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    <p><strong>gcc, in C mode:</strong></p> <p>Uninitialised globals which are not declared <code>extern</code> are treated as "common" symbols, not weak symbols.</p> <p>Common symbols are merged at link time so that they all refer to the same storage; if more than one object attempts to initialise such a symbol, you will get a link-time error. (If they aren't explicitly initialised anywhere, they will be placed in the BSS, i.e. initialised to 0.)</p> <p><strong>gcc, in C++ mode:</strong></p> <p>Not the same - it doesn't do the common symbols thing. "Uninitialised" globals which are not declared <code>extern</code> are implicitly initialised to a default value (0 for simple types, or default constructor).</p> <hr> <p>In either case, a weak symbol allows an initialised symbol to be overridden by a non-weak initialised symbol of the same name at link time.</p> <hr> <p>To illustrate (concentrating on the C case here), I'll use 4 variants of a main program, which are all the same except for the way that <code>global</code> is declared:</p> <ol> <li><p><strong>main_init.c</strong>:</p> <pre><code>#include &lt;stdio.h&gt; int global = 999; int main(void) { printf("%d\n", global); return 0; } </code></pre></li> <li><p><strong>main_uninit.c</strong>, which omits the initialisation:</p> <pre><code>#include &lt;stdio.h&gt; int global; int main(void) { printf("%d\n", global); return 0; } </code></pre></li> <li><p><strong>main_uninit_extern.c</strong>, which adds the <code>extern</code> keyword:</p> <pre><code>#include &lt;stdio.h&gt; extern int global; int main(void) { printf("%d\n", global); return 0; } </code></pre></li> <li><p><strong>main_weak_init.c</strong>, which initialises <code>global</code> and declares it to be a weak symbol:</p> <pre><code>#include &lt;stdio.h&gt; int global __attribute__((weak)) = 999; int main(void) { printf("%d\n", global); return 0; } </code></pre></li> </ol> <p>and <strong>another_def.c</strong> which initialises the same global:</p> <pre><code>int global = 1234; </code></pre> <hr> <p>Using <code>main_uninit.c</code> on its own gives 0:</p> <pre><code>$ gcc -o test main_uninit.c &amp;&amp; ./test 0 </code></pre> <p>but when <code>another_def.c</code> is included as well, <code>global</code> is explicitly initialised and we get the expected result:</p> <pre><code>$ gcc -o test main_uninit.c another_def.c &amp;&amp; ./test 1234 </code></pre> <p>(Note that this case fails instead if you're using C++.)</p> <p>If we try with both <code>main_init.c</code> and <code>another.def.c</code> instead, we have 2 initialisations of <code>global</code>, which won't work:</p> <pre><code>$ gcc -o test main_init.c another_def.c &amp;&amp; ./test /tmp/cc5DQeaz.o:(.data+0x0): multiple definition of `global' /tmp/ccgyz6rL.o:(.data+0x0): first defined here collect2: ld returned 1 exit status </code></pre> <hr> <p><code>main_uninit_extern.c</code> on its own won't work at all - the <code>extern</code> keyword causes the symbol to be an ordinary external reference rather than a common symbol, so the linker complains:</p> <pre><code>$ gcc -o test main_uninit_extern.c &amp;&amp; ./test /tmp/ccqdYUIr.o: In function `main': main_uninit_extern.c:(.text+0x12): undefined reference to `global' collect2: ld returned 1 exit status </code></pre> <p>It works fine once the initialisation from <code>another_def.c</code> is included:</p> <pre><code>$ gcc -o test main_uninit_extern.c another_def.c &amp;&amp; ./test 1234 </code></pre> <hr> <p>Using <code>main_init_weak.c</code> on its own gives the value we initialised the weak symbol to (999), as there is nothing to override it:</p> <pre><code>$ gcc -o test main_init_weak.c &amp;&amp; ./test 999 </code></pre> <p>But pulling in the other definition from <code>another_def.c</code> does work in this case, because the strong definition there overrides the weak definition in <code>main_init_weak.c</code>:</p> <pre><code>$ gcc -o test main_init_weak.c another_def.c &amp;&amp; ./test 1234 </code></pre>
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    1. POWhy uninitialized global variable is weak symbol?
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    1. USCiro Santilli 新疆改造中心 六四事件 法轮功
    1. USMatthew Slattery
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    1. CONice answer, thx. May I ask if it applies to C++?
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      1. USMaciej Hehl
    2. COGreat explanation with examples. Could you also include how weak interacts with functions?
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      1. USTomek
    3. CO@Maciej - good point, forgot this was tagged with both when writing my answer originally! No, it's not the same for C++, so I've updated my answer accordingly.
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      1. USMatthew Slattery
 

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