0

The following two lines of code are not returning the same value. Any reason for that?

int i;

i = 1;
i = i + i++; //Returns 2, expecting 3

And

i = 1;
i = i++ + i; //Returns 3

Semantically, this should be the same a + b = b + a right?

The same with decreasing i:

i = 1;
i = i - i--; //Returns 0, expecting 1

And

i = 1;
i = i-- - i; //Returns 1, expecting -1

What confuses me even more is the usage of post increment operators:

i = 1;
i = i + ++i; //Returns 3

And

i = 1;
i = ++i + i; //Returns 4, expecting 3

Same again with decreasing operator:

i = 1;
i = i - --i; //Returns 1

And

i = 1;
i = --i - i; //Returns 0, expecting -1

Last Question:

How are these two lines interpreted by the compiler?

i = i+++i; // is it i + ++i or i++ + i?
i = i---i; // is it i - --i or i-- - i?
Moslem Ben Dhaou
  • 6,897
  • 8
  • 62
  • 93
  • 2
    Almost a duplicate of http://stackoverflow.com/q/13516689/11683 and http://stackoverflow.com/a/3346729/11683 (last one is from Eric Lippert, must read). – GSerg Oct 07 '13 at 10:02
  • `i = 1; i = i + i++` Why do you expect `3` here? – Sriram Sakthivel Oct 07 '13 at 10:03
  • This just goes to show that you should never use `++` and/or `--` with a variable which is also being used elsewhere in the same expression. It's too difficult to read. – Matthew Watson Oct 07 '13 at 10:19
  • possible duplicate of [C# Pre- & Post Increment confusions](http://stackoverflow.com/questions/8573190/c-sharp-pre-post-increment-confusions) – nawfal Jul 20 '14 at 08:10

4 Answers4

2

There is a difference between pre-increment (++i) and post-increment (i++). The difference is:

Pre-increment will add the value before using the result. Post-increment will use the reslt.. then add the value. So, your first example:

int i;

i = 1;
i = i + i++; // First use equals 1, second use equals one. After this line though
             // i equals 2, because of your use of post-increment.

Likewise, your second example:

i = 1;
i = i++ + i; // First use is 1. After the first use.. it is incremented..
             // The second use it is 2. Therefore, 1 + 2 == 3.

As for your last question... why not put it into a console application and try it yourself?

Simon Whitehead
  • 63,300
  • 9
  • 114
  • 138
2
i = i + i++; //Returns 2, expecting 3

Know as post increment. Value will be used first and then incremented. It is equivalent to

i = i + i;
i = i+1;

and this is pre-increment. Value will be incremented first and then used.

i = i++ + i; //Returns 3

is equivalent to

i = i+1;
i = i + i;

i = i+++i; // is it i + ++i or i++ + i?

is interpretted as

i = i + 1; i = i + i;

and this

i = i---i; // is it i - --i or i-- - i?

is interpretted as

i= i-1;
i = i-i;
Ehsan
  • 31,833
  • 6
  • 56
  • 65
0

Precedence of increament or decreament operator is always higher than arithmetic operators like + - * /
Refer http://msdn.microsoft.com/en-us/library/aa691323(v=vs.71).aspx for details. Also the value of postfixed increament or decreament operator is effective only after the current statement. For eg:

i = i + i++ = 1 + 1++ = 1 + 1 = 2; // value of i is effective after increament is done
i = i++ + i = 1++ + 2 = 3; //Next value of i contains updated value i.e. 2
i = i + ++i = 1 + ++1 = 1 + 2 = 3; //Next value of i contains updated value i.e. 2
i = ++i + i = ++1 + i = 2 + 2; //Both value of i contains updated value i.e. 2 because of prefixed operator
i = i+++i = i++ + i;
i = i---i = i-- - i;
Adarsh Kumar
  • 1,134
  • 7
  • 21
0

I think the best way to try to understand this, is look at what the compiler makes of this.

See here for an overview of this for the case 

x=i-- - --i;

https://stackoverflow.com/a/8573429/959028

best regards

Community
  • 1
  • 1
Wim Reymen
  • 190
  • 2
  • 4