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POAssembly operator AND
 

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  1. POAssembly operator AND
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    <p>In order to continue this: <a href="https://stackoverflow.com/questions/19210032/debugging-c-program-int-declaration">Debugging C program (int declaration)</a> I decided to test more code and see how compiler reacts to it. So I decided to try this one to test local variables:</p> <pre><code>#include &lt;stdio.h&gt; main() { int a,b,c,d,e,f,g; a=0xbeef; b=0xdead; c=0x12; d=0x65; e=0xfed; f=0xaa; g=0xfaceb00c; a=a+b; printf("%d",a); } </code></pre> <p>Ok I did that int a,b,c... just to test the main's frame size and see the sub $0x10,%esp growing up, (I'm under linux so that is why maybe is sub), now to sub $0x30,%esp so here is the the gdb output with "disas main" command:</p> <pre><code> 0x0804841c &lt;+0&gt;: push %ebp 0x0804841d &lt;+1&gt;: mov %esp,%ebp 0x0804841f &lt;+3&gt;: and $0xfffffff0,%esp 0x08048422 &lt;+6&gt;: sub $0x30,%esp ;7 int vars 4-byte is 7*4=28. 30 is enough 0x08048425 &lt;+9&gt;: movl $0xbeef,0x14(%esp) 0x0804842d &lt;+17&gt;: movl $0xdead,0x18(%esp) 0x08048435 &lt;+25&gt;: movl $0x12,0x1c(%esp) 0x0804843d &lt;+33&gt;: movl $0x65,0x20(%esp) 0x08048445 &lt;+41&gt;: movl $0xfed,0x24(%esp) 0x0804844d &lt;+49&gt;: movl $0xaa,0x28(%esp) 0x08048455 &lt;+57&gt;: movl $0xfaceb00c,0x2c(%esp) 0x0804845d &lt;+65&gt;: mov 0x18(%esp),%eax 0x08048461 &lt;+69&gt;: add %eax,0x14(%esp) 0x08048465 &lt;+73&gt;: mov 0x14(%esp),%eax 0x08048469 &lt;+77&gt;: mov %eax,0x4(%esp) 0x0804846d &lt;+81&gt;: movl $0x8048510,(%esp) 0x08048474 &lt;+88&gt;: call 0x80482f0 &lt;printf@plt&gt; 0x08048479 &lt;+93&gt;: leave 0x0804847a &lt;+94&gt;: ret </code></pre> <p>This line: <code>0x0804841f &lt;+3&gt;:and $0xfffffff0,%esp</code> what is and operator and why is there a large number?</p> <p>And why the offset in movl commands isn't negative like: <code>movl $0xa,-0x4(%ebp)</code> So far I know is the AND is a logical operator like 1 and 1 is 1, 0 and 0 is 0, 1 and 0 is 0 etc... If it is the case, %esp has the ebp value that was the base frame address of who called the main function.</p> <p>can any of you explain why this is compiled like this?</p> <p>I think I'm missing something. Edit: I saw some "topics" on stackoverflow talking about this. Going to share: <a href="https://stackoverflow.com/questions/4175281/what-does-it-mean-to-align-the-stack">link1</a> <a href="https://stackoverflow.com/questions/4228261/understanding-the-purpose-of-some-assembly-statements">link2</a> <a href="https://stackoverflow.com/questions/4817473/understanding-stack-allocation-and-alignment">link3</a></p>
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    1. CO`and $0xfffffff0,%esp` is used to clear the lower 4 bits of `%esp`; i.e. to align it on a 16-byte boundary.
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    2. COhum gdb: `1: $esp = (void *) 0xbffff6f0 ; this is on +9 offset to main.` so is this moving to `movl $0xbeef,0x14(%esp)`
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    3. COI found this: [link](http://stackoverflow.com/questions/4228261/understanding-the-purpose-of-some-assembly-statements)
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