Is inserting map size as value into a map undefined behavior?

Question

edit: this question should have not been closed, if you look at the answers you will see they are totally different(old question has no mention of C++17).

I was reading a PVS blog post where they mention the following bug.

(reduced)

std::map<int,int> m;
m[7]=5;
auto val = 15;
if (!m.contains(val)){
    m[val] = m.size(); // bug here
}

According to blog post this is buggy. I always thought that operator [] call for map is a function call so .size() is sequenced before [] because functions act as sequence point.

So why is this a bug?

note: I know sequence points do not exist since C++11, but I use them since new wording is much harder for me to understand.

Answer 1

Pre C++17

§ 1.9 Program execution [intro.execution] (n3690 c++14 draft)

15. Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.

and 5.17 [expr.ass] doesn't mention any sequencing between the operands of built-in assignment. So the evaluation of the two operands of the built-in assignment operator = is unsequenced with regards to each other.

m[val] and m.size() can be evaluated in any order (can even overlap - interleaved the CPU instructions).

Considering:

• m[val] has a side effect of modifying the map's size (a scalar)

• the value computation of m.size() accesses the map's size

§ 1.9 Program execution [intro.execution] (n3690 c++14 draft)

15. [...] If a side effect on a scalar object is unsequenced relative to either [...] or a value computation using the value of the same scalar object, the behavior is undefined.

So yes, the behavior is indeed Undefined.

C++17

§8.5.18 Assignment and compound assignment operators [expr.ass] (n4713 C++17 draft)

1. The assignment operator (=) [...] The right operand is sequenced before the left operand.

So the behavior is defined. m.size() will be evaluated before m[val]