When i run the following example i get the output 0,2,1
class ZiggyTest2{
static int f1(int i) {
System.out.print(i + ",");
return 0;
}
public static void main(String[] args) {
int i = 0;
int j = 0;
j = i++; //After this statement j=0 i=1
j = j + f1(j); //After this statement j=0 i=1
i = i++ + f1(i); //i++ means i is now 2. The call f1(2) prints 2 but returns 0 so i=2 and j=0
System.out.println(i); //prints 2?
}
}
I don't understand why the output is 0,2,1 and not 0,2,2
i = i++ + f1(i);
i++ means i is now 2. The call f1(i) prints 2 but returns 0 so i=2 and j=0
before this i = 1 , now imagine f1() called and replaced with 0
so
i = i++ + 0;
now it would be
i = 1 + 0 // then it will increment i to 2 and then (1 +0) would be assigned back to `i`
In Simpler words (from here @ Piotr )
"i = i++" roughly translates to
int oldValue = i;
i = i + 1;
i = oldValue;
Another such Example :
If we expand i = i++ + f1(i) statement, we get something like following
save the value of i in a temp variable, say temp1 // temp1 is 1
increment i by one (i++) // i gets the value 2
execute f1(i), save return value in, say temp2 // temp2 is 0, print 2
assign temp1 + temp2 to i // i becomes 1 again
I guess main steps can be summarized like above.
The solution can be understood from this example
public static void main(String[] args) {
int i = 0;
i = i++;
System.out.println("i is" + i);
}
/* The output is "i is 0" */
Hence from this line,
i = i++ + f1(i);
Your i is still 1 and obviously the function is going to return 0. Which is again stored in i and hence the value 1. Instead of the updated value of i being stored in i, you are overriding it by the assignment operator.
In Post increment operator value will of operand will increase after use. Example
int k =1;
int l = k++;
System.out.println("...k..."+k+"...l.."+l);
First k (value = 1) is assigned with l after that the k value will increase
Similarly thing happens in the following line
i = i++ + f1(i);
Pre increment means: add one to variable and return incremented value; Post increment - first return i, then increment it;
int i, j, k;
i = 0; // 0
j = i++; // return i , then increment i
// j = 0; i = 1;
k = ++i; // first increment and return i
//k = 2; i = 2;
// now
++j == --k == --i // would be true => 1==1==1;
// but , using post increment would
// j++ == k-- == i-- // false because => 0 == 2 == 2;
// but after that statement j will be 1, k = 1, i = 1;
To get In-depth you need to see Expression and its Evaluation Order
Here's little explanation about equation i++ + f1(i) evaluation
In Equation Compiler get "i" which is Equals to 1 and put it on stack as first operand then increments "i", so its value would be 2, and calculates second operand by calling function which would be 0 and at the time of operation (+) execution operands would be 1 and 0.
Hope this explaination might be of some help :
j = i++; // Here since i is post incremented, so first i being 0 is assigned to j
// and after that assignment , i is incremented by 1 so i = 1 and j = 0.
i = i++ + f1(i); // here again since i is post incremented, it means, the initial value
// of i i.e. 1 as in step shown above is used first to solve the
// expression i = i(which is 1) + f1(1)**Since i is 1**
// after this step the value of i is incremented. so i now becomes 2
// which gets displayed in your last System.out.println(i) statement.
Do try this
i = ++i + f1(i); // here i will be first inremented and then that value will be used
// to solve the expression i = i + f1(i)'
So in short during post increment, expression is first solved and then value is incremented. But in pre increment, value is first incremented and then expression is solved.
But if you write only
i++;
++i;
Then both means the same thing.
Regards
来源：https://stackoverflow.com/questions/8615772/post-and-pre-increment-operators