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    <blockquote> <p>when that function calls, end of function that local variable x will be what?</p> </blockquote> <p>Nonexistent. Evaportated. Vapor. Gone.</p> <p><code>x</code> here is a variable with automatic lifetime. "Automatic" refers to the fact that when <code>x</code> goes out of scope, it will be destroyed. (Not exactly <em>deleted</em>, as that term implies calling <code>delete</code>)</p> <p>The same is true of all automatic variables, wherether they be integrals, strings, floats, or even pointers:</p> <pre><code>void f2() { int a=100; int* b=&amp;a; } </code></pre> <p>Here both <code>a</code> and <code>b</code> are automatic variables. They will both be destroyed at the end of their scope -- the end of <code>f2()</code>. However, all that will be destroyed is the varible itself. If an automatic variable is a pointer (eg, a <em>raw</em> pointer), then the <em>thing the pointer points to</em> will <strong>not</strong> be destroyed. Consider:</p> <pre><code>void f3() { char* p = new char [256]; } </code></pre> <p><code>p</code> here is still an automatic variable, of type pointer-to-char. The thing <code>p</code> points to is not automatic -- that is dynamically allocated (we used <code>new</code>).</p> <p>As written above, <code>p</code> will be destroyed but the memory it points to will not. This is a memory leak. In order to resolve this, you must <code>delete</code> it:</p> <pre><code>void f3() { char* p = new char [256]; delete [] p; } </code></pre> <p>Now the memory pointed to by <code>p</code> has been correctly destroyed, and the pointer itself (<code>p</code>) will be destroyed at the end of its scope since it's an automatic.</p> <p>The lifetime of <code>p</code> and what <code>p</code> points to are not connected in any way. Just because you <code>delete</code> <code>p</code> doesn't mean that <code>p</code> <em>itself</em> is now also destroyed. <code>p</code> is still an automatic variable.</p> <p>An example:</p> <pre><code>void f3() { char* p = new char [256]; cout &lt;&lt; (void*) p &lt;&lt; "\n" delete [] p; // p still exists, it just points to to nothing useable cout &lt;&lt; (void*) p &lt;&lt; "\n" } </code></pre> <p>After the <code>delete</code>, we print the value of <code>p</code> to the screen. If you run this, you'll see that the value of <code>p</code> doesn't change from before deleteing it. <code>p</code> itself was unaffected by <code>delete [] p;</code>. But at <code>f3()</code>'s closing brace <code>}</code>, it will fall out of scope and there be destroyed.</p>
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    1. COI dont understand, using delete will delete pointed value, but still there will be pointed address. so after delete[] that pointed address will be deleted becouse of end of scope?
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      1. USVito Carleone
    2. CO@VitoCarleone: Almost. Let me elaborate a bit further.
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    3. CO@VitoCarleone: Edited. Does that help?
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