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I came across an expression that states: vector<int> pair(n). As pair is an inbuilt structure in itself, can we use it as a variable name in c++ and why??

Alan Birtles
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Muskaan Singla 216
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    `using namespace std` can be confusing, yes. Please show a [mcve] – 463035818_is_not_an_ai Apr 26 '21 at 08:21
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    Here's the list of keywords you can't use: https://en.cppreference.com/w/cpp/keyword – m88 Apr 26 '21 at 08:21
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    there's no problem using names from the standard library as long as you aren't using `using namespace std` – Alan Birtles Apr 26 '21 at 08:23
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    You can even have a vector called vector: `std::vector vector{ 1,2,3 };` is perfectly valid code (if a little odd). – Adrian Mole Apr 26 '21 at 08:24
  • this very much depends on what headers you include and what using directive is in the code. – 463035818_is_not_an_ai Apr 26 '21 at 08:24
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    Very related, if not an actual duplicate: [Why is "using namespace std;" considered bad practice?](https://stackoverflow.com/q/1452721/10871073) – Adrian Mole Apr 26 '21 at 08:28
  • The question is a bit incomplete, so we can't assume `using namespace std;`. It might have said `using std::vector;`, in which case `std::pair` is not brought in. – MSalters Apr 26 '21 at 08:30
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    I don't understand the obsession with `using namespace std;` for this question. Even with that, you could do something like `vector vector;`, it doesn't suddenly prevent you from shadowing a type name with a variable name just because the type name is in scope. – chris Apr 26 '21 at 08:32
  • @chris it does if you want to use `std::vector` again in the same scope: https://godbolt.org/z/GWrf75nWd – Alan Birtles Apr 26 '21 at 08:52
  • @AlanBirtles, Yes, that's a property of shadowing, not of the using directive. – chris Apr 26 '21 at 08:52
  • @chris that code would have compiled fine if you used `std::vector` instead of using namespace std + unqualified vector – Aykhan Hagverdili Apr 26 '21 at 09:12

3 Answers3

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Yes, you can. The templates from the standard library are called std::pair and std::vector. They are not built-in, in the sense that they aren't part of the language, but part of the standard library. Hence this is not a problem:

#include <vector>  
std::vector<int> pair(n);

One can use using directives to drop the std:: prefix:

#include <vector>
using std::vector;
vector<int> pair(n);

Commonly used is also using namespace std;:

#include <vector>
using namespace std;
vector<int> pair(n);

Though, once you pulled the complete namespace into scope, the code can be rather confusing.

Read Why is “using namespace std;” considered bad practice? and try to avoid it. And give proper names. std::pair is rather common, so you better choose a better name for the vector, it certainly is not a pair.

463035818_is_not_an_ai
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As pair is an inbuilt structure in itself, can we use it as a variable name in c++ and why?

That is not correct. The Standard Template Library provides a class template called "pair" but that is not an inbuilt type and, furthermore, "pair" is neither a keyword nor a reserved word in the C++ language.

Furthermore, the aforementioned "pair" template is actually provided (in the <utility> header) in the std namespace, so its fully qualified name is std::pair. So, unless you both #include <utility> (explicitly or implicitly) and are using std::pair; (or using namespace std; – but see here), you can use the name, "pair," as an identifier, just like any other non-reserved word.

In fact, even if you do #include <utility> and are using std::pair; (or using namespace std;), you can still use "pair" as an identifier, as mentioned the comment by chris.

Adrian Mole
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  • I'd say "_The C++ standard library_" instead of "_The Standard Template Library_" though. – Ted Lyngmo Apr 26 '21 at 08:43
  • If you haven't included , that doesn't necessarily mean pair isn't defined in std namespace. It's possible that some other standard header you include includes or defines pair. unordered_map, for instance, certainly does. – Aykhan Hagverdili Apr 26 '21 at 09:08
  • @AyxanHaqverdili Notice that I mentioned inclusion of that header, "explicitly *or implicitly*." From what I can tell, if `std::pair` is defined, it will be defined in that header (but I'd need to thoroughly check the Standard, to be sure). – Adrian Mole Apr 26 '21 at 09:11
  • @AdrianMole it's very possible that there is a third header that only defines`pair` (not other things defines) and both the headers ( and some other header you used) include that. – Aykhan Hagverdili Apr 26 '21 at 09:13
  • @AyxanHaqverdili [This Draft C++17 Standard](http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2017/n4713.pdf) strongly suggests (Section 23.1, Table 34) that "pairs" are defined in the `` header. – Adrian Mole Apr 26 '21 at 09:19
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std::pair is not an "in-built structure". It's a qualified name from the Standard Library. Since it's qualified, it generally does not clash with your own names.

This is exactly why namespaces were added to the C++ language.

MSalters
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