In lesson 10.1 -- Introduction to dynamic memory allocation, we introduced the concept of naming collisions and namespaces. As a reminder, a naming collision occurs when two identical identifiers are introduced into the same scope, and the compiler can’t disambiguate which one to use. When this happens, compiler or linker will produce an error because they do not have enough information to resolve the ambiguity. As programs become larger, the number of identifiers increases linearly, which in turn causes the probability of a naming collision occurring to increase exponentially.
Let’s revisit an example of a naming collision, and then show how we can resolve it using namespaces. In the following example, foo.cpp and goo.cpp are the source files that contain functions that do different things but have the same name and parameters.
foo.cpp:
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// This doSomething() adds the value of its parameters int doSomething(int x, int y) { return x + y; } |
goo.cpp:
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// This doSomething() subtracts the value of its parameters int doSomething(int x, int y) { return x - y; } |
main.cpp:
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#include <iostream> int doSomething(int x, int y); // forward declaration for doSomething int main() { std::cout << doSomething(4, 3) << '\n'; // which doSomething will we get? return 0; } |
If this project contains only foo.cpp or goo.cpp (but not both), it will compile and run without incident. However, by compiling both into the same program, we have now introduced two different functions with the same name and parameters into the same scope (the global scope), which causes a naming collision. As a result, the linker will issue an error:
goo.cpp:3: multiple definition of `doSomething(int, int)'; foo.cpp:3: first defined here
Note that this error happens at the point of redefinition, so it doesn’t matter whether function doSomething is ever called.
One way to resolve this would be to rename one of the functions, so the names no longer collide. But this would also require changing the names of all the function calls, which can be a pain, and is subject to error. A better way to avoid collisions is to put your functions into your own namespaces. For this reason the standard library was moved into the std namespace.
Defining your own namespaces
C++ allows us to define our own namespaces via the namespace keyword. Namespaces that you create for your own declarations are called user-defined namespaces. Namespaces provided by C++ (such as the global namespace) or by libraries (such as namespace std) are not considered user-defined namespaces.
Namespace identifiers are typically non-capitalized.
Here is an example of the files in the prior example rewritten using namespaces:
foo.cpp:
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namespace foo // define a namespace named foo { // This doSomething() belongs to namespace foo int doSomething(int x, int y) { return x + y; } } |
goo.cpp:
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namespace goo // define a namespace named goo { // This doSomething() belongs to namespace goo int doSomething(int x, int y) { return x - y; } } |
Now doSomething() inside of foo.cpp is inside the foo namespace, and the doSomething() inside of goo.cpp is inside the goo namespace. Let’s see what happens when we recompile our program.
main.cpp:
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int doSomething(int x, int y); // forward declaration for doSomething int main() { std::cout << doSomething(4, 3) << '\n'; // which doSomething will we get? return 0; } |
The answer is that we now get another error!
ConsoleApplication1.obj : error LNK2019: unresolved external symbol "int __cdecl doSomething(int,int)" (?doSomething@@YAHHH@Z) referenced in function _main
In this case, the compiler was satisfied (by our forward declaration), but the linker could not find a definition for doSomething in the global namespace. This is because both of our versions of doSomething are no longer in the global namespace!
There are two different ways to tell the compiler which version of doSomething() to use, via the scope resolution operator, or via using statements (which we’ll discuss in a later lesson in this chapter).
For the subsequent examples, we’ll collapse our examples down to a one-file solution for ease of reading.
Accessing a namespace with the scope resolution operator (::)
The best way to tell the compiler to look in a particular namespace for an identifier is to use the scope resolution operator (::). The scope resolution operator tells the compiler that the identifier specified by the right-hand operand should be looked for in the scope of the left-hand operand.
Here is an example of using the scope resolution operator to tell the compiler that we explicitly want to use the version of doSomething() that lives in the foo namespace:
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#include <iostream> namespace foo // define a namespace named foo { // This doSomething() belongs to namespace foo int doSomething(int x, int y) { return x + y; } } namespace goo // define a namespace named goo { // This doSomething() belongs to namespace goo int doSomething(int x, int y) { return x - y; } } int main() { std::cout << foo::doSomething(4, 3) << '\n'; // use the doSomething() that exists in namespace foo return 0; } |
This produces the expected result:
7
If we wanted to use the version of doSomething() that lives in goo instead:
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#include <iostream> namespace foo // define a namespace named foo { // This doSomething() belongs to namespace foo int doSomething(int x, int y) { return x + y; } } namespace goo // define a namespace named goo { // This doSomething() belongs to namespace goo int doSomething(int x, int y) { return x - y; } } int main() { std::cout << goo::doSomething(4, 3) << '\n'; // use the doSomething() that exists in namespace goo return 0; } |
This produces the result:
1
The scope resolution operator is great because it allows us to explicitly pick which namespace we want to look in, so there’s no potential ambiguity. We can even do the following:
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#include <iostream> namespace foo // define a namespace named foo { // This doSomething() belongs to namespace foo int doSomething(int x, int y) { return x + y; } } namespace goo // define a namespace named goo { // This doSomething() belongs to namespace goo int doSomething(int x, int y) { return x - y; } } int main() { std::cout << foo::doSomething(4, 3) << '\n'; // use the doSomething() that exists in namespace foo std::cout << goo::doSomething(4, 3) << '\n'; // use the doSomething() that exists in namespace goo return 0; } |
This produces the result:
7 1
Using the scope resolution operator with no name prefix
The scope resolution operator can also be used in front of an identifier without providing a namespace name (e.g. ::doSomething). In such a case, the identifier (e.g. doSomething) is looked for in the global namespace.
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#include <iostream> void print() // this print lives in the global namespace { std::cout << " there\n"; } namespace foo { void print() // this print lives in the foo namespace { std::cout << "Hello"; } } int main() { foo::print(); // call print() in foo namespace ::print(); // call print() in global namespace (same as just calling print() in this case) return 0; } |
In the above example, the ::print() performs the same as if we’d called print() with no scope resolution, so use of the scope resolution operator is superfluous in this case. But the next example will show a case where the scope resolution operator with no namespace can be useful.
Identifier resolution from within a namespace
If an identifier inside a namespace is used and no scope resolution is provided, the compiler will first try to find a matching declaration in that same namespace. If no matching identifier is found, the compiler will then check each containing namespace in sequence to see if a match is found, with the global namespace being checked last.
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#include <iostream> void print() // this print lives in the global namespace { std::cout << " there\n"; } namespace foo { void print() // this print lives in the foo namespace { std::cout << "Hello"; } void printHelloThere() { print(); // calls print() in foo namespace ::print(); // calls print() in global namespace } } int main() { foo::printHelloThere(); return 0; } |
This prints:
Hello there
In the above example, print() is called with no scope resolution provided. Because this use of print() is inside the foo namespace, the compiler will first see if a declaration for foo::print() can be found. Since one exists, foo::print() is called.
If foo::print() had not been found, the compiler would have checked the containing namespace (in this case, the global namespace) to see if it could match a print() there.
Note that we also make use of the scope resolution operator with no namespace (::print()) to explicitly call the global version of print().
Multiple namespace blocks are allowed
It’s legal to declare namespace blocks in multiple locations (either across multiple files, or multiple places within the same file). All declarations within the namespace are considered part of the namespace.
circle.h:
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#ifndef CIRCLE_H #define CIRCLE_H namespace basicMath { inline constexpr double pi{ 3.14 }; } #endif |
growth.h:
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#ifndef GROWTH_H #define GROWTH_H namespace basicMath { // the constant e is also part of namespace basicMath inline constexpr double e{ 2.7 }; } #endif |
main.cpp:
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#include "circle.h" // for basicMath::pi #include "growth.h" // for basicMath::e #include <iostream> int main() { std::cout << basicMath::pi << '\n'; std::cout << basicMath::e << '\n'; return 0; } |
This works exactly as you would expect:
3.14 2.7
The standard library makes extensive use of this feature, as each standard library header file contains its declarations inside a namespace std block contained within that header file. Otherwise the entire standard library would have to be defined in a single header file!
Note that this capability also means you could add your own functionality to the std namespace. Doing so causes undefined behavior most of the time, because the std namespace has a special rule, prohibiting extension from user code.
Warning
Do not add custom functionality to the std namespace.
When you separate your code into multiple files, you’ll have to use a namespace in the header and source file.
add.h
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#ifndef ADD_H #define ADD_H namespace basicMath { // function add() is part of namespace basicMath int add(int x, int y); } #endif |
add.cpp
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#include "add.h" namespace basicMath { // define the function add() int add(int x, int y) { return x + y; } } |
main.cpp
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#include "add.h" // for basicMath::add() #include <iostream> int main() { std::cout << basicMath::add(4, 3) << '\n'; return 0; } |
If the namespace is omitted in the source file, the linker won’t find a definition of basicMath::add, because the source file only defines add (global namespace). If the namespace is omitted in the header file, “main.cpp” won’t be able to use basicMath::add, because it only sees a declaration for add (global namespace).
Nested namespaces
Namespaces can be nested inside other namespaces. For example:
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#include <iostream> namespace foo { namespace goo // goo is a namespace inside the foo namespace { int add(int x, int y) { return x + y; } } } int main() { std::cout << foo::goo::add(1, 2) << '\n'; return 0; } |
Note that because namespace goo is inside of namespace foo, we access add as foo::goo::add.
Since C++17, nested namespaces can also be declared this way:
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#include <iostream> namespace foo::goo // goo is a namespace inside the foo namespace (C++17 style) { int add(int x, int y) { return x + y; } } int main() { std::cout << foo::goo::add(1, 2) << '\n'; return 0; } |
Namespace aliases
Because typing the fully qualified name of a variable or function inside a nested namespace can be painful, C++ allows you to create namespace aliases, which allow us to temporarily shorten a long sequence of namespaces into something shorter:
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#include <iostream> namespace foo::goo { int add(int x, int y) { return x + y; } } int main() { namespace active = foo::goo; // active now refers to foo::goo std::cout << active::add(1, 2) << '\n'; // This is really foo::goo::add() return 0; } // The active alias ends here |
One nice advantage of namespace aliases: If you ever want to move the functionality within foo::goo to a different place, you can just update the active alias to reflect the new destination, rather than having to find/replace every instance of foo::goo.
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#include <iostream> namespace foo::goo { } namespace v2 { int add(int x, int y) { return x + y; } } int main() { namespace active = v2; // active now refers to v2 std::cout << active::add(1, 2) << '\n'; // We don't have to change this return 0; } |
It’s worth noting that namespaces in C++ were not originally designed as a way to implement an information hierarchy -- they were designed primarily as a mechanism for preventing naming collisions. As evidence of this, note that the entirety of the standard library lives under the singular namespace std:: (with some nested namespaces used for newer library features). Some newer languages (such as C#) differ from C++ in this regard.
In general, you should avoid deeply nested namespaces.
When you should use namespaces
In applications, namespaces can be used to separate application-specific code from code that might be reusable later (e.g. math functions). For example, physical and math functions could go into one namespace (e.g. math::). Language and localization functions in another (e.g. lang::).
When you write a library or code that you want to distribute to others, always place your code inside a namespace. The code your library is used in may not follow best practices -- in such a case, if your library’s declarations aren’t in a namespace, there’s an elevated chance for naming conflicts to occur. As an additional advantage, placing library code inside a namespace also allows the user to see the contents of your library by using their editor’s auto-complete and suggestion feature.
 6.3 -- Local variables |
 Index |
 6.1 -- Compound statements (blocks) |
This tutorial is really very helpful.Thanks Alex.
one comment on the header files included in the main.cpp
User defined header files should be in double quotes(ex: #include "a.h") otherwise compiler will throw an error.
Fantastic tutorial, I'm really enjoying it - thank you very much for sharing your knowledge, Alex. At this point of tutorial, I have a question:
I followed the examples in this section, and found out everything works fine when using scope resolution operator:
double dSin = 0.0;
double dCos = 0.0;
foo::GetSinCos(30.0, dSin, dCos);
cout << "The sin is " << dSin << endl;
cout << "The cos is " << dCos << endl;
But when I try the "using namespace" keyword method, I get "ambiguous call to overloaded function" error:
double dSin = 0.0;
double dCos = 0.0;
using namespace foo;
GetSinCos(30.0, dSin, dCos);
cout << "The sin is " << dSin << endl;
cout << "The cos is " << dCos << endl;
Does anyone have an idea why is this happening? Thanks :)
I'm confused! Consider "cin", for example. It is in the namespace std. Is there any other namespaces with object "cin"? Otherwise, why we need to declare the namespace std::cin.
In other words, would you please give me an example for two standard objects with the same name in different namespaces? (I mean an object that is not defined by user)
He used the 'std::cin' as an example to show you how it works.
also, while I'm not aware of such colliding cases, it might be possible that custom namespaces have colliding names, and this is when it might come useful.
nice work!!
well.. this didn't seem like the hardest section (pointers was a hard concept for someone new to cpp), yet for some reason I cant get this to work!
I'm also confused that you mention the header files, but then the functions that you use in the header files don't look like simple prototypes, since they have a return statements, they look like full functions ..hmm
The simplified example is below:
header.h:
#ifndef ...etc
namespace exF{ bool bX() }externalFn.cpp:
bool bX(){ return true;}so in my main.cpp, I should just be able to use:
main.cpp :
#include "header.h" int main(){ if (exF:bX() == true){ cout << "yay"; } retrun 0; }but the only error I get is: undefined reference to 'exf::bx()'
I dont understand why, should we actually be defining the full functions (not just the proto) in the header files now?
(The rest of this tutorial has been great by the way... Thank you so much for sharing your wealth of knowledge and teaching skills ;) )
I'm not certain why, but removing the inline option (learnt in the previous lesson) solved my problem.
my question
How to get the out put in GUI for C++ Programing?
In this function, the number of arguments is zero
but in
The number of arguments is not know[infinity].