可以将文章内容翻译成中文,广告屏蔽插件会导致该功能失效:

问题:

I understand the difference between this operations (post-increment, pre-increment). But question: I have an expression:

int x = 4;

long y = x * 4 - x++;

The highest priority has post-unary operator, than "*" and last "-". In my opinion it will be:

long y = x * 4 - x++;

1). x++ => return 4 (save x = 5)

2). final expression: 5 * 4 - 4 = 16

But when I compile this in IDE the answer is 12 ! What's a problem and where did I do smth. wrong?

回答1:

EDITED

Here are the steps to evaluate the expression:

Evaluate all the expression that has increment/decrement from left to right

Replace all variable with their true value

Evaluate the expression using PEMDAS rule

Example:

int x = 4;

int y = ++x * 5 / x-- + --x;

first we need to substitute all values of x before evaluating the expression (substitute from left to right)

++x -> since it is a post increment we will first increment x before substituting, thus x will be 5

5 * 5 / x-- + --x -> this will be the new equation

now we will substitute x in x--

x-- -> since it is a post decrement x here will be substituted with 5 and after the substitution decrement x, thus x will be 4

5 * 5 / 5 + --x -> this will be the new equation

now we will substitute x in --x

--x -> since it is a pre decrement we will first decrement x substituting, thus x will be 3

5 * 5 / 5 + 3 // this will be the new equation

Since there are no variables in the equation we will now evaluate the expression using PEMDAS

25 / 5 + 3 5 + 3

8

thus the result will be 8

回答2:

x evaluates to 4, and x++ evaluates to 4 as well, and then gets incremented to 5. So it's essentially 4 * 4 - 4 , which gives 12 as expected.

回答3:

Multiplication is resolved separately, so x == 4. Incrementation occurs afterwards, so it's actually 4 * 4 - 4 == 12 and after this operation x == 5.

回答4:

The left side of a subtraction is always computed before the right side because subtraction is left associative i.e. 5 - 2 - 1 is (5 - 2) - 1, not 5 - (2 - 1).

This is why the multiplication happens before the increment. Associativity determines what happens first here, not precedence.

回答5:

long y = 4 * 4 - 4;

x will be incremented after this assignment

Everytime x is "called" with x++ it gets incremented after E.g.

int x = 1;

System.out.println(x++ + x++ * x++ + x++);

// 1 + 2 * 3 + 4