Java에서 사후 증가 (i ++) 및 사전 증가 (++ i) 연산자는 어떻게 작동합니까?

Java에서 사후 증가 (i ++) 및 사전 증가 (++ i) 연산자는 어떻게 작동합니까?

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이 Java 코드의 출력을 설명해 주시겠습니까?

int a=5,i;

i=++a + ++a + a++;
i=a++ + ++a + ++a;
a=++a + ++a + a++;

System.out.println(a);
System.out.println(i);

두 경우 모두 출력은 20입니다.

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도움이 되나요?

a = 5;
i=++a + ++a + a++; =>
i=6 + 7 + 7; (a=8)

a = 5;
i=a++ + ++a + ++a; =>
i=5 + 7 + 8; (a=8)

요점은 ++a값 을 증가시키고 즉시 반환한다는 것입니다.

a++ 또한 값을 증가 시키지만 (백그라운드에서) 변수의 변경되지 않은 값을 반환합니다. 나중에 실행되는 것처럼 보입니다.

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++a증분하고 변수를 사용합니다.
a++변수를 사용한 다음 증분합니다.

당신이 가지고 있다면

a = 1;

그리고 당신은

System.out.println(a++); //You will see 1

//Now a is 2

System.out.println(++a); //You will see 3

codaddict는 특정 스 니펫을 설명 합니다.

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두 경우 모두 먼저 값을 계산하지만 사후 증분에서는 이전 값을 유지하고 계산 후 반환합니다.

++ a

1. a = a + 1;
2. 반환 a;

a ++

1. 온도 = a;
2. a = a + 1;
3. 반환 온도;

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i = ++a + ++a + a++;

이다

i = 6 + 7 + 7

Working : a에서 6까지 증가 (현재 값 6) + a에서 7까지 증가 (현재 값 7). 합계는 이제 13입니다. a (= 7)의 현재 값에 더한 다음 a를 8로 증가시킵니다. 합계는 20이고 할당 완료 후 a의 값은 8입니다.

i = a++ + ++a + ++a;

이다

i = 5 + 7 + 8

Working : a의 시작 값은 5입니다. 덧셈에 사용하고 6 (현재 값 6)으로 증가시킵니다. +의 다른 피연산자를 얻으려면 a를 현재 값 6에서 7로 증가시킵니다. 합계는 12이고 a의 현재 값은 7입니다. 다음으로 a를 7에서 8까지 증가시키고 (현재 값 = 8) 이전 합계 12에 더하여 20을 얻습니다.

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++aa평가되기 전에 증가 합니다. a++평가 a하고 증분합니다.

주어진 표현과 관련 :

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.

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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

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++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!

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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++; =>
                                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: 38
System.out.println(i);         //output: 29

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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.

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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

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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.

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pre-increment and post increment are equivalent if not in an expression

int j =0;
int r=0         
for(int v = 0; v<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

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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) => 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) => 7 + 6.then a=7 is provided to post increment => 7 + 7 + 6 =20 and a =8.

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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 <br>
i=a++ + ++a + ++a;
//this means i= 5+7+8=20 (this could be complicated, 
//but its working like this),<br>
a=++a + ++a + a++;
//as a is 6+7+7=20 (this is incremented like this)

참고URL : https://stackoverflow.com/questions/2371118/how-do-the-post-increment-i-and-pre-increment-i-operators-work-in-java