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2010-04-04T00:38:59.793
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The simple answer is: Yes. More specifically, we can differentiate between two causes for this. I'll call them implementation-dependent and implementation-independent. The one case that doesn't depend on your environment at all is that of static objects in C++, which was mentioned here. The following code dies before <code>main()</code>: <pre><code>#include <iostream> class Useless { public: Useless() { throw "You can't construct me!"; } }; static Useless object; int main() { std::cout << "This will never be printed" << std::endl; return 0; } </code></pre> More interesting are the platform-dependent causes. Some were mentioned here. One that was mentioned here a couple of times was the usage of dynamically linked libraries (DLLs in windows, SOs in Linux, etc.) - if your OS's loader loads them before <code>main()</code>, they might cause your application do die before <code>main()</code>. A more general version of this cause is talking about all the things your binary's entry point does before calling your entry point(<code>main()</code>). Usually when you build your binary there's a pretty serious block of code that's called when your operating system's loader starts to run your binary, and when it's done it calls your <code>main()</code>. One common thing this code does is initializing the C/C++ standard library. This code can fail for any number of reasons (shortage of any kind of system resource it tries to allocate for one). One interesting way on for a binary to execute code before <code>main()</code> on windows is using TLS callbacks (google will tell you more about them). This technique is usually found in malware as a basic anti-debugging trick (this trick used to fool ollydbg back then, don't know if it still does). The point is that your question is actually equivalent to "is there a way that loading a binary would cause user code to execute before the code in <code>main()</code>?", and the answer is hell, yeah!
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1. POIs there any way a C/C++ program can crash before main()?
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