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    <p>Turns out, the issue was caused by incorrect use of memory, as expected. I'm 99% sure the following example is accurate; it's pretty much pseudo code, so wont compile, obviously.</p> <h3>Update:</h3> <p>Just added a 3rd solution thanks to nusi.</p> <h3>The wrong way (with stack memory):</h3> <pre><code>std::map&lt;int, MyType1&gt; myMap; void firstFunctionRunFromThread1() { MyType1 mt1; mt1.Test = "Test 1"; myMap[0] = mt1; } void onlyFunctionRunFromThread2() { MyType1 mt1 = myMap[0]; // This actually does print "Test 1", so the memory is being accessed. std::cout &lt;&lt; mt1.Test &lt;&lt; endl; /* However, because we're using stack memory here, this value is lost * when we go back to thread #1. */ mt1.Test = "Test 2"; } void secondFunctionFromThread1() { MyType1 mt1 = myMap[0]; // This will actually print out "Test 1", where "Test 2" is expected! std::cout &lt;&lt; mt1.Test &lt;&lt; endl; } </code></pre> <h3>The <em>complicated</em>, correct method (using heap memory):</h3> <p>See also the simple method which uses stack memory.</p> <pre><code>std::map&lt;int, MyType1&gt; myMap; void firstFunctionRunFromThread1() { // Use heap memory so the memory stays allocated. MyType1 *mt1 = new MyType1(); mt1-&gt;Test = "Test 1"; myMap[0] = *mt1; } void onlyFunctionRunFromThread2() { /* Now, get a pointer to the object; we can't use stack memory * here because the values assigned would be lost as soon as * we try and access them from secondFunctionFromThread1() */ MyType1 *mt1 = &amp;myMap[0]; // As expected, this prints "Test 1" std::cout &lt;&lt; mt1-&gt;Test &lt;&lt; endl; /* Now, because we're assigning to memory on the heap, this will * stay assigned until the entire application exits, yay! */ mt1-&gt;Test = "Test 2"; } void secondFunctionFromThread1() { /* Not sure if using heap memory here is neccecary, but we'll do * it anwyay just to play on the safe side... let me know if this * is pointless... */ MyType1 *mt1 = &amp;myMap[0]; // Hurray, this prints "Test 2"!!! :) std::cout &lt;&lt; mt1-&gt;Test &lt;&lt; endl; } </code></pre> <h3>The <em>simple</em>, correct method (using stack memory correctly):</h3> <p>Thanks to nusi for <a href="https://stackoverflow.com/questions/724536/does-memory-stay-allocated-when-a-c-thread-exits/731025#731025">his answer</a>.</p> <pre><code>std::map&lt;int, MyType1&gt; myMap; void firstFunctionRunFromThread1() { MyType1 mt1; mt1.Test = "Test 1"; myMap[0] = mt1; } void onlyFunctionRunFromThread2() { /* Using the &amp; before the variable turns it into a reference, so * instead of using stack memory, we use the original memory. * NOTE: Is this explanation correct? */ MyType1 &amp;mt1 = myMap[0]; // This actually does print "Test 1", so the memory is being accessed. std::cout &lt;&lt; mt1.Test &lt;&lt; endl; // We're assigning to the reference, so this works. mt1.Test = "Test 2"; } void secondFunctionFromThread1() { MyType1 mt1 = myMap[0]; // Prints "Test 1" as expected. std::cout &lt;&lt; mt1.Test &lt;&lt; endl; } </code></pre>
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    1. COwriting "MyType1 &mt1 = myMap[0]" in onlyFunctionRunFromThread2() of your first implementation would have solved your problem.
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    2. COThanks for the tip, and thanks for the answer.
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