Why is it clear that a template function instantiation will not be inlined?

manpreet Best Answer 2 years ago

_x000D_ What he's saying (I think) is that the add function is not getting inlined. In other words, the compiler might inline do_op like this: int c = func_ptr(4, 5); but it won't also inline add like this: int c = 4 + 5; However, he might be wrong in that simple example. Generally, when you call a function through a pointer, the compiler can't know (at compile-time) what function you'll be calling, so it can't inline the function. Example: void f1() { ... } void f2() { ... } void callThroughPointer() { int i = arc4random_uniform(2); void (*f)() = i ? f2 : f1; f(); } Here, the compiler cannot know whether callThroughPointer will call f1 or f2, so there is no way for it to inline either f1 or f2 in callThroughPointer. However, if the compiler can prove at compile-time what function will be called, it is allowed to inline the function. Example: void f1() { ... } void f2() { ... } void callThroughPointer2() { int i = arc4random_uniform(2); void (*f)() = i ? f2 : f1; f = f1; f(); } Here, the compiler can prove that f will always be f1, so it's allowed to inline f1 into callThroughPointer2. (That doesn't mean it will inline f1…) Similarly, in the example you quoted in your post, the compiler can prove that func_ptr is always add in the call to do_op, so it's allowed to inline add. (That doesn't mean it will inline add…)