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    <p>No sir. That is not valid. You can't return a pointer to a local variable. Once <code>lab()</code> exits <code>k</code> does not exist and dereferencing a pointer to it causes undefined behavior.</p> <p>Think about where <code>k</code> is stored. Automatic variables that you take the address of are stored on the stack. The stack grows when functions are entered and shrinks when they exit. When <code>lab()</code> returns the stack space that was allocated to <code>k</code> is reclaimed and can be reused by the runtime, possibly for other local variables in some other functions.</p> <p>There are a couple of ways to fix this. The easiest is to have the caller provide a location to store the value in rather than having <code>lab()</code> try to find space. This eliminates the problem of <code>k</code> being deallocated when <code>lab()</code> returns.</p> <pre><code>int* lab(int* i) { *i = 9; return i; } int main(void) { int k; cout &lt;&lt; *lab(&amp;k) &lt;&lt; endl; return 0; } </code></pre> <p>Another way is to declare <code>k</code> as <code>static</code>. Static variables are stored in permanent storage somewhere, not on the stack, so their addresses remain valid throughout the lifetime of the program.</p> <pre><code>int* lab() { static int k=9; return &amp;k; } </code></pre> <p>And yet another way is to allocate memory on the heap using <code>new</code>.</p> <pre><code>int* lab() { int* i = new int; *i = 9; return i; } int main(void) { int* i = lab(); cout &lt;&lt; *i &lt;&lt; endl; delete i; return 0; } </code></pre>
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    1. USJohn Kugelman
    1. USJohn Kugelman
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    1. COoh... i see. i'm newbie in c++. pointers make me nuts, sorry :)
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      1. USskater_nex
    2. COThanks. Now I get it. The same problem with reference, right? I mean there is no difference between returning a pointer or reference in this case? Sorry for the noob question.
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    3. CO@skater_nex: In this case, there is a slight difference. If you pass an int in by reference, you can't reseat it to reference k. You can assign to it the value of k, but it will still be referencing the int that was passed in, and is thus perfectly safe to return. But if you declared a reference in the function, initialized with k, you could not, safely, return it by reference.
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      1. USBenjamin Lindley
 

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