Java x-- 或 x++ 在这里做什么？
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what does x-- or x++ do here?
提问by Nana
it is a silly Q for most of u - i know - but i one of the beginner here, and I can not understand why the output in here are 12 what does this (x--) do to the result ?
对于大多数人来说，这是一个愚蠢的 Q - 我知道 - 但我是这里的初学者之一，我不明白为什么这里的输出是 12 这 ( x--) 对结果有什么影响？
int x, y;
x = 7;
x-- ;
y = x * 2;
x = 3;
采纳答案by Jaydee
Just a cautionary note, sometimes the pre and post increment operators can have unexpected results
只是一个警告，有时前后增量运算符可能会产生意想不到的结果
Why does this go into an infinite loop?
and what does:
什么是：
x[i]=i++ + 1;
do
做
Read here: http://www.angelikalanger.com/Articles/VSJ/SequencePoints/SequencePoints.html
在这里阅读：http: //www.angelikalanger.com/Articles/VSJ/SequencePoints/SequencePoints.html
回答by NimChimpsky
x--subtracts/decrements the value of x by one.
x--将 x 的值减/减一。
Conversley x++adds/increments by one.
Conversleyx++加一/递增。
The plus or minus signs can either be before (--x) or after (x--) the variable name, prefixand postfix. If used in a expression the prefix will return the value after operation has been performed and the postfix will return the value before operation has been performed.
加号或减号可以在 ( --x)之前或之后 ( x--) 变量名、前缀和后缀。如果在表达式中使用，前缀将在操作执行后返回值，后缀将在操作执行前返回值。
int x = 0;
int y = 0;
y = ++x; // y=1, x=1
int x = 0;
int y = 0;
y = x++;// y=0, x=1
回答by darioo
x--will decrement value of xby 1. It is a postfix decrement operator, --xis a prefix decrement operator.
x--将值减x1。它是一个后缀自减运算符，--x是一个前缀自减运算符。
So, what's going on here?
那么，这里发生了什么？
int x, y; //initialize x and y
x = 7; //set x to value 7
x--; //x is decremented by 1, so it becomes 6
y = x * 2; //y becomes 6*2, therefore y becomes 12
x = 3; //x becomes 3
By analogy, the ++will increase a value by 1. It also has a prefix and postfix variant.
以此类推，the++会将值增加 1。它也有前缀和后缀变体。
回答by Nicolas Repiquet
x++ increments x after x being evaluated. ++x increments x before x being evaluated.
x++ 在计算 x 后增加 x。++x 在计算 x 之前增加 x。
int x = 0;
print(++x); // prints 1
print(x); // prints 1
int y = 0;
print(y++); // prints 0
print(y); // prints 1
The same goes for --
同样适用于——
回答by pyCoder
Example:
例子：
x = 7;
y = --x; /* prefix -- */
Here y = 6 (--x reduce x by 1)
这里 y = 6 (--x 将 x 减 1)
y = x--; /* postfix -- */
Here y = 6 (x-- use first the value of x in the expression and then reduce x by 1)
这里 y = 6 (x--首先使用表达式中 x 的值，然后将 x 减 1)
回答by Buhake Sindi
x++is essentially x = x + 1(the same applies for ++x). xis incremented by 1.
x++本质上是x = x + 1（同样适用于++x）。x增加 1。
x--is essentially x = x - 1(the same applies for --x). xis decremented by 1.
x--本质上是x = x - 1（同样适用于--x）。x减 1。
The difference is that how x++and ++xis used in the statement/expression: In ++x, xis incremented by 1 firstbefore being used while in x++, xis used (before incrementation) firstand once it's used, it gets incremented by 1.
不同的是，如何x++和++x在语句/表达被用于：在++x，x递增1第一，同时在被使用之前x++，x使用（增量之前）第一和一旦它的使用，它就会被加1。
回答by Powertieke
--is the 'decrement' operator. It simply means that the variable it operates on (in this case the xvariable) gets is decremented by 1.
--是“递减”运算符。它只是意味着它所操作的变量（在本例中为x变量）减 1。
Basically it is shorthand for :
基本上它是以下的简写：
x = x - 1;
So what the code does :
那么代码的作用是：
int x,y ; # Define two variables that will hold an integer
x=7; # Set variable X to value 7
x-- ; # Decrement x by one : so x equals 7 - 1 = 6
y= x * 2; # Multiply x by two and set the result to the y variable: 6 times 2 equals 12
x=3; # set x to value 3 (I do not know why this is here).
