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
++a increments and also then uses the variable.a++ uses and also then increments the variable.
You are watching: Pre increment vs post increment java
If you have
a = 1;and you do
System.out.println(a++); //You will view 1//Now a is 2System.out.println(++a); //You will view 3codaddict describes your details snippet.
Does this help?
a = 5;i=++a + ++a + a++; =>i=6 + 7 + 7; (a=8)a = 5;i=a++ + ++a + ++a; =>i=5 + 7 + 8; (a=8)The main suggest is the ++a increments the value and immediately return it.
a++ also increments the value (in the background) however returns unchanged value the the variable - what looks choose it is executed later.
In both instances it an initial calculates value, however in post-increment that holds old value and after calculating returns it
++aa = a + 1;return a;
a++temp = a;a = a + 1;return temp;
i = ++a + ++a + a++;is
i = 6 + 7 + 7Working: increment a come 6 (current worth 6) + increment a to 7 (current value 7). Amount is 13 now include it to present value that a (=7) and then increment a come 8. Sum is 20 and also value the a ~ the assignment completes is 8.
i = a++ + ++a + ++a;is
i = 5 + 7 + 8Working: at the start value of a is 5. Use it in the enhancement and climate increment it come 6 (current worth 6). Increment a from present value 6 to 7 to obtain other operand of +. Amount is 12 and current value of a is 7. Next increment a native 7 to 8 (current worth = 8) and include it come previous amount 12 to get 20.
++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 endi = ((a++) + (++a) + (++a)) == ((5) + (7) + (8)); // a is 8 at the endThe parenteses ns used above are implicitly offered by Java. If girlfriend look at the state this way you can quickly see, that they are both the exact same as they are commutative.
In the above example
int a = 5,i;i=++a + ++a + a++; //Ans: ns = 6 + 7 + 7 = 20 climate a = 8 i=a++ + ++a + ++a; //Ans: ns = 8 + 10 + 11 = 29 then a = 11a=++a + ++a + a++; //Ans: a = 12 + 13 + 13 = 38System.out.println(a); //Ans: a = 38System.out.println(i); //Ans: i = 29
++a is prefix increment operator:
a++ is postfix increment operator:the change is provided first,then the an outcome is calculated and also stored.
Once girlfriend remember the rules, EZ for ya to calculation everything!
I think however if girlfriend combine all of your statements and also run it in Java 8.1 friend will gain a various answer, at least that"s what my experience says.
The code will work-related like this:
int a=5,i;i=++a + ++a + a++; /*a = 5; i=++a + ++a + a++; => i=6 + 7 + 7; (a=8); i=20;*/i=a++ + ++a + ++a; /*a = 5; i=a++ + ++a + ++a; => i=8 + 10 + 11; (a=11); i=29;*/a=++a + ++a + a++; /*a=5; a=++a + ++a + a++; => a=12 + 13 + 13; a=38;*/System.out.println(a); //output: 38System.out.println(i); //output: 29
Presuming the 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 evaluate to:
i = (6, a is currently 6) + (7, a is currently 7) + (7, a is currently 8)so i is 6 + 7 + 7 = 20 and so 20 is printed.
i = (5, a is currently 6) + (7, a is now 7) + (8, a is now 8)so ns is 5 + 7 + 8 = 20 and also so 20 is printed again.
a = (6, a is currently 6) + (7, a is currently 7) + (7, a is currently 8)and after all of the appropriate hand next is evaluated (including setting a to 8) then a is collection to 6 + 7 + 7 = 20 and also so 20 is printed a final time.
when a is 5, then a++ offers a 5 come the expression and also 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 friend calculate
i = 6 + 7 + 7i = 5 + 7 + 8
Pre-increment method that the change is incremented prior to it"s evaluated in the expression. Post-increment method that the change is incremented ~ it has been evaluated for use in the expression.
Therefore, look at carefully and you"ll watch that all 3 assignments room arithmetically equivalent.
pre-increment and also post increment are indistinguishable if not in an expression
int j =0;int r=0 for(int v = 0; v
a=5; i=++a + ++a + a++;is
i = 7 + 6 + 7Working: pre/post increment has actually "right come left" Associativity , and also pre has actually precedence over short article , so first of all pre increment will certainly be fix as (++a + ++a) => 7 + 6 . Then a=7 is detailed to post increment => 7 + 6 + 7 =20 and also a =8.
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a=5; i=a++ + ++a + ++a;is
i=7 + 7 + 6Working: pre/post increment has actually "right come left" Associativity , and also pre has actually precedence over write-up , so an initial of every pre increment will certainly be fix as (++a + ++a) => 7 + 6.then a=7 is detailed to write-up increment => 7 + 7 + 6 =20 and also a =8.
I believe you space executing every these statements in different way executing with each other will an outcome => 38 ,29
int a=5,i;i=++a + ++a + a++;//this means i= 6+7+7=20 and when this result is save in i,//then last *a* will be incremented i=a++ + ++a + ++a;//this means i= 5+7+8=20 (this can be complicated, //but that working prefer this),a=++a + ++a + a++;//as a is 6+7+7=20 (this is incremented choose this)
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