Object-Oriented Programming (OOP) Lecture 33: Templates and User-Defined Types, Slides of Object Oriented Programming

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Object-Oriented Programming

(OOP)

Lecture No. 33

Recap

• Templates are generic abstractions
• C++ templates are of two kinds
  • Function Templates
  • Class Templates
• A general template can be specialized to

specifically handle a particular type

User-Defined Types

• Besides primitive types, user-defined

types can also be passed as type

arguments to templates

• Compiler performs static type checking to

diagnose type errors

…User-Defined Types

• Consider the String class without

overloaded operator “==“

class String {

char* pStr; … // Operator “==“ not defined

};

…User-Defined Types

class String {

char* pStr; … friend bool operator ==( const String&, const String& ); };

… User-Defined Types

bool operator ==( const String& x,

const String& y ) { return strcmp(x.pStr, y.pStr) == 0;

}

Overloading vs Templates

• Different data types, similar operation

Needs function overloading

• Different data types, identical operation

Needs function templates

Example

Overloading vs Templates

• ‘+’ operation is overloaded for different

operand types

• A single function template can calculate

sum of array of many types

…Example

Overloading vs Templates

String operator +( const char * str1, const String& y ) { String tmp; tmp.pStr = new char[ strlen(str1) + strlen(y.pStr) + 1 ]; strcpy( tmp.pStr, str1 ); strcat( tmp.pStr, y.pStr ); return tmp;

}

…Example

Overloading vs Templates

template< class T >

T sum( T* array, int size ) {

T sum = 0;

for (int i = 0; i < size; i++) sum = sum + array[i];

return sum;

}

Example – Policy

• Write a function that compares two given

character strings

• Function can perform either case-sensitive

or non-case sensitive comparison

First Solution

int caseSencompare( char* str1,

char* str2 )

{

for (int i = 0; i < strlen( str1 ) && i < strlen( str2 ); ++i) if ( str1[i] != str2[i] ) return str1[i] - str2[i];

return strlen(str1) - strlen(str2);

}

Second Solution

int compare( char* str1, char* str2, bool caseSen )

{

for (int i = 0; i < strlen( str1 ) && i < strlen( str2 ); i++) if ( … ) return str1[i] - str2[i];

return strlen(str1) - strlen(str2);

} Docsity.com

…Second Solution

// if condition:

(caseSen && str1[i] != str2[i])

|| (!caseSen &&

toupper(str1[i])

toupper(str2[i]))