Fix #15023 FN redundantInitialization with braces or parentheses

[chrchr-github]

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[danmar]

This is an AI review. Take it with a grain of salt and feel free to reject it by resolving the comment.

For the new ( / { initialization forms, tok->str() is ( or {, so this looks for operator( / operator{ and the custom-assignment-operator check never triggers. Result with this PR:

int g();
struct A { A(int); A& operator=(int); A& operator=(const A&); };
void f1() { A a(g()); a = 1; }   // redundantInitialization (conclusive)
void f2() { A a = g(); a = 1; }  // redundantInitialization (inconclusive)

I'd expect both to be inconclusive. What matters is the later a = 1, so maybe:

Suggested change

	const std::string op = "operator" + tok->str();
	const std::string op = isInitialization && Token::Match(tok, "[{(]") ? "operator=" : "operator" + tok->str();

With that, f1 becomes inconclusive like f2, P p{ g(), g() }; p = P{1, 2}; for a plain struct is still reported conclusively, and TestOther passes.

FYI, this PR and #8847 restructure the same block in checkRedundantAssignment(), so whichever is merged second will need a rebase.

[chrchr-github]

• "operator" + tok->str();should always be "operator="
• the inconclusive logic is flawed already, there is a conclusive warning for S f() { S s = g(); s = 1; return s; }