Output Explanation C++

[ichihgo]

Int i ,x;I=4;X= ++i + ++i + ++i ;

Output = 19

int i ,x,y,z;i=4;x= ++i ;y= ++i ;z= ++i ;cout<<x+y+z;

Output = 18

 

why ?? 

[keja]

The C standard says that a variable should only be assigned at most once between two sequence points. A semi-colon for instance is a sequence point. 

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

and so on violate that rule. The standard also says that behavior is undefined and not unspecified. Some compilers do detect these and produce some result but this is not per standard.

 

http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1256.pdf @ 3.4.4

[Hatsune Miku 「 」]

nevermind.

[Ciccioo]

X = ++i + ++i + ++i;

this piece of code invokes undefined behaviour, so you're not guaranteed to get any consistent result

the problem is that you're modifying the value of i more than once in that single line, but the standard doesn't guarantee anything in doing so. apparently, i is incremented twice (to 6) and those values get summed (12), then i is incremented again (7) and added to the previous result (19)

read more

[Midnight]

It depends on the compiler. It's bad practice to add pre-increments anyway.

[Tigerbomb8]

This is that code in assembly