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Operator precedence and Associativity in C/C++

Please note, that this has nothing to do with Operator Precedence.. () and ++ , Undefined behavior and sequence points , Why are these constructs (using ++) undefined behavior? and the hundreds similar questions about this here


Shortly: is the Associativity guaranteed by the standard?

Detailed example: from Wikipedia's article for operator precedence, operator* and operator/ have the same priority and they are Left-to-right operators. Does this mean, that the standard guarantees, that this:

int res = x / y * z / t;

will be evaluated as

int res = ( ( x / y ) * z ) / t;

or it's implementation defined?

If it's guaranteed, could you quote?


It's just out of curiosity, I always write brackets in these cases.
Ready to delete the question, if there's such one.

like image 793
Kiril Kirov Avatar asked Dec 20 '22 17:12

Kiril Kirov


1 Answers

From the latest publicly available draft

5.6 Multiplicative operators [expr.mul]

1 The multiplicative operators *, /, and % group left-to-right.

multiplicative-expression:
pm-expression
multiplicative-expression * pm-expression
multiplicative-expression / pm-expression
multiplicative-expression % pm-expression

So parsing will go like:

int res = x / y * z / t;
int res = (x / y * z) / t;
int res = ((x / y) * z) / t;
like image 138
TemplateRex Avatar answered Dec 24 '22 00:12

TemplateRex