Checking If I understood this concept right (associativity)

Question

I have the following expression in C. I'm asked whether we can predict the value of z

int x,y,z;
z= (x=2) + (y=x)

I know that the () operator has left associativity. Does this mean that the left parenthesis will be evaluated before the ones on the right?

I'm trying to understand if I got this concept correctly

Thanks in advance

not an expert but the expression seems to miss a sequence point. — Jean-François Fabre, Mar 09 '18 at 20:01
I'm not sure parentheses have associativity in the first place. — ForceBru, Mar 09 '18 at 20:02
There's an associativity/precedence table. Parentheses have the highest level of precedence and left-to-right associativity — Francisco José Letterio, Mar 09 '18 at 20:03
why would you bother to try to figure out what is the output of such an horrible code? Shoot the guy who wrote it, and rewrite it properly. — Jean-François Fabre, Mar 09 '18 at 20:09
Because ot's an exercise to understand how the compiler works — Francisco José Letterio, Mar 09 '18 at 20:10
Good programming avoids the compiler having to make any such decisions. Such code is of negative use to future SO users/visitors.. — Martin James, Mar 09 '18 at 20:26
Parentheses in a precedence table represent the *function call operator*, which does have a meaningful precedence. The high precedence of `()` is why `*f()` means `*(f())` instead of `(*f)()` — , Mar 09 '18 at 20:50
@Francisco - The only thing to learn here is never to do this again. *"You can never guess what this code does"* is the opposite of what you do as a professional. — Bo Persson, Mar 10 '18 at 01:28

score 0 · Answer 1 · answered Mar 09 '18 at 20:01

0

No. Associativity says that x + (y + z) == (x + y) + z == x + y + z; the parentheses don't matter when chaining 2 or more of the same operator.

In your code, the compiler is free to evaluate the operands of + in either order. You don't know if the result will be

x=2; y=x; z = x + y;

or

y=x; x = 2; z + x + y;

answered Mar 09 '18 at 20:01

chepner

But that's the associativity of the addition, not the parentheses, isn't it? – Francisco José Letterio Mar 09 '18 at 20:04
No, parentheses aren't operators. They group expressions to modify *precedence*, since `z = x = 2 + y = x` would otherwise be parsed as ... I'm not even sure off the top of my head. Maybe `z = (x = (2 + (y = x)))`? – chepner Mar 09 '18 at 20:08

1 Answers1