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What is the declaration of overloaded pre-increment operator implemented as member function?
The pre-increment operator accepts no parameters and returns the same object (by reference) after incrementing. The post-increment operator accepts an unused (dummy) integer parameter and returns a copy of the object (by value) that is made immediately prior to incrementing the object. Note that it is good practice to always use the pre-increment operator even if a post-increment operator exists. The only time you should ever use a post-increment is when you actually store the return value. If you don't store the return value then you will end up making an unnecessary copy, which is highly inefficient. With primitive data types that are less than or equal in length to a pointer this isn't a major issue, but it's good practice nonetheless. If you do it for primitives then you're far more likely to remember to do it for class instances as well. The following example emulates an integer type with pre-increment and post-increment operators implemented: class Simple { public: // Construction: Simple(int data = 0):m_data(data){} Simple(const Simple& simple):m_data(simple.m_data){} public: // Assignment: Simple& operator= (const Simple& simple) { m_data = simple.m_data; return( *this ); } // pre-increment: Simple& operator++ () { // no parameters! ++m_data; // increment this object return( *this ); } // return a reference to this object // post-increment: Simple operator++(int) { // int parameter (not used)! Simple copy( *this ); // call the copy constructor ++m_data; // increment this object return( copy ); } // return the copy (by value) private: int m_data; };
What are applications of pre-increment and post-increment operators why these operators behaviour are made like this is there any reason for such behaviour?
All operators must return a value, and it is this return value (the evaluation of the operation) that differentiates the pre-increment and post-increment operators. If the return value is of no concern, then the pre-increment operator is the preferred version, particularly when working with objects. This is because the post-increment operator's implementation will generally make a copy of the original object which it will return after incrementing the original object. If the return value is of no concern, then making a copy adds an expensive overhead (this doesn't apply to primitive data types since CPU optimisation can pre-empt the return value without requiring a copy to be made). Applications for post-increment and pre-increment are many and varied, but ultimately the goal is to increment the value. Whether you use post-increment or pre-increment is largely down to whether you need to use the return value or not, and whether you need to use the original value or the incremented value. Although the same thing can be achieved with longhand code, it is more efficient to use the appropriate post-increment or pre-increment operators. For instance: int x=1; int y=x; ++x; Can be implemented more efficiently with: int x=1; int y= x++; Whereas: int x=1; ++x; int y=x; Can be implemented more efficiently with: int x=1; int y= ++x; Use caution when pre-incrementing the same value in a compound statement. For example: int x=0; int y= ++x * ++x; The expected result is 2, but the actual result is 4. This is because both expressions (++x) return the same value: a reference to x. And since x is incremented twice, the expression evaluates to 2*2, not 1*2. To get the expected result of 2, each operation must be evaluated separately: int x=0; int y= ++x; y*= ++x;
Why there is only 2 plus in c plus plus languagewhy not 3 or 4 plus?
I assume by 2 plus you really mean ++. This is the increment operator which is used to increment the operand. If placed before the operand, the operator evaluates the incremented operand (prefix increment). If placed after the operand, the operator evaluates the non-incremented operand (postfix increment). +++ and ++++ are meaningless but are assumed to mean incrementing an increment. If you wish to increment an increment, you must use the compound expression ++(++) or (++)++. Thus for the variable x, prefix incrementing twice would be achieved with ++(++x), while postfix incrementing twice would be achieved with (x++)++. You can also mix the two, such as ++(x++) or (++x)++, both of which would increment x twice but would evaluate the increment of x. If postfix increment is not a requirement, it would be much easier to use the compound expression x+=n, where n is the amount you wish to increment. This is the same as saying x=x+n.
How what operator is most efficient you plus plus or plus plus you?
Efficiency is the same; the difference is when the "++" is evaluated, before or after other operations. For example: a = b++ // This will first copy b to a, then increment b. a = ++b // This will first increment b, then copy it to a. If you have the "++" operator by itself, it makes no different if you use prefix or postfix. a++ is the same as ++a.
What does C plus C mean?
In terms of the C++ programming language itself, C++ literally means "the successor to C". In terms of the operator, operator++ is the increment operator. It has two forms: prefix (++c) and postfix (c++). Both do exactly the same thing and are effectively shorthand for the more verbose c = c + 1 or the more concise c += 1. The difference between the the prefix and postfix versions is in their evaluation. ++c will evaluate to c + 1 while c++ evaluates to c (the value of before the increment). However, the evaluations are only of importance when used in compound expressions: int a, b, c = 42; a = c++; // a==42, c==43 b = ++c; // b==44, c==44 Where the evaluation is of no importance, we can use either form. However, for user-defined types, the postfix operator usually incurs an overhead. This can be demonstrated by the following minimal example: struct foo { int i; foo& operator++ () { // prefix increment ++i; // increment this->i return *this; // return the incremented object } foo& operator++ (int) { // postfix increment foo f {*this}; // copy this object ++i; // increment this->i return f; // return the pre-incremented object } // ... }; Note that in both cases, the object itself is incremented, the only difference is in what value is returned to the caller (the newly incremented object or the original non-incremented object). However, because postfix increment requires that we copy the original value, we incur an overhead. The more complex the object, the greater that overhead will be. Built-in types do not suffer the copy overhead because the copy can be factored away by the compiler. Even so, it's a good idea to get into the habit of using prefix increment for built-in types unless you have good reason to use postfix increment. Similarly, it is usually a good idea to omit the postfix operator from user-defined classes unless we have good reason to provide it. Note that the postfix operator has an unused int argument in order to differentiate it from the prefix operator. The way to remember which is which is that the prefix operator has no argument, which is in common with all the other unary member operators. The postfix operator is the anomaly, thus it gets the dummy argument. The prefix and postfix decrement operators (operator--) work in a similar way, except they decrement the operand rather than increment it.
What is a 'post fix expression' in java programming?
Postfix expressions are expressions where the operator is at the end of the expression. These include the "++" (increment) and "--" (decrement) operators. Most Java expressions use in-fix notation (e.g. "a + b") but the increment and decrement operators can be postfix ("e.g. "a++" to increment variable a) or even prefix (e.g. "++a").
When a pre increment is followed by a post increment?
You cannot follow a prefix increment with a postfix increment. int x = 40; int y = ++x++; // error This has to be an error because the postfix operator has higher precedence than the prefix operator. Thus the above code is equivalent to: int x = 40; int y = ++(x++); // error The expression (x++) is evaluated first. This increments x to 41 but the expression evaluates to 40 (the original value of x). Thus the prefix operator is essentially trying to evaluate the expression ++40 rather than ++x. This cannot work because the value 40 is not a modifiable lvalue. It has to be modifiable because we want to increment the value and return a reference to the modified value. But there's nothing to refer to here. The value is temporary and will fall from scope immediately after we use it. That is, the prefix operator may well be able to increment the temporary value to 41, but that value immediately falls from scope. With nothing to refer to, the prefix expression cannot be evaluated. The only way we can use both operators together is if we reverse the precedence using parenthesis: int x = 40; int y = (++x)++; Now the prefix operator is evaluated first, returning a reference to x which (now) holds the value 41. The postfix operator then increments x to 42 but returns the original value of 41 which is then assigned to y. Thus when all statements have been executed, y holds the value 41 while x holds the value 42.
What is the difference between post and pre increment unary operators in c with example?
They both increment the variable. But the value returned by the pre-increment operator is the value of the variable after it has been incremented, while the value returned by the post-increment operator is the value before it has been incremented.For example:int a = 1;int b = ++a;// Now a is 2 and b is also 2.int a = 1;int b = a++;// Now a is 2 but b is 1.
Why are there 2 plus sign in c plus plus and Why not a single plus sign?
In C and in C++, the ++ operator means to increment. C++ was intended to be the next version, i.e. the incremental next step, of C, hence the use of the ++ operator.
What is the advantage of auto indexing?
If you are referring to the use of the auto-increment/decrement operator, it is expressed more simply, and in some hardware architectures it is a hard-wired machine instruction, resulting in a faster program.
What are the safety requirements for using a manlift?
The safety requirements for using a manlift include but are not limited to ensuring the operator is using fall protection and checking that the operator is alert and in proper condition to use the equipment. It is also important to be certain the operator is properly trained for using a manlift as incorrect use may comprise the safety of many people.
Why you write c plus plus not plus plus c?
C++ uses the postfix increment operator whereas ++C uses the prefix increment operator. Both do exactly the same thing; they increment C (the same as C=C+1 increments C). The difference is only in the return value. ++C returns a reference to C, whereas C++ returns the original value of C.
