Question about conditional operators ( || vs |)

**patishi** · 10-06-2013

I have a little question regarding this..
I will give an example: let's say i have this condition

Code:

int x = 5;
if(x ==2 || x ==5 || x ==1 || x ==8 || ....){
     return 1;
}

Here I pretty much sure (correct me if I am wrong) that the code will stop at: x == 5 and will return 1, and will not continue to check all the rest of the numbers.

But when i use the operator | instead, like this:

Code:

unsigned short y = 316; // sorry for not using Hex number 
unsigned short mask = 1;
if(((y <<1) & mask) | ((y << 2)& mask) | ((y >> 1) & mask) ...){
     return 1;
}

let's assume that the first check will result in 1 (e.g. true) will the code stop and return 1 immediately like in the first example or it will continue to check until the end?
Thx!

**Salem** · 10-06-2013

There is no short-circuit left-to-right evaluation with the | operator.
The compiler has to evaluate the whole expression, and it's free to do so in any order it chooses that is compatible with the precedence of the operators involved.

So for example, the sub-expression ((y >> 1) & mask) may be evaluated before ((y << 2)& mask)

Your analysis of the || case is correct.

**laserlight** · 10-06-2013

Originally Posted by patishi

Here I pretty much sure (correct me if I am wrong) that the code will stop at: x == 5 and will return 1, and will not continue to check all the rest of the numbers.

Yes, you are correct.

Originally Posted by patishi

let's assume that the first check will result in 1 (e.g. true) will the code stop and return 1 immediately like in the first example or it will continue to check until the end?

It will continue to check until the end. Note that the order of evaluation is unspecified, so the "first check" could well be ((y << 2) & mask) and "the end" could be ((y << 1) & mask), or perhaps all the shifts are evaluated first.

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