confused with operators ++ --

[o3hC5Di1]

Thanks Disch for your elaborate explanation - helped me understand the issue too.

All the best,

NwN

[Kronolynx]

--a*b(original value)/b(original value): 5/5
That 2nd 'b' may or may not be the original value. It may be the incremented value. This behavior is not defined by the C++ standard and you must never do this in actual code.

@disch i have one more question. how should it be done if i want the 2nd 'b' to keep the original value ?

thanks again for the help.

[Peter87]

Just split it on more lines

c=--a*b/b;
++b;

It's probably a good idea to always avoid side effects in the expressions to be honest. Trying to do everything in one line is often bad for readability. Much easier to follow:

--a;
c=a*b/b;
++b;

[Kronolynx]

thanks you very much :)