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> As far as I understand it, this limitation is also the only thing preventing tail-call elimination in C/C++. Nothing "prevents" tail-call elimination in C. C

> As far as I understand it, this limitation is also the only thing preventing tail-call elimination in C/C++. Nothing "prevents" tail-call elimination in C. Compilers do it all the time. > Tail-recursive functions which are only used internally can also be optimised to re-use the same stack frame and iterate with a conditional jump. If such functions are also exposed in the interface, it's pretty trivial to expose a dummy entry point which invokes the optimised version (if we want to) Can you illustrate with a concrete example what you mean? If the function doesn't need a stack frame, there will be no stack frame either way. If the function needs a stack frame, that stack frame must be allocated in the function's start block either way, and tail recursive calls will of course not jump to that start block but to its successor. I don't see how being "exposed in the interface" (in C speak, static vs. non-static) would make a difference here.

heavenlyblue 6y ago
The issue is basically that tail-recursive functions need to have this information recorded in their signature (for external interfaces), but this is not recorded in C and thus you can't provide this guarantee at the interface boundaries.
- tom_mellior 6y ago
  This is not concrete in any way. What guarantee do you mean? What "this information" should be "recorded in their signature"? What does this murky "information" have to do with enabling tail recursion? Here's a tail recursive function C function that is compiled to tail recursive machine code without any problems: https://gcc.godbolt.org/z/dc886a https://gcc.godbolt.org/z/dc886a This function can be called from external compilation units without any problems. Whether or not it is compiled to tail recursive code is an implementation detail that the caller doesn't need to know about.

heavenlyblue 6y ago

The issue is basically that tail-recursive functions need to have this information recorded in their signature (for external interfaces), but this is not recorded in C and thus you can't provide this guarantee at the interface boundaries.

tom_mellior 6y ago
This is not concrete in any way. What guarantee do you mean? What "this information" should be "recorded in their signature"? What does this murky "information" have to do with enabling tail recursion? Here's a tail recursive function C function that is compiled to tail recursive machine code without any problems: https://gcc.godbolt.org/z/dc886a https://gcc.godbolt.org/z/dc886a This function can be called from external compilation units without any problems. Whether or not it is compiled to tail recursive code is an implementation detail that the caller doesn't need to know about.