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A separate definition of max() is required for each overload declaration that has a unique set of parameters When a function name is declared more than once in a particular scope, the compiler interprets the second (and subsequent) declarations as follows
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If the parameter lists of the two functions differ in either the number or type of their parameters, the two functions are considered to be overloaded For example:
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// overloaded functions void print( const string & ); void print( vector<int> & );
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If both the return type and the parameter list of the two function declarations match exactly, the second declaration is treated as a redeclaration of the first For example:
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// declares the same function void print( const string &str ); void print( const string & );
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The parameter names are irrelevant when parameter lists are compared
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If the parameter lists of the two functions match exactly but the return types differ, the second declaration is treated as an erroneous redeclaration of the first and is flagged at compile-time as an error For example:
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unsigned int max( int i1, int i2 ); int max( int , int ); // error: only return type is different
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A function's return type is not enough to distinguish between two overloaded functions
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If the parameter lists of the two functions differ only in their default arguments, the second declaration is treated as a redeclaration of the first For example:
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// declares the same function int max( int *ia, int sz ); int max( int *, int = 10 );
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A typedef name provides an alternative name for an existing data type; it does not create a new data type Therefore, two function parameter lists that differ only in that one uses a typedef and the other uses the type to which the typedef corresponds are not different parameter lists The following two function declarations of calc() are treated as having exactly the same parameter list The second declaration results in a compile-time error, because, although it declares the same parameter list, it declares a different return type from the first declaration
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// typedef does not introduce a new type typedef double DOLLAR; // error: same parameter list, different return types extern DOLLAR calc( DOLLAR ); extern int calc( double );
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When a parameter type is const or volatile, the const or volatile qualifier is not taken into account when the declarations of different functions are identified For example, the following two delarations declare the same function:
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// declares the same function void f( int ); void f( const int );
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The fact that the parameter is const is relevant only within the definition of function: it means that expressions in the function body cannot change the value of the parameter However, for a pass-by-value argument, this is completely transparent to a user of the function: a user never sees the modifications applied by a function to a pass-by-value argument (Pass-by-value arguments and other methods of argument passing are discussed in Section 73) Declaring a parameter const when the argument is passed by value does not change in any way the kind of arguments that can be passed to the function Any argument of type int can be used in a call to the function f(int) as well as the function f(const int) Because both functions accept the same set of arguments, the declarations just shown do not declare an overloaded function The function f() can be defined as
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void f( int i ) { }
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or as
void f( const int i ) { }
Providing both of these definitions in the same program is an error, however, because these definitions define the same function twice However, if const or volatile applies to the type to which a pointer or reference parameter refers, then the const or volatile qualifier is taken into account when the declarations of different functions are identified
// declares different functions void f( int* ); void f( const int* ); // also declares different functions void f( int& ); void f( const int& );