Could C++ standards gurus please enlighten me:
Since which C++ standard version has this statement failed because (v) seems to be equivalent to (*&v)?
I.e. for example the code:
#define DEC(V) ( ((V)>0)? ((V)-=1) : 0 )
...{...
register int v=1;
int r = DEC(v) ;
...}...
This now produces warnings under -std=c++17 like:
cannot take address of register variable
left hand side of operand must be lvalue
Many C macros enclose ALL macro parameters in parentheses, of which the above is meant only to be a representative example.
The actual macros that produce warnings are for instance
the RTA_* macros in /usr/include/linux/rtnetlink.h.
Short of not using/redefining these macros in C++, is there any workaround?
If you look at the revision summary of the latest C++1z draft, you'd see this in [diff.cpp14.dcl.dcl]
[dcl.stc]
Change: Removal of register storage-class-specifier.
Rationale: Enable repurposing of deprecated keyword in future
revisions of this International Standard.
Effect on original feature: A valid C++ 2014 declaration utilizing the register
storage-class-specifier is ill-formed in this International Standard.
The specifier can simply be removed to retain the original meaning.
The warning may be due to that.
register is no longer a storage class specifier, you should remove it. Compilers may not be issuing the right error or warnings but your code should not have register to begin with
The following is a quote from the standard informing people about what they should do with regards to register in their code (relevant part emphasized), you probably have an old version of that file
C.1.6 Clause 10: declarations [diff.dcl]
Change: In C++, register is not a storage class specifier.
Rationale: The storage class specifier had no effect in C++.
Effect on original feature: Deletion of semantically well-defined feature.
Difficulty of converting: Syntactic transformation.
How widely used: Common.
Your worry is unwarranted since the file in question does not actually contain the register keyword:
grep "register" /usr/include/linux/rtnetlink.h
outputs nothing. Either way, you shouldn't be receiving the warning since:
System headers don't emit warnings by default, at least in GCC
It isn't wise to try to compile a file that belongs to a systems project like the linux kernel in C++ mode, as there may be subtle and nasty breaking changes
Just include the file normally or link the C code to your C++ binary. Report a bug if you really are getting a warning that should normally be suppressed to your compiler vendor.
Related
I am using visual studio 2017. In C++, I tried to assign a pointer to 'this' pointer. It showed compiler error as "assignment to 'this' (anachronism)". Anachronism means adding something into the period it can't exist like roman emperor checks computer. So is the compiler warning also like this. Or is there any specific meaning in coding for the word "anachronism"?.
A good while ago, this pointer could be assigned values. I met such assignments in the code of the Cfront compiler. I wrote about it in this note: Celebrating the 30-th anniversary of the first C++ compiler: let's find the bugs in it. Examples:
expr.expr(TOK ba, Pexpr a, Pexpr b)
{
register Pexpr p;
if (this) goto ret;
....
this = p;
....
}
inline toknode.~toknode()
{
next = free_toks;
free_toks = this;
this = 0;
}
Anachronism is something that was OK long time ago but not anymore.
Starting from C++98, this is an rvalue and as such cannot be assigned (i.e. cannot appear on the left from an assignment operator).
See §9.3.2 The this pointer:
In the body of nonstatic (9.3) member function, the keyword this is a non-lvalue expression whose value is the address of the object for which the function is called.
Starting from C++11, this is specified as a prvalue.
From google:
a thing belonging or appropriate to a period other than that in which it exists, especially a thing that is conspicuously old-fashioned.
Which basically means that this used to be allowed in the past (probably by pre standards compliant compilers) but isn't allowed anymore.
This question tapped my curiosity, so I dug into my copies of all the ratified C++ standards and technical corrigenda I have (which works out to everything between C++98 and C++17 inclusive). None of them contain the word "anachronism" in any form.
The ARM (The C++ Annotated Reference Manual) by Ellis and Stroustrup, 1990 (a base document written to guide development of the C++ standard) has Section 18.3 entitled "Anachronisms". The first paragraph of that section says
The extensions provided here may be provided by an implementation to ease the use of C programs as C++ programs or to provide continuity from earlier C++ implementations. Note that each of these features has undesirable aspects. An implementation providing them should also provide a way for the user to ensure that they do not occur in a source file. A C++ implementation is not obliged to provide these features.
Is an implementation allowed to issue a diagnostic message for a well-formed program?
For example some compilers issue a warning about unused expression result when compiling the following well-formed program:
int main() { 0; }
Are such compilers allowed to consider that warning a diagnostic message?
It's perfectly fine to issue a diagnostic, as long as the rules below are met in any corresponding scenario. §1.4/2:
Although this International Standard states only requirements on C ++
implementations, those requirements are often easier to understand if
they are phrased as requirements on programs, parts of programs, or
execution of programs. Such requirements have the following meaning:
If a program contains no violations of the rules in this International Standard, a conforming implementation shall, within
its resource limits, accept and correctly execute that program.
If a program contains a violation of any diagnosable rule or an occurrence of a construct described in this Standard as
“conditionally-supported” when the implementation does not support
that construct, a conforming implementation shall issue at least one
diagnostic message.
If a program contains a violation of a rule for which no diagnostic is required, this International Standard places no requirement on
implementations with respect to that program.
"Accepting" solely addresses the acknowledgment of the implementation that this is a well-formed program, not the absence of any diagnostics. After all, despite any warnings issued in the process, implementations still yield the object file you asked for.
However, there is one rule concerning templates that does require that there be no diagnostic issued; §14.6/8:
No diagnostic shall be issued for a template for which a valid
specialization can be generated.
An implementation can issue (1)any number of diagnostics it wants, as long as it does issue the required diagnostics.
It must accept correct programs, to the degree that it's able to,
C++14 §1.4/2:
” If a program contains no violations of the rules in this International Standard, a conforming imple-
mentation shall, within its resource limits, accept and correctly execute that program"
but it can issue diagnostics about it.
The C++ standard does not differentiate between error messages and warning messages, but this is a de facto standard. An error message means (by convention) that no binary is produced, because the problem is too severe. A warning message means (by convention) that there is a potential problem, but not a direct violation of language rules, and so a binary is produced unless there are also errors.
Sometimes the lines are a bit blurred, where implementations incorrectly but for pragmatic reasons accept invalid code, with only warnings or even no diagnostics. For new code one may therefore ask the compiler to treat every warning as an error, and aim for completely clean compiles. And as I understand it that's now absolutely not uncommon.
With some compilers, e.g. Visual C++, it can however be problematic, because the compiler issues too many silly-warnings, warnings about perfectly legitimate and non-problematic constructs. Then one has to somehow suppress those warnings. E.g. via #pragma directives, if possible, or by code rewrites.
Happily for Visual C++ there exists a header with such #pragma directives that turn off sillywarnings, compiled about five years ago from a community effort in the comp.lang.c++ Usenet group. And happily, for the community edition of Visual Studio 2015 there is an extension that provides a project template with that header included. These are both by me.
For the code in question,
int main() { 0; }
… instead of suppressing the warning, which generally is a useful one, you should rewrite the code to express your intent explicitly:
int main() { (void)0; }
The (void) cast tells the compiler that it's your intent to discard the value of that expression.
In the case of using this construct for an otherwise unused function argument, you can additionally declare an incomplete class of the same name, to prevent inadvertent use of the name:
(void)arg_name; struct arg_name;
But since it's unconventional it may trip up other programmers – with the compilers I use the error message for later use of the name is not exactly intuitive.
(1) Except as noted by Columbo in his answer, C++14 §14.6/8 “No diagnostic shall be issued for a template for which a valid specialization can be generated.”.
I'm testing some C++ code with -std=c++11. I noticed a warning I had not seen before:
'register' storage class specifier is deprecated
What does the standard say about this (other than its deprecated)?
Is it implementation defined?
Will compilers take it as a hint and try to put the value in a register?
Will this eventually lead to a compile failure?
Perhaps something else?
C++11, [dcl.stc]:
3 - A register specifier is a hint to the implementation that the variable so declared will be heavily used. [ Note: The hint can be ignored and in most implementations it will be ignored if the address of the variable
is taken. This use is deprecated (see D.2). — end note ]
There is a proposal to remove the register keyword as a storage specifier, while reserving it as a keyword: Remove Deprecated Use of the register Keyword (n4340). This may or may not be implemented in C++1z (tentative C++17); it would pose challenges for compatibility with C, where register still has semantic effect (a C register variable or parameter cannot have its address taken or be subject to array-to-pointer decay).
Is it implementation defined?
It always was.
Will compilers take it as a hint and try to put the value in a register?
It's implementation-defined.
Will this eventually lead to a compile failure?
No.
EDIT Perhaps I misunderstood the question about the eventual compile failure. I took it to mean an eventual failure in the current compilation. If the question is about the future of the register keyword, anything is possible: I don't care to fortune-tell.
This question already has answers here:
Why does flowing off the end of a non-void function without returning a value not produce a compiler error?
(11 answers)
Closed 6 years ago.
I found this in one of my libraries this morning:
static tvec4 Min(const tvec4& a, const tvec4& b, tvec4& out)
{
tvec3::Min(a,b,out);
out.w = min(a.w,b.w);
}
I'd expect a compiler error because this method doesn't return anything, and the return type is not void.
The only two things that come to mind are
In the only place where this method is called, the return value isn't being used or stored. (This method was supposed to be void - the tvec4 return type is a copy-and-paste error)
a default constructed tvec4 is being created, which seems a bit unlike, oh, everything else in C++.
I haven't found the part of the C++ spec that addresses this. References (ha) are appreciated.
Update
In some circumstances, this generates an error in VS2012. I haven't narrowed down specifics, but it's interesting, nonetheless.
This is undefined behavior from the C++11 draft standard section 6.6.3 The return statement paragraph 2 which says:
[...] Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function. [...]
This means that the compiler is not obligated provide an error nor a warning usually because it can be difficult to diagnose in all cases. We can see this from the definition of undefined behavior in the draft standard in section 1.3.24 which says:
[...]Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).[...]
Although in this case we can get both gcc and clang to generate a wanring using the -Wall flag, which gives me a warning similar to this:
warning: control reaches end of non-void function [-Wreturn-type]
We can turn this particular warning into an error using the -Werror=return-type flag. I also like to use -Wextra -Wconversion -pedantic for my own personal projects.
As ComicSansMS mentions in Visual Studio this code would generate C4716 which is an error by default, the message I see is:
error C4716: 'Min' : must return a value
and in the case where not all code paths would return a value then it would generate C4715, which is a warning.
Maybe some elaboration on the why part of the question:
As it turns out, it is actually quite hard† for a C++ compiler to determine whether a function exits without a return value. In addition to the code paths that end in explicit return statements and the ones that fall off the end of the function, you also have to consider potential exception throws or longjmps in the function itself, as well as all of its callees.
While it is quite easy for a compiler to identify a function that looks like it might be missing a return, it is considerably harder to prove that it is missing a return. In order to lift compiler vendors of this burden, the standard does not require this to generate an error.
So compiler vendors are free to generate a warning if they are quite sure that a function is missing a return and the user is then free to ignore/mask that warning in those rare cases where the compiler was actually wrong.
†: In the general case, this is equivalent to the halting problem, so it is actually impossible for a machine to decide this reliably.
Compile your code with -Wreturn-type option:
$ g++ -Wreturn-type source.cpp
This will give you warning. You can turn the warning into error if you use -Werror too:
$ g++ -Wreturn-type -Werror source.cpp
Note that this will turn all warnings into errors. So if you want error for specific warning, say -Wreturn-type, just type return-type without -W part as:
$ g++ -Werror=return-type source.cpp
In general you should always use -Wall option which includes most common warnings — this includes missing return statement also. Along with -Wall, you can use -Wextra also, which includes other warnings not included by -Wall.
Maybe some additional elaboration on the why part of the question.
C++ was designed so that a very large body of pre-existing body of C code compiles with minimum amount of changes. Unfortunately, C itself was paying a similar duty to earliest pre-standard C which did not even have the void keyword and instead relied on a default return type of int. C functions usually did return values, and whenever code superficially similar to Algol/Pascal/Basic procedures was written without any return statements, the function was, under the hood, returning whichever garbage was left on the stack. Neither the caller nor the callee assigns the value of the garbage in a reliable way. If the garbage is then ignored by every caller, everything is fine and C++ inherits the moral obligation to compile such code.
(If the returned value is used by the caller, the code may behave non-deterministically, similar to processing of an uninitialized variable. Could the difference be reliably identified by a compiler, in a hypothetical successor language to C? This is hardly possible. The caller and the callee may be in different compilation units.)
The implicit int is just a part of the C legacy involved here. A "dispatcher" function might, depending on a parameter, return a variety of types from some code branches, and return no useful value from other code branches. Such a function would generally be declared to return a type long enough to hold any of the possible types and the caller might need to cast it or extract it from a union.
So the deepest cause is probably the C language creators' belief that procedures that do not return any value are just an unimportant special case of functions that do; this problem got aggravated by the lack of focus on type safety of function calls in the oldest C dialects.
While C++ did break compatibility with some of the worst aspects of C (example), the willingness to compile a return statement without a value (or the implicit value-less return at the end of a function) was not one of them.
As already mentioned, this is undefined behavior and will give you a compiler warning. Most places I've worked require you to turn on compiler settings to treat warnings as errors - which enforces that all your code must compile with 0 errors and 0 warnings. This is a good example of why that is a good idea.
This is more of the standard C++ rule/feature which tends to be flexible with things and which tends to be more close to C.
But when we talk of the compilers, GCC or VS, they are more for professional usage and for variety of development purposes and hence put more strict development rules as per your needs.
That makes sense also, my personal opinion, because the language is all about features and its usage whereas compiler defines the rules for optimal and best way of using it as per your needs.
As mentioned in above post, compiler sometimes gives the error, sometimes gives warning and also it has the option of skipping these warning etc, indicating the freedom to use the language and its features in a way that suits us best.
Along with this there are several other questions mentioning this behaviour of returning a result without having a return statement. One simple example would be:
int foo(int a, int b){ int c = a+b;}
int main(){
int c = 5;
int d = 5;
printf("f(%d,%d) is %d\n", c, d, foo(c,d));
return 0;
}
Could this anomaly be due stack properties and more specifically:
Zero-Address Machines
In zero-address machines, locations of both operands are assumed to be at a default location.
These machines use the stack as the source of the input operands and the result goes back into
the stack. Stack is a LIFO (last-in-first-out) data structure that all processors support, whether
or not they are zero-address machines. As the name implies, the last item placed on the stack
is the first item to be taken out of the stack. All operations on this type of machine assume that the required input operands are the top
two values on the stack. The result of the operation is placed on top of the stack.
In addition to that, for accessing memory to read and write data same registers are used as data source and destination(DS (data segment) register), that store first the variables needed for the calculation and then the returned result.
Note:
with this answer I would like to discuss one possible explanation of the strange behaviour at machine (instruction) level as it has already a context and its covered in adequately wide range.
This question already has answers here:
Why does flowing off the end of a non-void function without returning a value not produce a compiler error?
(11 answers)
Closed 6 years ago.
I found this in one of my libraries this morning:
static tvec4 Min(const tvec4& a, const tvec4& b, tvec4& out)
{
tvec3::Min(a,b,out);
out.w = min(a.w,b.w);
}
I'd expect a compiler error because this method doesn't return anything, and the return type is not void.
The only two things that come to mind are
In the only place where this method is called, the return value isn't being used or stored. (This method was supposed to be void - the tvec4 return type is a copy-and-paste error)
a default constructed tvec4 is being created, which seems a bit unlike, oh, everything else in C++.
I haven't found the part of the C++ spec that addresses this. References (ha) are appreciated.
Update
In some circumstances, this generates an error in VS2012. I haven't narrowed down specifics, but it's interesting, nonetheless.
This is undefined behavior from the C++11 draft standard section 6.6.3 The return statement paragraph 2 which says:
[...] Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function. [...]
This means that the compiler is not obligated provide an error nor a warning usually because it can be difficult to diagnose in all cases. We can see this from the definition of undefined behavior in the draft standard in section 1.3.24 which says:
[...]Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).[...]
Although in this case we can get both gcc and clang to generate a wanring using the -Wall flag, which gives me a warning similar to this:
warning: control reaches end of non-void function [-Wreturn-type]
We can turn this particular warning into an error using the -Werror=return-type flag. I also like to use -Wextra -Wconversion -pedantic for my own personal projects.
As ComicSansMS mentions in Visual Studio this code would generate C4716 which is an error by default, the message I see is:
error C4716: 'Min' : must return a value
and in the case where not all code paths would return a value then it would generate C4715, which is a warning.
Maybe some elaboration on the why part of the question:
As it turns out, it is actually quite hard† for a C++ compiler to determine whether a function exits without a return value. In addition to the code paths that end in explicit return statements and the ones that fall off the end of the function, you also have to consider potential exception throws or longjmps in the function itself, as well as all of its callees.
While it is quite easy for a compiler to identify a function that looks like it might be missing a return, it is considerably harder to prove that it is missing a return. In order to lift compiler vendors of this burden, the standard does not require this to generate an error.
So compiler vendors are free to generate a warning if they are quite sure that a function is missing a return and the user is then free to ignore/mask that warning in those rare cases where the compiler was actually wrong.
†: In the general case, this is equivalent to the halting problem, so it is actually impossible for a machine to decide this reliably.
Compile your code with -Wreturn-type option:
$ g++ -Wreturn-type source.cpp
This will give you warning. You can turn the warning into error if you use -Werror too:
$ g++ -Wreturn-type -Werror source.cpp
Note that this will turn all warnings into errors. So if you want error for specific warning, say -Wreturn-type, just type return-type without -W part as:
$ g++ -Werror=return-type source.cpp
In general you should always use -Wall option which includes most common warnings — this includes missing return statement also. Along with -Wall, you can use -Wextra also, which includes other warnings not included by -Wall.
Maybe some additional elaboration on the why part of the question.
C++ was designed so that a very large body of pre-existing body of C code compiles with minimum amount of changes. Unfortunately, C itself was paying a similar duty to earliest pre-standard C which did not even have the void keyword and instead relied on a default return type of int. C functions usually did return values, and whenever code superficially similar to Algol/Pascal/Basic procedures was written without any return statements, the function was, under the hood, returning whichever garbage was left on the stack. Neither the caller nor the callee assigns the value of the garbage in a reliable way. If the garbage is then ignored by every caller, everything is fine and C++ inherits the moral obligation to compile such code.
(If the returned value is used by the caller, the code may behave non-deterministically, similar to processing of an uninitialized variable. Could the difference be reliably identified by a compiler, in a hypothetical successor language to C? This is hardly possible. The caller and the callee may be in different compilation units.)
The implicit int is just a part of the C legacy involved here. A "dispatcher" function might, depending on a parameter, return a variety of types from some code branches, and return no useful value from other code branches. Such a function would generally be declared to return a type long enough to hold any of the possible types and the caller might need to cast it or extract it from a union.
So the deepest cause is probably the C language creators' belief that procedures that do not return any value are just an unimportant special case of functions that do; this problem got aggravated by the lack of focus on type safety of function calls in the oldest C dialects.
While C++ did break compatibility with some of the worst aspects of C (example), the willingness to compile a return statement without a value (or the implicit value-less return at the end of a function) was not one of them.
As already mentioned, this is undefined behavior and will give you a compiler warning. Most places I've worked require you to turn on compiler settings to treat warnings as errors - which enforces that all your code must compile with 0 errors and 0 warnings. This is a good example of why that is a good idea.
This is more of the standard C++ rule/feature which tends to be flexible with things and which tends to be more close to C.
But when we talk of the compilers, GCC or VS, they are more for professional usage and for variety of development purposes and hence put more strict development rules as per your needs.
That makes sense also, my personal opinion, because the language is all about features and its usage whereas compiler defines the rules for optimal and best way of using it as per your needs.
As mentioned in above post, compiler sometimes gives the error, sometimes gives warning and also it has the option of skipping these warning etc, indicating the freedom to use the language and its features in a way that suits us best.
Along with this there are several other questions mentioning this behaviour of returning a result without having a return statement. One simple example would be:
int foo(int a, int b){ int c = a+b;}
int main(){
int c = 5;
int d = 5;
printf("f(%d,%d) is %d\n", c, d, foo(c,d));
return 0;
}
Could this anomaly be due stack properties and more specifically:
Zero-Address Machines
In zero-address machines, locations of both operands are assumed to be at a default location.
These machines use the stack as the source of the input operands and the result goes back into
the stack. Stack is a LIFO (last-in-first-out) data structure that all processors support, whether
or not they are zero-address machines. As the name implies, the last item placed on the stack
is the first item to be taken out of the stack. All operations on this type of machine assume that the required input operands are the top
two values on the stack. The result of the operation is placed on top of the stack.
In addition to that, for accessing memory to read and write data same registers are used as data source and destination(DS (data segment) register), that store first the variables needed for the calculation and then the returned result.
Note:
with this answer I would like to discuss one possible explanation of the strange behaviour at machine (instruction) level as it has already a context and its covered in adequately wide range.