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    <p>In object oriented programming languages, an array of objects is really just an array of pointers to the objects. So if a pointer takes 4 bytes, and an object takes 5, an array of 10 objects will actually be 4*10 bytes long (plus overhead).</p> <pre><code>MyClass[] my_variable = new MyClass[10]; // this will allocate 10*pointersize bytes, plus overhead, you still need to allocate more space for the objects themselves </code></pre> <p>In C, an array of structures is an array of structures. If a structure takes 8 bytes, an array of 10 of them takes 80 bytes. You don't need to allocate even more space. It's already there.</p> <pre><code>struct data_t my_variable[10]; // this will allocate 10*sizeof(data_t) bytes </code></pre> <p>So the easy answer is:</p> <pre><code>int function2(struct Person *list, int *index) { char* nameInput; // User input code goes here ... // IMPORTANT: the user input code must allocate the space for nameInput // if it doesn't, then a copy of the buffer must be made // instead of the direct assignment below // the memory for this instance of the person was already created when the // array was created, so just save the values list[*index].name = nameInput; list[*index].age = ageInput; *index += 1; return 0; } </code></pre> <p>When your code executes this line:</p> <pre><code>list[(*index)++] = *prsn; </code></pre> <p>It is quite literally copying 8 bytes (on a 32 bit machine, sizeof(int) + sizeof(char*)), as I think you noticed something was up with your commment "// !Why use the dereferencing *prsn here?"</p> <p>The address to the memory allocated with malloc isn't being copied in that line, but the contents of the memory are. That's why it is a memory leak, the address is lost when the function exits because it was never put into the array.</p>
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Detail views are divided into sections. All the information in the data section comes from columns in the selected row. The other sections display data from other, related rows.

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