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    <p>The stack is used for the local variables but also for intermediate values that the function need. Here, your <code>foo()</code> function calls <code>bar()</code> by giving it two pointers, one to the string (<code>argv[1]</code>) and the other to the <code>buf</code> variable; those pointer values are pushed on the stack, namely in the <code>mov %eax,0x4(%esp)</code> and <code>mov %edx,(%esp)</code> opcodes. Thus, <code>foo()</code> needs a bit more than 256 bytes of stack space.</p> <p>In more details:</p> <pre><code>0x08048473 &lt;foo+0&gt;:push %ebp 0x08048474 &lt;foo+1&gt;:mov %esp,%ebp </code></pre> <p>That's the standard function prologue: the function will use <code>%ebp</code> to points to the stack elements <em>before</em> it was called (i.e. its arguments).</p> <pre><code>0x08048476 &lt;foo+3&gt;:sub $0x118,%esp </code></pre> <p>Some space is reserved on the stack, mainly (but not only) for <code>buf[]</code>.</p> <pre><code>0x0804847c &lt;foo+9&gt;:mov 0x8(%ebp),%eax 0x0804847f &lt;foo+12&gt;:add $0x4,%eax 0x08048482 &lt;foo+15&gt;:mov (%eax),%edx </code></pre> <p><code>0x8(%ebp)</code> is the <code>argv</code> function argument; these opcodes fetch the pointer from <code>argv[1]</code> and store the result in <code>%edx</code>. This will become the first argument to <code>bar()</code>.</p> <pre><code>0x08048484 &lt;foo+17&gt;:lea -0x100(%ebp),%eax </code></pre> <p>This stores in <code>%eax</code> the address of <code>buf[]</code> -- the compiler decided that <code>buf[]</code> was to be located in the upper 256 bytes of the stack space it reserved with <code>sub</code>.</p> <pre><code>0x0804848a &lt;foo+23&gt;:mov %eax,0x4(%esp) 0x0804848e &lt;foo+27&gt;:mov %edx,(%esp) </code></pre> <p>The two arguments for <code>bar()</code> are pushed on the stack (actually, written in the two top stack positions, <code>%esp</code> having been already adjusted).</p> <pre><code>0x08048491 &lt;foo+30&gt;:call 0x8048454 &lt;bar&gt; </code></pre> <p><code>bar()</code> is called.</p> <pre><code>0x08048496 &lt;foo+35&gt;:leave 0x08048497 &lt;foo+36&gt;:ret </code></pre> <p><code>leave</code> undoes the prologue (it is equivalent to <code>mov %ebp, %esp; pop %ebp</code>). <code>ret</code> exits the function.</p> <p>GCC is known to overallocate a bit on the stack; here, it could have reserved 264 bytes instead of 280. This seems to be an artefact of its internal register allocation optimizer (the extra stack slots <em>were</em> used to store intermediate values, but the optimizer finally found ways to keep the corresponding values in registers only).</p>
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    1. POCan someone explain the assembly code please?
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    1. POCan someone explain the assembly code please?
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    1. COThank you. That was very clear. So the first two lines is where the function return value is stored then? Isn't the return value stored after the function parameters and before the called function's local variables?
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    2. COThe return value is normally exchanged in a conventional register (%eax on x86 processors). Here you have no `return` so the assembly code does nothing about `%eax`: the caller will find that `foo()` "returns" whatever value `%eax` contains at that point.
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    3. COSorry meant to ask about the point where bar() returns to - the point where foo() continues execution once bar returns. Where is that stored?
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