CODEWITHSUNDEEP
c#
Salar Askar
Salar Askar

Reputation: 126

Why in C# the answer of (++x - --x) is always 1 for every value of x? Why not 0?

I don't understand why in C#, ++x - --x is 1 for every value of x.

int x = 0;

Console.WriteLine(++x - --x); // gives 1

I believe the answer should be 0, as variable x involves both sides. And x is incremented and decrimented before the processing of substraction. The same equation in C++ returns 0, that make sense to me.

Can anyone give a clue what's actually happening here?

Upvotes: 3

Views: 231

Answers (1)

paxdiablo
paxdiablo

Reputation: 882166

This is actually undefined behaviour in C and C++. For example, C states that reading and modifying an object without an intervening sequence point is undefined.

However, C# locks down the behaviour so that behaviour is defined. From the online docs:

Unrelated to operator precedence and associativity, operands in an expression are evaluated from left to right. The following examples demonstrate the order in which operators and operands are evaluated:

OPERAND EVALUATION
Expression Order of evaluation
a + b a, b, +
a + b * c a, b, c, *, +
a / b + c * d a, b, /, c, d, *, +
a / (b + c) * d a, b, c, +, /, d, *

And, from the operator precedence table, you can see that ++x (unary category) has a higher precedence than x + y (additive category).

In your case, the sub-expression ++x is evaluated first and results in x being pre-incremented so the value of that sub-expression is the original x plus one.

Then the sub-expression --x is evaluated and results in x (already incremented from the previous step) being pre-decremented so the value of that sub-expression is the original x.

And, since x + 1 - x == 1 for all but the darkest corners of the math universe, the result is always one. I still wouldn't do something like this, but it at least is well defined.

Upvotes: 8

Related Questions