What can be a subexpression? - c++

Question

I read "C++.Primer plus. Stephen Prata"(6th edition).
On page 209 was:
y = (4 + x++) + (6 + x++);
The expression 4 + x++ is not a full
expression, so C++ does not guarantee that x will be incremented
immediately after the subexpression 4 + x++ is evaluated. Here the
full expression is the entire assignment statement, and the semicolon
marks the sequence point, so all that C++ guarantees is that x will
have been incremented twice by the time the program moves to the
following statement. C++ does not specify whether x is incremented
after each subexpression is evaluated or only after all the
expressions have been evaluated, which is why you should avoid
statements of this kind.
And I read "Sequence Points and Expression Evaluation" Visual Systems Journal, August 2002. Klaus Kreft & Angelika Langer.
There was:
x[i]=i++ + 1;
Let's assume variable i has the value 1 before we enter the statement.
What will be the result of evaluation of this expression? The correct
answer is: we don't know. However, programmers ever too often believe
that they know what this program fragment does. Typical answers
include: "x[1] will have the value 2", or "x[2] will have the value
2", or even "x[1] will have the value 3".
The third option is definitely wrong. This will not happen because i++
is a postfix increment and returns i's initial value 1; hence the
value of the right hand side of the assignment is 2, and definitely
not 3. [...] So far so good, but we do not know
which entry of the array x will be modified. Will the index be 1 or 2
when the right hand side value will be assigned to x[i]?
There is no definite answer to this question. It fully depends on the
order in which the compiler evaluates the subexpressions. If the
compiler starts on the right hand side of the assignment and evaluates
i++ + 1 before it figures out which position in the array x must be
assigned to then x[2] will be modified because i will have already
been incremented in the course of evaluating the subexpression i++.
Conversely, if the compiler starts on the left hand side and figures
out that it must assign to position i in array x, which at that time
will still be position 1, before it evaluates the right hand side then
we'll end up with a modification of x[1]. Both outcomes are equally
likely and equally correct. "
How understand where is subexpression?
4 + x++ and 6 + x++ are subexpressions, because they are into round brackets?
x[i] and i++ + 1 are subexpression? Why?
I'm interested in this, because I want to understand where side effect can happened in the hypothesis.

Answer 1

Breaking this down, the line
y = (4 + x++) + (6 + x++);
is an expression-statement. Such a thing consists of an expression followed by a ;, so
y = (4 + x++) + (6 + x++)
is an expression.
Since this expression is not part of another expression (but only of an expression-statement), it is a full-expression. A sub-expression on the other hand is an expression that is part of another expression. In the following, I will use capital letters to name expressions, rather than C++ identifiers.
The full-expression above is an assignment-expression of the form:
y = A
where A is the remaining additive-expression
(4 + x++) + (6 + x++)
An additive expression is of the form X + Y, so we break this down into two expressions
(4 + x++)
(6 + x++)
The first one consists of an expression of the form (Z), where Z is 4 + x++. And 4 + x++ consists of two expressions 4 and x++. And so on. All of these expressions are part of
y = (4 + x++) + (6 + x++)
and hence they are sub-expressions of the above expression.

Answer 2

What can be a subexpression?
Any expression can be a subexpression. Although, some expressions may not be subexpressions of certain other expressions.
4 + x++ and 6 + x++ are subexpressions
Correct. Both of those are arithmetic expressions, additions to be more specific.
because they are into round brackets?
Well, sort of. Being inside parentheses, they are indeed the subexpression of the parenthesized expression.
† In general, they're subexpressions because they're expressions, but are also part of another expression.
x[i] and i++ + 1 are subexpression? Why?
Yes, they are. See †.
Here is a handy list of all possible expressions in c++.
Let's find the subexpressions in y = (4 + x++) + (6 + x++);. The first expression that has no subexpressions is 4. It is a literal. It is a subexpression of 4 + x++ which is an addition. Additions have the form A + B. In this case, subexpression A is 4 an subexpressiond B is x++, which is a post increment. Oh, but that contains a subexpression too: x. It is an identifier, and contains no subexpressions. 4 + x++ is a subexpression of the parenthesized expression (4 + x++). That is a subexpression of (4 + x++) + (6 + x++) which is a subexpression of y = (4 + x++) + (6 + x++); which is an assignment. The assignement is a full-expression - not a subexpression. I left some of the subexpressions unexplored, and I shall leave them as an exercise for the reader.