,1), %edx this instruction is use to calculate the address (indexed