fix: passing ResultError to Error thru operator<<
It is typical to pass error to callers like following;
if (!result) {
return Error() << result.error();
}
To transfer errno(or ResultError#code()), Error defines a specialization
operator<<(const ResultError&).
This change fixes so that ResultError is properly handled
Bug: 132145659
Test: atest libbase_test
Change-Id: Ib35457da2d4b923d8e652c54ac510a75546cf918
diff --git a/include/android-base/result.h b/include/android-base/result.h
index 897c48f..4a8e1ef 100644
--- a/include/android-base/result.h
+++ b/include/android-base/result.h
@@ -81,8 +81,7 @@
struct ResultError {
template <typename T>
- ResultError(T&& message, int code)
- : message_(std::forward<T>(message)), code_(code) {}
+ ResultError(T&& message, int code) : message_(std::forward<T>(message)), code_(code) {}
template <typename T>
operator android::base::expected<T, ResultError>() {
@@ -122,18 +121,16 @@
template <typename T>
Error& operator<<(T&& t) {
+ if constexpr (std::is_same_v<std::remove_cv_t<std::remove_reference_t<T>>, ResultError>) {
+ errno_ = t.code();
+ return (*this) << t.message();
+ }
int saved = errno;
ss_ << t;
errno = saved;
return *this;
}
- Error& operator<<(const ResultError& result_error) {
- (*this) << result_error.message();
- errno_ = result_error.code();
- return *this;
- }
-
const std::string str() const {
std::string str = ss_.str();
if (append_errno_) {