StackOverflow2013

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primarykey
Id
2468168
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0
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0
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0
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CreationDate
2010-03-18T07:37:47.660
FavoriteCount
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2010-03-18T07:37:47.660
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0
OwnerUserId
28169
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2468152
PostTypeId
2
Score
6
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LastEditorDisplayName
text
Body
The <code>this</code> pointer is not a member of the class. It's just a construct that is used in methods belonging to the class to refer to the current instance. If you have a class like this: <pre><code>class IntPair { public: IntPair(int a, int b) : _a(a), _b(b) { } int sum() const { return _a + _b; } public: int _a; int _b; }; </code></pre> This class only needs space for two instances of <code>int</code> for each instance. Once you've created an instance and are running the <code>sum()</code> method, that method is called with a pointer to the instance, but that pointer always comes from somewhere else, it isn't stored in the object instance. For example: <pre><code>IntPair *fib12 = new IntPair(89, 144); cout << fib12->sum(); </code></pre> Notice how the variable that becomes the <code>this</code> pointer is stored outside the object, in the scope that created it. You could, in fact, always transform a method like the one above into: <pre><code>static int sum2(const IntPair* instance) { return instance->_a + instance->_b; } </code></pre> If the above is defined inside the class (so it can access the private members), there's no difference. In fact, this is how methods are implemented behind the scene; the <code>this</code> pointer is just a hidden argument to all member methods. The call would become: <pre><code>IntPair* fib12 = new IntPair(89, 144); cout << IntPair::sum2(fib12); </code></pre>
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