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Puzzling C++ syntax using "&" [duplicate]

This question already has answers here: Closed 11 years ago.

Possible Duplicate:

What is the meaning of & in c++?

Here is a C++ class declaration snippet containing an ampersand that doesn't 开发者_Python百科make sense to me:

class rimage : public matrix {  
public:  
    rimage( void );  
    rimage( int, int );  
    rimage( rimage & );  
    rimage & operator=( const matrix & A );  
    rimage & operator=( const rimage & A );  
    rimage & operator/=( int );  
...

What does rimage( rimage & ); mean? Why is the & at the end?

This snippet was from a text book on PCNNs

This class declaration is giving me compilation problems so I need to figure this out.


The declaration

rimage ( rimage & )

Might be easier to parse if you give the argument a name:

rimage ( rimage & A )

From this, it's a bit clearer that this defines a copy constructor that takes in another rimage by non-const reference.

More generally, when you leave the names of parameters out of function declarations, they can sometimes be a bit harder to read. For example, this function

void DoSomething (int (int))

Can be hard to read, but if you add in a parameter name like this:

void DoSomething (int function(int))

Now you can see a bit more clearly that it's a function that takes another function in as a parameter.

For something even harder, consider

void DoSomething (int (&)[5])

With a parameter it's a bit easier to see what this is:

void DoSomething (int (&array)[5])

This is a function that takes in a reference to an array of five integers.

Hope this helps!


The & signifies that it is an argument being passed by reference. In this case, the rimage(rimage&) is the declaration for a copy constructor.


It's a reference to an 'rimage' object

https://isocpp.org/wiki/faq/references

It's probably just confusing you because it's unnamed in this declaration.

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