Const-qualify operator()s. NFC. (#1638)

As a general rule, `operator()` should be const-qualified.
The exceptions are few and far between.

I was actually looking for a Guideline on that subject;
I grepped for `operator()` and found that not only is there
no such Guideline yet, the doc actually contained these
places that (needlessly) violated the general rule.
pull/1645/head
Arthur O'Dwyer 6 years ago committed by GitHub
parent c379d3f6ea
commit 567331ab4e
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@ -6550,7 +6550,7 @@ It's a standard-library requirement.
using result_type = size_t;
using argument_type = My_type;
size_t operator() (const My_type & x) const
size_t operator()(const My_type & x) const
{
size_t xs = x.s.size();
if (xs < 4) throw Bad_My_type{}; // "Nobody expects the Spanish inquisition!"
@ -14523,7 +14523,7 @@ It is harder to ensure absence of errors in detached threads (and potentially de
void f() { std::cout << "Hello "; }
struct F {
void operator()() { std::cout << "parallel world "; }
void operator()() const { std::cout << "parallel world "; }
};
int main()
@ -14537,7 +14537,7 @@ It is harder to ensure absence of errors in detached threads (and potentially de
void f() { std::cout << "Hello "; }
struct F {
void operator()() { std::cout << "parallel world "; }
void operator()() const { std::cout << "parallel world "; }
};
int main()

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