As an example I know that
void foo(int *num)
{
num++;
}
foo(&num);
will pass num by reference. However, why do people use
void foo(int &num)
{
num++;
}
I've heard that you should use that syntax over what I posted above, why is this?
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BobbyT432
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2https://stackoverflow.com/a/57492/12243443 – Manziel Mar 11 '20 at 16:43
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`int *num` does not pass `num` by reference. – Mooing Duck Mar 11 '20 at 16:44
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3Sounds like you could use a [good C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) – NathanOliver Mar 11 '20 at 16:44
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1Did you meant `void foo(int *num) { (*num)++; }`? – Jarod42 Mar 11 '20 at 16:45
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`int *num`, `foo(&num);` This passes a pointer to `num` by value. – eerorika Mar 11 '20 at 16:46
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"*Pass-by-reference*" *might* refer to "*pass-by-pointer*" in the context of C, but in C++ these are certainly different things, see linked duplicate. – walnut Mar 11 '20 at 16:46
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It's the same reason people use both hammers and screwdrivers. They do different things, and each one has its own use cases where the other one isn't convenient. – Sam Varshavchik Mar 11 '20 at 16:50
1 Answers
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Your first example passes by pointer, not reference.
The second example void foo(int &num) is a function that takes its argument by reference.
Jesper Juhl
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