This might be a stupid question, but I am confused. I had a feeling that an immediate (consteval) function has to be executed during compile time and we simply cannot see its body in the binary.
This article clearly supports my feeling:
This has the implication that the [immediate] function is only seen at compile time. Symbols are not emitted for the function, you cannot take the address of such a function, and tools such as debuggers will not be able to show them. In this matter, immediate functions are similar to macros.
The similar strong claim might be found in Herb Sutter's publication:
Note that draft C++20 already contains part of the first round of reflection-related work to land in the standard: consteval functions that are guaranteed to run at compile time, which came from the reflection work and are designed specifically to be used to manipulate reflection information.
However, there is a number of evidences that are not so clear about this fact.
From cppreference:
consteval - specifies that a function is an immediate function, that is, every call to the function must produce a compile-time constant.
It does not mean it has to be called during compile time only.
From the P1073R3 proposal:
There is now general agreement that future language support for reflection should use constexpr functions, but since "reflection functions" typically have to be evaluated at compile time, they will in fact likely be immediate functions.
Seems like this means what I think, but still it is not clearly said. From the same proposal:
Sometimes, however, we want to express that a function should always produce a constant when called (directly or indirectly), and a non-constant result should produce an error.
Again, this does not mean the function has to be evaluated during compile time only.
From this answer:
your code must produce a compile time constant expression. But a compile time constant expression is not an observable property in the context where you used it, and there are no side effects to doing it at link or even run time! And under as-if there is nothing preventing that
Finally, there is a live demo, where consteval function is clearly called during runtime. However, I hope this is due to the fact consteval is not yet properly supported in clang and the behavior is actually incorrect, just like in Why does a consteval function allow undefined behavior?
To be more precise, I'd like to hear which of the following statements of the cited article are correct:
An immediate function is only seen at compile time (and cannot be evaluated at run time)
Symbols are not emitted for an immediate function
Tools such as debuggers will not be able to show an immediate function
To be more precise, I'd like to hear which of the following statements of the cited article are correct:
An immediate function is only seen at compile time (and cannot be evaluated at run time)
Symbols are not emitted for an immediate function
Tools such as debuggers will not be able to show an immediate function
Almost none of these are answers which the C++ standard can give. The standard doesn't define "symbols" or what tools can show. Almost all of these are dealer's choice as far as the standard is concerned.
Indeed, even the question of "compile time" vs. "run time" is something the standard doesn't deal with. The only question that concerns the standard is whether something is a constant expression. Invoking a constexpr function may produce a constant expression, depending on its parameters. Invoking a consteval function in a way which does not produce a constant expression is il-formed.
The one thing the standard does define is what gets "seen". Though it's not really about "compile time". There are a number of statements in C++20 that forbid most functions from dealing in pointers/references to immediate functions. For example, C++20 states in [expr.prim.id]/3:
An id-expression that denotes an immediate function shall appear only
as a subexpression of an immediate invocation, or
in an immediate function context.
So if you're not in an immediate function, or you're not using the name of an immediate function to call another immediate function (passing a pointer/reference to the function), then you cannot name an immediate function. And you can't get a pointer/reference to a function without naming it.
This and other statements in the spec (like pointers to immediate function not being valid results of constant expressions) essentially make it impossible for a pointer/reference to an immediate function to leak outside of constant expressions.
So statements about the visibility of immediate functions are correct, to some degree. Symbols can be emitted for immediate functions, but you cannot use immediate functions in a way that would prevent an implementation from discarding said symbols.
And that's basically the thing with consteval. It doesn't use standard language to enforce what must happen. It uses standard language to make it impossible to use the function in a way that will prevent these things from happening. So it's more reasonable to say:
You cannot use an immediate function in a way that would prevent the compiler from executing it at compile time.
You cannot use an immediate function in a way that would prevent the compiler from discarding symbols for it.
You cannot use an immediate function in a way that would force debuggers to be able to see them.
Quality of implementation is expected to take things from there.
It should also be noted that debugging builds are for... debugging. It would be entirely reasonable for advanced compiler tools to be able to debug code that generates constant expressions. So a debugger which could see immediate functions execute is an entirely desirable technology. This becomes moreso as compile-time code grows more complex.
The proposal mentions:
One consequence of this specification is that an immediate function never needs to be seen by a back end.
So it is definitely the intention of the proposal that calls are replaced by the constant. In other words, that the constant expression is evaluated during translation.
However, it does not say it is required that it is not seen by the backend. In fact, in another sentence of the proposal, it just says it is unlikely:
It also means that, unlike plain constexpr functions, consteval functions are unlikely to show up in symbolic debuggers.
More generally, we can re-state the question as:
Are compilers forced to evaluate constant expressions (everywhere; not just when they definitely need it)?
For instance, a compiler needs to evaluate a constant expression if it is the number of elements of an array, because it needs to statically determine the total size of the array.
However, a compiler may not need to evaluate other uses, and while any decent optimizing compiler will try to do so anyway, it does not mean it needs to.
Another interesting case to think about is an interpreter: while an interpreter still needs to evaluate some constant expressions, it may just do it lazily all the time, without performing any constant folding.
So, as far as I know, they aren't required, but I don't know the exact quotes we need from the standard to prove it (or otherwise). Perhaps it is a good follow-up question on its own, which would answer this one too.
For instance, in [expr.const]p1 there is a note that says they can, not that they are:
[Note: Constant expressions can be evaluated during translation. — end note]
Related
constexpr permits expressions which can be evaluated at compile time to be ... evaluated at compile time.
Why is this keyword even necessary? Why not permit or require that compilers evaluate all expressions at compile time if possible?
The standard library has an uneven application of constexpr which causes a lot of inconvenience. Making constexpr the "default" would address that and likely improve a huge amount of existing code.
It already is permitted to evaluate side-effect-free computations at compile time, under the as-if rule.
What constexpr does is provide guarantees on what data-flow analysis a compliant compiler is required to do to detect1 compile-time-computable expressions, and also allow the programmer to express that intent so that they get a diagnostic if they accidentally do something that cannot be precomputed.
Making constexpr the default would eliminate that very useful diagnostic ability.
1 In general, requiring "evaluate all expressions at compile time if possible" is a non-starter, because detecting the "if possible" requires solving the Halting Problem, and computer scientists know that this is not possible in the general case. So instead a relaxation is used where the outputs are { "Computable at compile-time", "Not computable at compile-time or couldn't decide" }. And the ability of different compilers to decide would depend on how smart their test was, which would make this feature non-portable. constexpr defines the exact test to use. A smarter compiler can still pre-compute even more expressions than the Standard test dictates, but if they fail the test, they can't be marked constexpr.
Note: despite the below, I admit to liking the idea of making constexpr the default. But you asked why it wasn't already done, so to answer that I will simply elaborate on mattnewport's last comment:
Consider the situation today. You're trying to use some function from the standard library in a context that requires a constant expression. It's not marked as constexpr, so you get a compiler error. This seems dumb, since "clearly" the ONLY thing that needs to change for this to work is to add the word constexpr to the definition.
Now consider life in the alternate universe where we adopt your proposal. Your code now compiles, yay! Next year you decide you to add Windows support to whatever project you're working on. How hard can it be? You'll compile using Visual Studio for your Windows users and keep using gcc for everyone else, right?
But the first time you try to compile on Windows, you get a bunch of compiler errors: this function can't be used in a constant expression context. You look at the code of the function in question, and compare it to the version that ships with gcc. It turns out that they are slightly different, and that the version that ships with gcc meets the technical requirements for constexpr by sheer accident, and likewise the one that ships with Visual Studio does not meet those requirements, again by sheer accident. Now what?
No problem you say, I'll submit a bug report to Microsoft: this function should be fixed. They close your bug report: the standard never says this function must be usable in a constant expression, so we can implement however we want. So you submit a bug report to the gcc maintainers: why didn't you warn me I was using non-portable code? And they close it too: how were we supposed to know it's not portable? We can't keep track of how everyone else implements the standard library.
Now what? No one did anything really wrong. Not you, not the gcc folks, nor the Visual Studio folks. Yet you still end up with un-portable code and are not a happy camper at this point. All else being equal, a good language standard will try to make this situation as unlikely as possible.
And even though I used an example of different compilers, it could just as well happen when you try to upgrade to a newer version of the same compiler, or even try to compile with different settings. For example: the function contains an assert statement to ensure it's being called with valid arguments. If you compile with assertions disabled, the assertion "disappears" and the function meets the rules for constexpr; if you enable assertions, then it doesn't meet them. (This is less likely these days now that the rules for constexpr are very generous, but was a bigger issue under the C++11 rules. But in principle the point remains even today.)
Lastly we get to the admittedly minor issue of error messages. In today's world, if I try to do something like stick in a cout statement in constexpr function, I get a nice simple error right away. In your world, we would have the same situation that we have with templates, deep stack-traces all the way to the very bottom of the implementation of output streams. Not fatal, but surely annoying.
This is a year and a half late, but I still hope it helps.
As Ben Voigt points out, compilers are already allowed to evaluate anything at compile time under the as-if rule.
What constexpr also does is lay out clear rules for expressions that can be used in places where a compile time constant is required. That means I can write code like this and know it will be portable:
constexpr int square(int x) { return x * x; }
...
int a[square(4)] = {};
...
Without the keyword and clear rules in the standard I'm not sure how you could specify this portably and provide useful diagnostics on things the programmer intended to be constexpr but don't meet the requirements.
#include <iostream>
int foo() {
std::cout<<"foo() is called\n";
return 9;
}
int bar() {
std::cout<<"bar() is called\n";
return 18;
}
int main() {
std::cout<<foo()<<' '<<bar()<<' '<<'\n';
}
// Above program's behaviour is unspecified
// clang++ evaluates function arguments from left to right: http://melpon.org/wandbox/permlink/STnvMm1YVrrSRSsB
// g++ & MSVC++ evaluates function arguments from right to left
// so either foo() or bar() can be called first depending upon compiler.
Output of above program is compiler dependent. Order in which function arguments are evaluated is unspecified. The reason I've read about this is that it can result in highly optimized code. How not specify an exact order of evaluation of function argument helps compiler to generate optimized code?
AFAIK, the order of evaluation is strictly specified in languages such as Java, C#, D etc.
I think the whole premise of the question is wrong:
How not specify an exact order of evaluation of function argument helps C & C++ compiler to generate optimized code?
It is not about optimizing code (though it does allow that). It is about not penalizing compilers because the underlying hardware has certain ABI constraints.
Some systems depend on parameters being pushed to stack in reverse order while others depend on forward order. C++ runs on all kinds of systems with all kinds on constraints. If you enforce an order at the language level you will require some systems to pay a penalty to enforce that order.
The first rule of C++ is "If you don't use it then you should not have to pay for it". So enforcing an order would be a violation of the prime directive of C++.
It doesn't. At least, it doesn't today. Maybe it did in the past.
A proposal for C++17 suggests defining left-right evaluation order for function calls, operator<< and so on.
As described in Section 7 of that paper, this proposal was tested by compiling the Windows NT kernel, and it actually led to a speed increase in some benchmarks. The authors' comment:
It is worth noting that these results are for the worst case scenario where the optimizers have not yet been updated to be aware of, and take advantage of the new evaluation rules and they are blindly forced to evaluate function calls from left to right.
suggests that there is further room for speed improvement.
The order of evaluation is related to the way arguments are passed. If stack is used to pass the arguments, evaluating right to left helps performance, since this is how arguments are pushed into the stack.
For example, with the following code:
void foo(bar(), baz());
Assuming calling conevention is 'passing arguments through the stack', C calling convention requires arguments to be pushed into stack starting from the last one - so that when callee function reads it, it would pop the first argument first and be able to support variadic functions. If order of evaluation was left to right, a result of bar() would have to be saved in the temporary, than baz() called, it's result pushed, following by temporary push. However, right-to-left evaluation allows compiler to avoid the temporary.
If arguments are passed through registers, order of evaluation is not overly important.
The original reason that the C and C++ standards didn't specify and an order of evaluation for function arguments is to provide more optimization opportunities for the compiler. Unfortunately, this rationale has not been backed up by extensive experimentation at the time these languages were initially designed. But it made sense.
This issue has been raised in the past few years. See this blog post by Herb Sutter and don't forget to go through the comments.
Proposal P0145R1 suggests that it's better to specify an order of evaluation for function arguments and for other operators. It says:
The order of expression evaluation, as
it is currently specified in the standard, undermines advices,
popular programming idioms, or the relative safety of standard library
facilities. The traps aren’t just for novices or the careless
programmer. They affect all of us indiscriminately, even when we know
the rules.
You can find more information about how this affects optimization opportunities in that document.
In the past few months, there has been a very extensive discussion about how this change in the language affects optimization, compatibility and portability. The thread begins here and continues here. You can find there numerous examples.
At the very end of Scott Schurr's talk "Introducing constexpr" at CppCon, he asks "Is there a way to poison a function"? He then explains that this can be done (albeit in a non-standard way) by:
Putting a throw in a constexpr function
Declaring an unresolved extern const char*
Referencing the unresolved extern in the throw
I sense that I'm a bit out of my depth here, but I'm curious:
What does it mean to "poison a function"?
What is the significance/usefulness of the technique he outlines?
In general it refers to making a function unusable, e.g. if you want to ban the use of dynamic allocation in a program you could "poison" the malloc function so it can't be used.
In the video he's using it in a more specific way, which is clear if you read the slide that is displayed when he talks about poisoning the function, which says "A way to force compile-time only?"
So he is talking about "poisoning" the function to make it uncallable at run-time, so it's only callable in constant expressions. The technique is to have a branch in the function which is never taken when called in a compile-time context, and to make that branch contain something that will cause an error.
A throw expression is allowed in a constexpr function, as long as it is never reached during compile-time invocations of the function (because you can't throw an exception at compile-time, it's an inherently dynamic operation, like allocating memory). So a throw expression that refers to an undefined symbol will not be used during compile-time invocations (because that would fail to compile) and cannot be used at run-time, because the undefined symbol causes a linker error.
Because the undefined symbol is not "odr-used" in the compile-time invocations of the function, in practice the compiler will not create a reference to the symbol, so it's OK that it's undefined.
Is that useful? He's demonstrating how to do it, not necessarily saying it's a good idea or widely useful. If you have a need to do it for some reason then his technique might solve your problem. If you don't have a need for it, you don't need to worry about it.
One reason it might be useful is when the compile-time version of some operation is not as efficient as it could be. There are restrictions on the kind of expressions allowed in a constexpr function (especially in C++11, some restrictions were removed in C++14). So you might have two versions of a function for performing a calculation, one that is optimal, but uses expressions that aren't allowed in a constexpr function, and one that is a valid constexpr function, but would perform poorly if called at run-time. You could poison the sub-optimal one to ensure it is never used for run-time calls, ensuring the more efficient (non-constexpr) version is used for run-time calls.
N.B. The performance of a constexpr function used at compile-time is not really important, because it has no run-time overhead anyway. It might slow down your compilation by making the compiler do extra work, but it won't have any run-time performance cost.
'Poisoning' an identifier means that any reference to the identifier after the 'poisoning' is a hard compiler error. This technique may be used, for instance, for hard deprecation (function IS deprecated, never use it!).
In GCC traditionally there was a pragma for this: #pragma GCC poison.
This question already has answers here:
Why do we need to mark functions as constexpr?
(4 answers)
Closed 2 years ago.
As far as I understand it, constexpr can be seen as a hint to the compiler to check whether given expressions can be evaluated at compile-time and do so if possible.
I know that it also imposes some restriction on the function or initialization declared as constexpr but the final goal is compile-time evaluation, isn't it?
So my question is, why can't we leave that at the compiler? It is obviously capable of checking the pre-conditions, so why doesn't it do for each expression and evaluate at compile-time where possible?
I have two ideas on why this might be the case but I am not yet convinced that they hit the point:
a) It might take too long during compile-time.
b) Since my code can use constexpr functions in locations where normale functions would not be allowed the specifier is also kind of part of the declaration. If the compiler did everything by itself, one could use a function in a C-array definition with one version of the function but with the next version there might be a compiler-error, because the pre-conditions for compile-time evaluation are no more satisfied.
constexpr is not a "hint" to the compiler about anything; constexpr is a requirement. It doesn't require that an expression actually be executed at compile time; it requires that it could.
What constexpr does (for functions) is restrict what you're allowed to put into function definition, so that the compiler can easily execute that code at compile time where possible. It's a contract between you the programmer and the compiler. If your function violates the contract, the compiler will error immediately.
Once the contract is established, you are now able to use these constexpr functions in places where the language requires a compile time constant expression. The compiler can then check the elements of a constant expression to see that all function calls in the expression call constexpr functions; if they don't, again a compiler error results.
Your attempt to make this implicit would result in two problems. First, without an explicit contract as defined by the language, how would I know what I can and cannot do in a constexpr function? How do I know what will make a function not constexpr?
And second, without the contract being in the compiler, via a declaration of my intent to make the function constexpr, how would the compiler be able to verify that my function conforms to that contract? It couldn't; it would have to wait until I use it in a constant expression before I find that it isn't actually a proper constexpr function.
Contracts are best stated explicitly and up-front.
constexpr can be seen as a hint to the compiler to check whether given expressions can be evaluated at compile-time and do so if possible
No, see below
the final goal is compile-time evaluation
No, see below.
so why doesn't it do for each expression and evaluate at compile-time where possible?
Optimizers do things like that, as allowed under the as-if rule.
constexpr is not used to make things faster, it is used to allow usage of the result in context where a runtime-variable expression is illegal.
This is only my evaluation, but I believe your (b) reason is correct (that it forms part of the interface that the compiler can enforce). The interface requirement serves both for the writer of the code and the client of the code.
The writer may intend something to be usable in a compile-time context, but not actually use it in this way. If the writer violates the rules for constexpr, they might not find out until after publication when clients who try to use it constexpr fail. Or, more realistically, the library might use the code in a constexpr sense in version 1, refactor this usage out in version 2, and break constexpr compatibility in version 3 without realizing it. By checking constexpr-compliance, the breakage in version 3 will be caught before deployment.
The interface for the client is more obvious --- an inline function won't silently become constexpr-required because it happened to work and someone used that way.
I don't believe your (a) reason (that it could take too long for the compiler) is applicable because (1) the compiler has to check much of the constexpr constraints anyway when the code is marked, (2) without the annotation, the compiler would only have to do the checking when used in a constexpr way (so most functions wouldn't have to be checked), and (3) IIUC the D programming language actually does allow functions to be compile-time evaluated if they meet requirements without any declaration assistance, so apparently it can be done.
I think I remember watching an early talk by Bjarne Stroustrup where he mentioned that programmers wanted fine grained control on this "dangerous" feature, from which I understand that they don't want things "accidentally" executed at compile time without them knowing. (Even if that sound like a good thing.)
There can be many reasons for that, but the only valid one is ultimatelly compilation speed I think ( (a) in your list ).
It would be too much burden on the compiler to determine for every function if it could be computed at compile time.
This argument is weaker as compilation times in general go down.
Like many other features of C++ what end up happening is that we end up with the "wrong defaults".
So you have to tell when you want constexpr instead of when you don't want constexpr (runtimeexpr); you have to tell when you want const intead of where you want mutable, etc.
Admitedly, you can imagine functions that take an absurd amount of time to run at compile time and that cannot be amortized (with other kind of machine resources) at runtime.
(I am not aware that "time-out" can be a criterion in a compiler for constexpr, but it could be so.)
Or it could be that one is compiling in a system that is always expected to finish compilation in a finite time but an unbounded runtime is admissible (or debuggable).
I know that this question is old, but time has illuminated that it actually makes sense to have constexpr as default:
In C++17, for example, you can declare a lambda constexpr but more importantly they are constexpr by default if they can be so.
https://learn.microsoft.com/en-us/cpp/cpp/lambda-expressions-constexpr
Note that lambda has all the "right" (opposite) defaults, members (captures) are const by default, arguments are templates by default auto, and now these functions are constexpr by default.
I have question when I compile an inline function in C++.
Can a recursive function work with inline. If yes then please describe how.
I am sure about loop can't work with it but I have read somewhere recursive would work, If we pass constant values.
My friend send me some inline recursive function as constant parameter and told me that would be work but that not work on my laptop, no error at compile time but at run time display nothing and I have to terminate it by force break.
inline f(int n) {
if(n<=1)
return 1;
else {
n=n*f(n-1);
return n;
}
}
how does this work?
I am using turbo 3.2
Also, if an inline function code is too large then, can the compiler change it automatically in normal function?
thanks
This particular function definitely can be inlined. That is because the compiler can figure out that this particular form of recursion (tail-recursion) can be trivially turned into a normal loop. And with a normal loop it has no problem inlining it at all.
Not only can the compiler inline it, it can even calculate the result for a compile-time constant without generating any code for the function.
With GCC 4.4
int fac = f(10);
produced this instruction:
movl $3628800, 4(%esp)
You can easily verify when checking assembly output, that the function is indeed inlined for input that is not known at compile-time.
I suppose your friend was trying to say that if given a constant, the compiler could calculate the result entirely at compile time and just inline the answer at the call site. c++0x actually has a mechanism for this called constexpr, but there are limits to how complex the code is allowed to be. But even with the current version of c++, it is possible. It depends entirely on the compiler.
This function may be a good candidate given that it clearly only references the parameter to calculate the result. Some compilers even have non-portable attributes to help the compiler decide this. For example, gcc has pure and const attributes (listed on that page I just linked) that inform the compiler that this code only operates on the parameters and has no side effects, making it more likely to be calculated at compile time.
Even without this, it will still compile! The reason why is that the compiler is allowed to not inline a function if it decides. Think of the inline keyword more of a suggestion than an instruction.
Assuming that the compiler doesn't calculate the whole thing at compile time, inlining is not completely possible without other optimizations applied (see EDIT below) since it must have an actual function to call. However, it may get partially inlined. In that case the compiler will inline the initial call, but also emit a regular version of the function which will get called during recursion.
As for your second question, yes, size is one of the factors that compilers use to decide if it is appropriate to inline something.
If running this code on your laptop takes a very long time, then it is possible that you just gave it very large values and it is simply taking a long time to calculate the answer... The code look ok, but keep in mind that values above 13! are going to overflow a 32-bit int. What value did you attempt to pass?
The only way to know what actually happens is to compile it an look at the assembly generated.
PS: you may want to look into a more modern compiler if you are concerned with optimizations. For windows there is MingW and free versions of Visual C++. For *NIX there is of course g++.
EDIT: There is also a thing called Tail Recursion Optimization which allows compilers to convert certain types of recursive algorithms to iterative, making them better candidates for inlining. (In addition to making them more stack space efficient).
Recursive function can be inlined to certain limited depth of recursion. Some compilers have an option that lets you to specify how deep you want to go when inlining recursive functions. Basically, the compiler "flattens" several nested levels of recursion. If the execution reaches the end of "flattened" code, the code calls itself in usual recursive fashion and so on. Of course, if the depth of recursion is a run-time value, the compiler has to check the corresponding condition every time before executing each original recursive step inside the "flattened" code. In other words, there's nothing too unusual about inlining a recursive function. It is like unrolling a loop. There's no requirement for the parameters to be constant.
What you mean by "I am sure about loop can't work" is not clear. It doesn't seem to make much sense. Functions with a loop can be easily inlined and there's nothing strange about it.
What are you trying to say about your example that "displays nothing" is not clear either. There is nothing in the code that would "display" anything. No wonder it "displays nothing". On top of that, you posted invalid code. C++ language does not allow function declarations without an explicit return type.
As for your last question, yes, the compiler is completely free to implement an inline function as "normal" function. It has nothing to do with function being "too large" though. It has everything to do with more-or-less complex heuristic criteria used by that specific compiler to make the decision about inlining a function. It can take the size into account. It can take other things into account.
You can inline recursive functions. The compiler normally unrolls them to a certain depth- in VS you can even have a pragma for this, and the compiler can also do partial inlining. It essentially converts it into loops. Also, as #Evan Teran said, the compiler is not forced to inline a function that you suggest at all. It might totally ignore you and that's perfectly valid.
The problem with the code is not in that inline function. The constantness or not of the argument is pretty irrelevant, I'm sure.
Also, seriously, get a new compiler. There's modern free compilers for whatever OS your laptop runs.
One thing to keep in mind - according to the standard, inline is a suggestion, not an absolute guarantee. In the case of a recursive function, the compiler would not always be able to compute the recursion limit - modern compilers are getting extremely smart, a previous response shows the compiler evaluating a constant inline and simply generating the result, but consider
bigint fac = factorialOf(userInput)
there's no way the compiler can figure that one out........
As a side note, most compilers tend to ignore inlines in debug builds unless specifically instructed not to do so - makes debugging easier
Tail recursions can be converted to loops as long as the compiler can satisfactorily rearrange the internal representation to get the recursion conditional test at the end. In this case it can do the code generation to re-express the recursive function as a simple loop
As far as issues like tail recursion rewrites, partial expansions of recursive functions, etc, these are usually controlled by the optimization switches - all modern compilers are capable of pretty signficant optimization, but sometimes things do go wrong.
Remember that the inline key word merely sends a request, not a command to the compiler. The compliler may ignore yhis request if the function definition is too long or too complicated and compile the function as normal function.
in some of the cases where inline functions may not work are
For functions returning values, if a loop, a switch or a goto exists.
For functions not returning values, if a return statement exists.
If function contains static variables.
If in line functions are recursive.
hence in C++ inline recursive functions may not work.