Post-increment and Pre-increment concept

Question

I don t understand the concept of postfix and prefix increment or decrement  Can anyone give a better explanation

User · Answer

From the C99 standard  C   should be the same  barring strange overloading   6 5 2 4 Postfix increment and decrement operators Constraints 1 The operand of the postfix increment or decrement operator shall have qualified or unqualified real or pointer type and shall be a modifiable lvalue  Semantics 2 The result of the postfix    operator is the value of the operand  After the result is obtained  the value of the operand is incremented   That is  the value 1 of the appropriate type is added to it   See the discussions of additive operators and compound assignment for information on constraints  types  and conversions and the effects of operations on pointers  The side effect of updating the stored value of the operand shall occur between the previous and the next sequence point  3 The postfix -- operator is analogous to the postfix    operator  except that the value of the operand is decremented  that is  the value 1 of the appropriate type is subtracted from it   6 5 3 1 Prefix increment and decrement operators Constraints 1 The operand of the prefix increment or decrement operator shall have qualified or unqualified real or pointer type and shall be a modifiable lvalue  Semantics 2 The value of the operand of the prefix    operator is incremented  The result is the new value of the operand after incrementation  The expression   E is equivalent to  E  1   See the discussions of additive operators and compound assignment for information on constraints  types  side effects  and conversions and the effects of operations on pointers  3 The prefix -- operator is analogous to the prefix    operator  except that the value of the operand is decremented

User · Answer

The difference between the postfix increment  x    and the prefix increment    x  is precisely in how the two operators evaluate their operands  The postfix increment conceptually copies the operand in memory  increments the original operand and finally yields the value of the copy  I think this is best illustrated by implementing the operator in code   int operator     int amp  n      postfix increment       int tmp   n      n   n   1      return tmp      The above code will not compile because you can t re-define operators for primitive types  The compiler also can t tell here we re defining a postfix operator rather than prefix  but let s pretend this is correct and valid C    You can see that the postfix operator indeed acts on its operand  but it returns the old value prior to the increment  so the result of the expression x   is the value prior to the increment  x  however  is incremented   The prefix increment increments its operand as well  but it yields the value of the operand after the increment   int amp  operator     int amp  n        n   n   1      return n      This means that the expression   x evaluates to the value of x after the increment   It s easy to think that the expression   x is therefore equivalent to the assignmnet  x x 1   This is not precisely so  however  because an increment is an operation that can mean different things in different contexts  In the case of a simple primitive integer  indeed   x is substitutable for  x x 1   But in the case of a class-type  such as an iterator of a linked list  a prefix increment of the iterator most definitely does not mean  adding one to the object

User · Answer

The pre increment is before increment value    e g       v  or 1   v   The post increment is after increment the value    e g     rmv    or rmv   1   Program   int rmv   10  vivek   10  cout  lt  lt   rmv        lt  lt  rmv    lt  lt  endl     the value is 10 cout  lt  lt     vivek      lt  lt    vivek     the value is 11

User · Answer

Post increment a     If  int b   a   then this means   int b   a   a   a 1    Here we add 1 to the value  The value is returned before the increment is made    For eg  a   1  b   a     Then b 1 and a 2  Pre-increment    a   If int b     a  then this means  a a 1   int b a     Pre-increment  This will add 1 to the main value  The value will be  returned after the increment is made  For a   1  b     a   Then b 2 and a 2

User · Answer

int i   1  int j   1   int k   i       post increment int l     j     pre increment  std  cout  lt  lt  k     prints 1 std  cout  lt  lt  l     prints 2   Post increment implies the value i is incremented after it has been assigned to k  However  pre increment implies the value j is incremented before it is assigned to l   The same applies for decrement

User · Answer

All four answers so far are incorrect  in that they assert a specific order of events   Believing that  urban legend  has led many a novice  and professional  astray  to wit  the endless stream of questions about Undefined Behavior in expressions   So   For the built-in C   prefix operator     x   increments x and produces  as the expression s result  x as an lvalue  while  x     increments x and produces  as the expression s result  the original value of x   In particular  for x   there is no no time ordering implied for the increment and production of original value of x  The compiler is free to emit machine code that produces the original value of x  e g  it might be present in some register  and that delays the increment until the end of the expression  next sequence point    Folks who incorrectly believe the increment must come first  and they are many  often conclude from that certain expressions must have well defined effect  when they actually have Undefined Behavior

User · Answer

Post-increment   int x  y  z   x   1  y   x      this means  y is assigned the x value first  then increase the value of x by 1  Thus y is 1  z   x    the value of x in this line and the rest is 2 because it was increased by 1 in the above line  Thus z is 2    Pre-increment   int x  y  z   x   1  y     x    this means  increase the value of x by 1 first  then assign the value of x to y  The value of x in this line and the rest is 2  Thus y is 2  z   x    the value of x in this line is 2 as stated above  Thus z is 2

User · Answer

Since we now have inline javascript snippets I might as well add an interactive example of pre and pos increment  It s not C   but the concept stays the same    x000D   x000D  let A   1  x000D  let B   1  x000D   x000D  console log  A       2   A       2   x000D  console log    B     2     B     2   x000D   x000D   x000D

User · Answer

No one has answered the question  Why is this concept confusing   As an undergrad Computer Science major it took me awhile to understand this because of the way I read the code   The following is not correct     x   y    X is equal to y post increment  Which would logically seem to mean X is equal to the value of Y after the increment operation is done  Post meaning after   or  x     y  X is equal to y pre-increment  Which would logically seem to mean X is equal to the value of Y before the increment operation is done  Pre meaning before     The way it works is actually the opposite  This concept is confusing because the language is misleading  In this case we cannot use the words to define the behavior  x   y is actually read as X is equal to the value of Y after the increment  x y   is actually read as X is equal to the value of Y before the increment   The words pre and post are backwards with respect to semantics of English  They only mean where the    is in relation Y   Nothing more   Personally  if I had the choice I would switch the meanings of   y and y    This is just an example of a idiom that I had to learn   If there is a method to this madness I d like to know in simple terms   Thanks for reading

User · Answer

int i  x   i   2  x     i     now i   3  x   3  i   2  x   i        now i   3  x   2    Post  means after - that is  the increment is done after the variable is read   Pre  means before - so the variable value is incremented first  then used in the expression

User · Answer

You should also be aware that the behaviour of postincrement decrement operators is different in C C   and Java   Given     int a 1    in C C   the expression   a     a     a     evaluates to 3  while in Java it evaluates to 6  Guess why     This example is even more confusing   cout  lt  lt  a     a     a    lt  lt    lt - gt    lt  lt  a     a       prints 9 lt - 2    This is because the above expression is equivalent to   operator lt  lt      operator lt  lt        operator lt  lt   cout  a     a             lt - gt           a     a     a

User · Answer

It s pretty simple  Both will increment the value of a variable  The following two lines are equal   x      x    The difference is if you are using the value of a variable being incremented   x   y    x     y    Here  both lines increment the value of y by one  However  the first one assigns the value of y before the increment to x  and the second one assigns the value of y after the increment to x   So there s only a difference when the increment is also being used as an expression  The post-increment increments after returning the value  The pre-increment increments before

[c++] Post-increment and Pre-increment concept?

Examples related to c++

Examples related to conceptual

Examples related to post-increment

Examples related to pre-increment