How do the post increment i and pre increment i operators work in Java

Question

Can you explain to me the output of this Java code   int a 5 i   i   a     a   a    i a       a     a  a   a     a   a     System out println a   System out println i     The output is 20 in both cases

User · Answer

a increments a before it is evaluated  a   evaluates a and then increments it   Related to your expression given   i       a       a     a          6     7     7       a is 8 at the end i     a         a       a        5     7     8       a is 8 at the end   The parenteses I used above are implicitly used by Java  If you look at the terms this way you can easily see  that they are both the same as they are commutative

User · Answer

a 5  i   a     a   a      is  i   7   6   7   Working  pre post increment has  right to left  Associativity   and pre has precedence over post   so first of all pre increment will be solve as    a     a    gt  7   6   then a 7 is provided to post increment    7   6   7  20 and a  8   a 5  i a       a     a    is  i 7   7   6   Working  pre post increment has  right to left  Associativity   and pre has precedence over post   so first of all pre increment will be solve as    a     a    gt  7   6 then a 7 is provided to post increment    7   7   6  20 and a  8

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In the above example   int a   5 i   i   a     a   a             Ans  i   6   7   7   20 then a   8   i a       a     a           Ans  i   8   10   11   29 then a   11  a   a     a   a             Ans  a   12   13   13   38  System out println a        Ans  a   38  System out println i        Ans  i   29

User · Answer

a increments and then uses the variable  a   uses and then increments the variable   If you have  a   1    and you do  System out println a       You will see 1    Now a is 2  System out println   a     You will see 3   codaddict explains your particular snippet

User · Answer

i     a     a   a      is  i   6   7   7   Working  increment a to 6  current value 6    increment a to 7  current value 7   Sum is 13 now add it to current value of a   7  and then increment a to 8  Sum is 20 and value of a after the assignment completes is 8   i   a       a     a    is  i   5   7   8   Working  At the start value of a is 5  Use it in the addition and then increment it to 6  current value 6   Increment a from current value 6 to 7 to get other operand of    Sum is 12 and current value of a is 7  Next increment a from 7 to 8  current value   8  and add it to previous sum 12 to get 20

User · Answer

I believe you are executing all these statements differently   executing together will result    38  29   int a 5 i  i   a     a   a      this means i  6 7 7 20 and when this result is stored in i    then last  a  will be incremented  lt br gt  i a       a     a    this means i  5 7 8 20  this could be complicated     but its working like this   lt br gt  a   a     a   a      as a is 6 7 7 20  this is incremented like this

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Does this help   a   5  i   a     a   a      gt  i 6   7   7   a 8   a   5  i a       a     a    gt  i 5   7   8   a 8    The main point is that   a increments the value and immediately returns it    a   also increments the value  in the background  but returns unchanged value of the variable - what looks like it is executed later

User · Answer

pre-increment and post increment are equivalent if not in an expression   int j  0  int r 0          for int v   0  v lt 10    v                 r            j              System out println j     r          1 1    2 2    3 3         4 4  5 5  6 6  7 7  8 8  9 9 10 10

User · Answer

when a is 5  then a   gives a 5 to the expression and increments a afterwards  while   a increments a before passing the number to the expression  which gives a 6 to the expression in this case    So you calculate  i   6   7   7 i   5   7   8

User · Answer

Presuming that you meant  int a 5  int i   i   a     a   a     System out println i    a 5   i a       a     a   System out println i    a 5   a   a     a   a     System out println a     This evaluates to   i    6  a is now 6     7  a is now 7     7  a is now 8    so i is 6   7   7   20 and so 20 is printed   i    5  a is now 6     7  a is now 7     8  a is now 8    so i is 5   7   8   20 and so 20 is printed again   a    6  a is now 6     7  a is now 7     7  a is now 8    and after all of the right hand side is evaluated  including setting a to 8  THEN a is set to 6   7   7   20 and so 20 is printed a final time

User · Answer

Pre-increment means that the variable is incremented BEFORE it s evaluated in the expression  Post-increment means that the variable is incremented AFTER it has been evaluated for use in the expression   Therefore  look carefully and you ll see that all three assignments are arithmetically equivalent

User · Answer

a is prefix increment operator    the result is calculated and stored first  then the variable is used    a   is postfix increment operator    the variable is used first  then the result is calculated and stored    Once you remember the rules  EZ for ya to calculate everything

User · Answer

In both cases it first calculates value  but in post-increment it holds old value and after calculating returns it    a   a   a   1  return a    a     temp   a  a   a   1  return temp

User · Answer

I believe however if you combine all of your statements and run it in Java 8 1 you will get a different answer  at least that s what my experience says   The code will work like this   int a 5 i   i   a     a   a                 a   5                                  i   a     a   a      gt                                  i 6   7   7   a 8   i 20     i a       a     a              a   5                                  i a       a     a    gt                                  i 8   10   11   a 11   i 29     a   a     a   a                 a 5                                  a   a     a   a      gt                                  a 12   13   13   a 38     System out println a            output  38 System out println i             output  29

[java] How do the post increment (i++) and pre increment (++i) operators work in Java?

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