Who defines C operator precedence and associativity?

Question

Introduction

In every textbook on C/C++, you\'ll find an operator precedence and associativity table such as the following:

Answer 1

Precedence and associativity are defined in the standard, and they decide how to build the syntax tree. Precedence works by operator type(1+2*3 is 1+(2*3) and not (1+2)*3) and associativity works by operator position(1+2+3 is (1+2)+3 and not 1+(2+3)).

Order of evaluation is different - it does not define how to build the syntax tree - it defines how to evaluate the nodes of operators in the syntax tree. Order of evaluation is defined not to be defined - you can never rely on it because compilers are free to choose any order they see fit. This is done so compilers could try to optimize the code. The idea is that programmers write code that shouldn't be affected by order of evaluation, and yield the same results no matter the order.

Answer 2

Left-to-right associativity means that f() - g() - h() means (f() - g()) - h(), nothing more. Suppose f returns 1. Suppose g returns 2. Suppose h returns 3. Left-to-right associativity means the result is (1 - 2) - 3, or -4: a compiler is still permitted to first call g and h, that has nothing to do with associativity, but it is not allowed to give a result of 1 - (2 - 3), which would be something completely different.

Answer 3

One way to think about precedence and associativity is to imagine that the language only allows statements containing an assignment and one operator, rather than multiple operators. So a statement like:

a = f1() * f2() + f3();

would not be allowed, since it has 5 operators: 3 function calls, multiplication, and addition. In this simplified language, you would have to assign everything to temporaries and then combine them:

temp1 = f1();
temp2 = f2();
temp3 = temp1 * temp2;
temp4 = f3();
a = temp3 + temp4;

Associativity and precedence specify that the last two statements must be performed in that order, since multiplication has higher precedence than addition. But it doesn't specify the relative order of the first 3 statements; it would be just as valid to do:

temp4 = f3();
temp2 = f2();
temp1 = f1();
temp3 = temp1 * temp2;
a = temp3 + temp4;

sftrabbit gave an example where associativity of function call operators is relevant:

a = f()();

When simplifying it as above, this becomes:

temp = f();
a = temp();

Answer 4

Operator precedence is defined in the appropriate standard. The standards for C and C++ are the One True Definition of what exactly C and C++ are. So if you look closely, the details are there. In fact, the details are in the grammar of the language. For example, take a look at the grammar production rule for + and - in C++ (collectively, additive-expressions):

additive-expression:
  multiplicative-expression
  additive-expression + multiplicative-expression
  additive-expression - multiplicative-expression

As you can see, a multiplicative-expression is a subrule of an additive-expression. This means that if you have something like x + y * z, the y * z expression is a subexpression of x + y * z. This defines the precedence between these two operators.

We can also see that the left operand of an additive-expression expands to another additive-expression, which means that with x + y + z, x + y is a subexpression of it. This defines the associativity.

Associativity determines how adjacent uses of the same operator will be grouped. For example, + is left-to-right associative, which means that x + y + z will be grouped like so: (x + y) + z.

Don't mistake this for order of evaluation. There is absolutely no reason why the value of z could not be computed before x + y is. What matters is that it is x + y that is computed and not y + z.

For the function call operator, left-to-right associativity means that f()() (which could happen if f returned a function pointer, for example) is grouped like so: (f())() (of course, the other direction wouldn't make any sense).

Now let's consider the example you were looking at:

f1() + f2() * f3()

The * operator has higher precedence than the + operator, so the expressions are grouped like so:

f1() + (f2() * f3())

We don't even have to consider associativity here, because we don't have any of the same operator adjacent to each other.

Evaluation of the functions call expressions is, however, completely unsequenced. There's no reason f3 couldn't be called first, then f1, and then f2. The only requirement in this case is that operands of an operator are evaluated before the operator is. So that would mean f2 and f3 have to be called before the * is evaluated and the * must be evaluated and f1 must be called before the + is evaluated.

Some operators do, however, impose a sequencing on the evaluation of their operands. For example, in x || y, x is always evaluated before y. This allows for short-circuiting, where y does not need to be evaluated if x is known already to be true.

The order of evaluation was previously defined in C and C++ with the use of sequence points, and both have changed terminology to define things in terms of a sequenced before relationship. For more information, see Undefined Behaviour and Sequence Points.

Answer 5

The precedence of operators in the C Standard is indicated by the syntax.

(C99, 6.5p3) "The grouping of operators and operands is indicated by the syntax. 74)"

74) "The syntax specifies the precedence of operators in the evaluation of an expression"

C99 Rationale also says

"The rules of precedence are encoded into the syntactic rules for each operator."

and

"The rules of associativity are similarly encoded into the syntactic rules."

Also note that associativity has nothing to do with evaluation order. In:

f1() * f2() + f3()

function calls are evaluated in any order. The C syntactic rules says that f1() * f2() + f3() means (f1() * f2()) + f3() but the evaluation order of the operands in the expression is unspecified.