| Summary: | pointer to function type mistaken when template functions are involved | ||
|---|---|---|---|
| Product: | gcc | Reporter: | Joaquín M López Muñoz <joaquin> |
| Component: | c++ | Assignee: | Not yet assigned to anyone <unassigned> |
| Status: | RESOLVED FIXED | ||
| Severity: | normal | CC: | gcc-bugs |
| Priority: | P3 | ||
| Version: | 3.2 | ||
| Target Milestone: | 3.3.3 | ||
| Host: | Target: | ||
| Build: | Known to work: | ||
| Known to fail: | Last reconfirmed: | ||
Fixed (at least) in 3.3.3. |
The following snippet: #include <iostream> void foo(){} template<typename T>void bar(){} template<typename T> void test(const T&) { std::cout<<typeid(T).name()<<std::endl; } int main() { test(&foo); test(&bar<int>); } produces this oputput on gcc version 3.2 20020927 (prerelease) under Cygwin: PFvvE FvvE But the two lines should be identical --in both cases, the deduced T should be void(*)(). Although it is not immediately apparent in the testcase provided, what happens is that the second type is taken as void(), i.e. the pointer somehow is dropped. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo