Disable unwanted overload for tolower

Question

While researching, I found Why “transform(s.begin(),s.end(),s.begin(),tolower)” can't be complied successfully? , and the solution is to use `::tolower`. As an exercise, I want to see if I can disable that (in locale header file) overload and use `std::tolower`. First, we start simple: ``` template charT tolower(charT c, std::locale const& loc) = delete; int main() { std::vector chars = { 'a', 'D', 'c', 'b' }; std::transform(chars.begin(), chars.end(), chars.begin(), std::tolower); } ``` The compiler cannot deduce the correct overload. So we provide the char template: ``` std::transform(chars.begin(), chars.end(), chars.begin(), std::tolower); ``` So we're back to square one. I know it is undefined behavior, but I thought I could put it into namespace std: ``` namespace std { template charT tolower(charT c, std::locale const& loc) = delete; } ``` Now the compiler says I'm redefining the function. ``` using std::tolower; template

Accepted Answer

You can't disable the other overload, but there are lots of ways to ensure the correct one is chosen:

```
int (*lower)(int) = std::tolower;
std::transform(chars.begin(), chars.end(), chars.begin(), lower);
```

or

```
std::transform(chars.begin(), chars.end(), chars.begin(), static_cast<int(*)(int)>(std::tolower));
```

or

```
std::transform(chars.begin(), chars.end(), chars.begin(), [](int c) { return std::tolower(c); });
```

or

```
struct ToLower {
  int operator()(int c) const { return std::tolower(c); }
};
std::transform(chars.begin(), chars.end(), chars.begin(), ToLower{});
```

The first two forms tell the compiler how to choose the right overload from the overload set, because only one of the overloads can be converted to a function pointer of that type.

The second two forms create a wrapper that invokes `tolower` so that normal overload resolution finds the right one, and you don't need to take the address of `tolower`.

Disable unwanted overload for tolower

Solution

Problem

Related problems