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What is the difference between prefix and postfix increment operator in c plus plus?
Wiki User
∙ 13y ago
Both the prefix and the postfix increment operators increment the operand. The difference is what is the value of the expression during the evaluation of the expression. In the prefix form, the value is already incremented. In the postfix form, it is not.
int a = 1;
int b = ++a;
// both a and b are now equal to 2
int a = 1;
int b = a++;
// a is equal to 2 and b is equal to 1
Wiki User
∙ 13y ago
Wiki User
∙ 10y agoThe prefix operator increments the operand before returning the value of the (incremented) operand to the expression, while the postfix operator returns the value first, then increments the operand.
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How can you differentiate overloading of pre-fix and post-fix increment operator?
The prefix increment operator is overloaded as operator++() while the postfix increment operator is overloaded as operator++(int).
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.
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.
What is difference between plus plus j and j plus plus in programming?
++j is the prefix increment operator while j++ is the postfix increment operator. The same applies to --j and j++, the prefix and postfix decrement operators. The difference between prefix and postfix increment is not the operation itself -- they both increment the operand, which is j. The difference is in the evaluation of that operation -- the value that is returned by the operation. Prefix will increment j and return a reference to j. Postfix essentially copies j, increments j, and returns a reference to the copy. In other words: If j is 1, ++j increments j (so j is 2) and returns a reference to j (which is now 2). If j is 1, j++ copies j (which is 1), increments j (making it 2) and returns the copy (which is still 1). When j is a primitive (such as int), no copy is actually made (the return value is pre-emptied by the CPU). Thus there is no difference in terms of performance, the only difference is the return value. However, when j is a non-primitive, such as a class type, the postfix operator must physically copy the class instance, increment the original instance and then return the copy. As such there will be a performance penalty. If you have no use for the return value, performance will be improved if you use prefix rather than postfix. The more complex the class, the more obvious the difference will become when repeatedly incrementing a class in a loop. although there is no difference in performance with primitives, it is good practice to use the prefix increment at all times, and use postfix only when you actually need the original value of the operand. Once you get into the habit of using prefix by default, you'll avoid making unnecessary copies of classes other than when you actually need to. In other words, use for( int x=0; x< 10; ++x ) rather than for( int x=0; x<10; x++ ), regardless of whether x is a primitive or a class type. In both cases, the return value is ignored, but it still exists. To capture the return value, you must assign it to another variable. Such as int y = x++, or use the operator in a compound expression (where the result is used as part of a larger expression). Postifx increment is often used with pointers and arrays, to assign a value to the current element being pointed to before advancing the pointer to the next element: int x[2]; int * p = x; *p++ = 1; // assigns x[0] and then advances to x[1]. *p = 2; // assigns x[1]. Thus x[0] is 1 while x[1] is 2.
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.
How can you differentiate overloading of pre-fix and post-fix increment operator?
The prefix increment operator is overloaded as operator++() while the postfix increment operator is overloaded as operator++(int).
What is i plus 1 operator in C plus plus?
The increment operator in C++ is defined by operator++(). All arithmetic types (char, int, float, double, long, short, long long and long double) and all pointer types except void* are supported by operator++(). User-defined types can overload operator++() to provide support where required. operator++() has two versions, prefix increment and postfix increment. Prefix increment behaves as one would expect, incrementing the operand by 1 and returning the modified value. Postfix increment also increments the operand, however, the return value is the pre-incremented value. To understand the difference between prefix and postfix, consider the following: int i = 0; int j = ++i; // i=1, j=1 int i = 0; int j = i++; // i=1, j=0
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").
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.
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.
What do the two plus stand for in C plus plus?
The ++ in C++ refers to the postfix increment operator (operator++()). It's literal meaning is "the successor to C", in reference to the C language upon which the C++ language is based.
Distinguish between operator overloading and function overloading?
in C++ there is no real difference as operators are overloaded by implementing them as functions. However, while we differentiate between function overloads by the function signature (the number and type of parameters), operator overloads are distinguished only by the parameter types. The parameters are interpreted as operands, and the number of operands will depend upon whether the operator is unary, binary or ternary. That is, for any given operator, the number of operands will be the same for each overload you implement. The only exceptions are the unary increment (++) and decrement (--) operators as they each have postfix and prefix variants. In order to differentiate their signatures, an unreferenced or dummy parameter must be passed to the postfix variants.
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.
What is difference between plus plus j and j plus plus in programming?
++j is the prefix increment operator while j++ is the postfix increment operator. The same applies to --j and j++, the prefix and postfix decrement operators. The difference between prefix and postfix increment is not the operation itself -- they both increment the operand, which is j. The difference is in the evaluation of that operation -- the value that is returned by the operation. Prefix will increment j and return a reference to j. Postfix essentially copies j, increments j, and returns a reference to the copy. In other words: If j is 1, ++j increments j (so j is 2) and returns a reference to j (which is now 2). If j is 1, j++ copies j (which is 1), increments j (making it 2) and returns the copy (which is still 1). When j is a primitive (such as int), no copy is actually made (the return value is pre-emptied by the CPU). Thus there is no difference in terms of performance, the only difference is the return value. However, when j is a non-primitive, such as a class type, the postfix operator must physically copy the class instance, increment the original instance and then return the copy. As such there will be a performance penalty. If you have no use for the return value, performance will be improved if you use prefix rather than postfix. The more complex the class, the more obvious the difference will become when repeatedly incrementing a class in a loop. although there is no difference in performance with primitives, it is good practice to use the prefix increment at all times, and use postfix only when you actually need the original value of the operand. Once you get into the habit of using prefix by default, you'll avoid making unnecessary copies of classes other than when you actually need to. In other words, use for( int x=0; x< 10; ++x ) rather than for( int x=0; x<10; x++ ), regardless of whether x is a primitive or a class type. In both cases, the return value is ignored, but it still exists. To capture the return value, you must assign it to another variable. Such as int y = x++, or use the operator in a compound expression (where the result is used as part of a larger expression). Postifx increment is often used with pointers and arrays, to assign a value to the current element being pointed to before advancing the pointer to the next element: int x[2]; int * p = x; *p++ = 1; // assigns x[0] and then advances to x[1]. *p = 2; // assigns x[1]. Thus x[0] is 1 while x[1] is 2.
What is the correct way to add 1 to the count variable?
A counter variable is "incremented" (the step number, 1 in this case, is added to it) in any of the following four ways: $counter = $counter + 1;$counter += 1; //this is shorthand for the above $counter++; //postfix increment operator $counter = 0;echo $counter++;The output would be 0++$counter; //prefix increment operator $counter = 0; echo ++$counter;The output is 1
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.
