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I'm reversing my C array's programs where I'm storing numbers in arrays like this...

int marks[5];
marks[0] = 1;
marks[1] = 3;
marks[2] = 2;
marks[3] = 7;
marks[4] = 9;

so the compiler showing me the disassembly...

enter image description here

Actually, my question is why it is storing like this? by using imul first before storing 1 then before storing 7, and using shl for storing 3, 2, and 9. Can anyone please help? Fell free to edit it*

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4
  • i think you have already been asked to post text not images
    – blabb
    Feb 15 at 7:35
  • Actually, in visual studio 2017 I can't copy the assembly code as there's no option there so that's why I've use an image to present my point.
    – Ramesses II
    Feb 15 at 10:03
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    @RamessesII unless you have some strange VS, you can for sure copy assembly from there.
    – Paweł Łukasik
    Feb 15 at 14:43
  • 1
    Lack of copy and paste (if true, which is unlikely) is not an excuse for posting an image of text. You can re-type it in a code block. It's not that much.
    – nobody
    Feb 15 at 15:51
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those are pointer arithmetics marks is an <<<;ADDRESS;>>>>> assume 0x10000000
it points to an integer whose size is 4 in 32 bit machine so the next integer will be at 0x10000004 , and the next will be at 0x10000008 and so on

&marks[0] = 0x10000000
&marks[1] = 0x10000004
&marks[2] = 0x10000008,c,10,14,18,.....nn

each of the 4 in the disassembly is sizeof(int) == 4
they are multiplied by 0 , 2^0 , 2^1,3 and 2^2 to result in 0,1,2,3,4 as indexes

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