下标是否在赋值运算符的右侧之前计算

Question

I'm trying to do something like

arr[getchar()-'a'] = getchar();

is it guaranteed that the subscript here evaluates before anything else?

Answer 1

Before C++17, no guaranty of order

Since C++17: (from C++'s order of evaluation )

17) In a subscript expression E1[E2], every value computation and side-effect of E1 is sequenced before every value computation and side effect of E2

20) In every simple assignment expression E1=E2 and every compound assignment expression E1@=E2, every value computation and side-effect of E2 is sequenced before every value computation and side effect of E1

So in your case:

   arr[getchar()-'a'] = getchar();
// (2)  (3)              (1)

Answer 2

No, it's not guaranteed. (at least for pre-C++17 code, as others mentioned ).

As by the order of evaluation rules (emphasis mine):

Order of evaluation of any part of any expression, including order of evaluation of function arguments is unspecified [...]. The compiler can evaluate operands and other subexpressions in any order, and may choose another order when the same expression is evaluated again.

There is no concept of left-to-right or right-to-left evaluation in C++. This is not to be confused with left-to-right and right-to-left associativity of operators: the expression a() + b() + c() is parsed as (a() + b()) + c() due to left-to-right associativity of operator+, but the function call to c may be evaluated first, last, or between a() or b() at run time

I'm quoting the C++ reference here, but the same holds for C .