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what does end() do in <set>? [SOLVED]
In the following code end() prints the no. Of elements in the set and not the last element but according to geeksforgeeks end() - Returns an iterator to the theoretical element that follows last element in the set. What does that mean? I thought it would print the last element in the set https://code.sololearn.com/cmtW1Yii3UdT/?ref=app
7 Respostas
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end() returns an iterator that points to the first element *after* the container (e.g a set)
http://www.cplusplus.com/reference/iterator/end/
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daniel just like in array, in this case 33 corresponds to the 4th index value
so *a.end() prints the next index value
that is 5
am i correct?
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Not exactly. Sets dont have indexes by any manner. It returns an iterator (kind of a pointer) to the element that is after the last element. By coincidence that may point to the place that the set.size() is stored.
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Saurabh Tiwari thats exactly what i ment. You accessed the place the set.size() is stored (And not a garbage value). It makes sense that the size is stored right after the set itself.
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*a.end() is undefined behaviour.
It's the same as if you had an array with 5 elements and tried to do array[5].
In this case the memory layout for std::set<int> is something like:
10 6D 79 00 00 00 00 00 00 00 00 00 00 00 00 00
A0 6D 79 00 00 00 00 00 00 6E 79 00 00 00 00 00
30 6E 79 00 00 00 00 00 05 00 00 00 00 00 00 00
The last part 05 00 00 00 00 00 00 00 is the variable that holds the size of the set.
*a.end() just happens to access that variable.
I could edit this to an 8 for example and *a.end() and a.size() will print an 8.
unsigned char* ptr = reinterpret_cast<unsigned char*>( &a ); // where a is the set
ptr[40] = 8;
But the layout could be different depending on compiler.
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Dennis whoa!!! that was some next level stuff
every single word went above my head
*feels demotivated
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daniel no. i dont think it is a coincidence
try changing the no. of elements
it always prints the no. of elements