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c#operatorsincrementdecrement
Moslem Ben Dhaou
Moslem Ben Dhaou

Reputation: 7005

Incrementing and decreasing i is not working properly

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?

Upvotes: 1

Views: 3369

Answers (4)

Adarsh Kumar
Adarsh Kumar

Reputation: 1150

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;

Upvotes: 0

Wim Reymen
Wim Reymen

Reputation: 200

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

Upvotes: 0

Ehsan
Ehsan

Reputation: 32661

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;

Upvotes: 3

Simon Whitehead
Simon Whitehead

Reputation: 65049

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?

Upvotes: 2

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