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Deducing return type of map generator helper (#1576)

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* Deduce map return type implicitly

Giving the first template argument to map generator function to deduce
return type is now optional even if the return type is different from
the type generated by mapped generator.
This commit is contained in:
Omer Ozarslan
2019-03-24 09:44:22 -05:00
committed by Martin Hořeňovský
parent 296d447452
commit 54089c4c8c
9 changed files with 220 additions and 14 deletions
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5
projects/SelfTest/UsageTests/Generators.tests.cpp
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@@ -144,6 +144,11 @@ TEST_CASE("Generators -- adapters", "[generators][generic]") {
auto i = GENERATE(map<std::string>([] (int val) { return std::to_string(val); }, values({ 1, 2, 3 })));
REQUIRE(i.size() == 1);
}
SECTION("Different deduced type") {
// This takes a generator that returns ints and maps them into strings
auto i = GENERATE(map([] (int val) { return std::to_string(val); }, values({ 1, 2, 3 })));
REQUIRE(i.size() == 1);
}
}
SECTION("Repeating a generator") {
// This will return values [1, 2, 3, 1, 2, 3]