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Where should I prefer using macros and where should I prefer constexpr?
Aren't they basically the same?
#define MAX_HEIGHT 720
vs
constexpr unsigned int max_height = 720;
Aren't they basically the same?
No. Absolutely not. Not even close.
Apart from the fact your macro is an int and your constexpr unsigned is an unsigned, there are important differences and macros only have one advantage.
Scope
A macro is defined by the preprocessor and is simply substituted into the code every time it occurs. The preprocessor is dumb and doesn't understand C++ syntax or semantics. Macros ignore scopes such as namespaces, classes or function blocks, so you can't use a name for anything else in a source file. That's not true for a constant defined as a proper C++ variable:
#define MAX_HEIGHT 720
constexpr int max_height = 720;
class Window {
// ...
int max_height;
};
It's fine to have a member variable called max_height because it's a class member and so has a different scope, and is distinct from the one at namespace scope. If you tried to reuse the name MAX_HEIGHT for the member then the preprocessor would change it to this nonsense that wouldn't compile:
class Window {
// ...
int 720;
};
This is why you have to give macros UGLY_SHOUTY_NAMES to ensure they stand out and you can be careful about naming them to avoid clashes. If you don't use macros unnecessarily you don't have to worry about that (and don't have to read SHOUTY_NAMES).
If you just want a constant inside a function you can't do that with a macro, because the preprocessor doesn't know what a function is or what it means to be inside it. To limit a macro to only a certain part of a file you need to #undef it again:
int limit(int height) {
#define MAX_HEIGHT 720
return std::max(height, MAX_HEIGHT);
#undef MAX_HEIGHT
}
Compare to the far more sensible:
int limit(int height) {
constexpr int max_height = 720;
return std::max(height, max_height);
}
Why would you prefer the macro one?
A real memory location
A constexpr variable is a variable so it actually exists in the program and you can do normal C++ things like take its address and bind a reference to it.
This code has undefined behaviour:
#define MAX_HEIGHT 720
int limit(int height) {
const int& h = std::max(height, MAX_HEIGHT);
// ...
return h;
}
The problem is that MAX_HEIGHT isn't a variable, so for the call to std::max a temporary int must be created by the compiler. The reference that is returned by std::max might then refer to that temporary, which doesn't exist after the end of that statement, so return h accesses invalid memory.
That problem simply doesn't exist with a proper variable, because it has a fixed location in memory that doesn't go away:
int limit(int height) {
constexpr int max_height = 720;
const int& h = std::max(height, max_height);
// ...
return h;
}
(In practice you'd probably declare int h not const int& h but the problem can arise in more subtle contexts.)
Preprocessor conditions
The only time to prefer a macro is when you need its value to be understood by the preprocessor, for use in #if conditions, e.g.
#define MAX_HEIGHT 720
#if MAX_HEIGHT < 256
using height_type = unsigned char;
#else
using height_type = unsigned int;
#endif
You couldn't use a variable here, because the preprocessor doesn't understand how to refer to variables by name. It only understands basic very basic things like macro expansion and directives beginning with # (like #include and #define and #if).
If you want a constant that can be understood by the preprocessor then you should use the preprocessor to define it. If you want a constant for normal C++ code, use normal C++ code.
The example above is just to demonstrate a preprocessor condition, but even that code could avoid using the preprocessor:
using height_type = std::conditional_t<max_height < 256, unsigned char, unsigned int>;
Generally speaking, you should use constexpr whenever you may, and macros only if no other solution is possible.
Rationale:
Macros are a simple replacement in the code, and for this reason, they often generate conflicts (e.g. windows.h max macro vs std::max). Additionally, a macro which works may easily be used in a different way which can then trigger strange compilation errors. (e.g. Q_PROPERTY used on structure members)
Due to all those uncertainties, it is good code style to avoid macros, exactly like you'd usually avoid gotos.
constexpr is semantically defined, and thus typically generates far less issues.
Great answer by Jonathon Wakely. I'd also advise you to take a look at jogojapan's answer as to what the difference is between const and constexpr before you even go about considering the usage of macros.
Macros are dumb, but in a good way. Ostensibly nowadays they're a build-aid for when you want very specific parts of your code to only be compiled in the presence of certain build parameters getting "defined". Usually, all that means is taking your macro name, or better yet, let's call it a Trigger, and adding things like, /D:Trigger, -DTrigger, etc. to the build tools being used.
While there's many different uses for macros, these are the two I see most often that aren't bad/out-dated practices:
Hardware and Platform-specific code sections
Increased verbosity builds
So while you can in the OP's case accomplish the same goal of defining an int with constexpr or a MACRO, it's unlikely the two will have overlap when using modern conventions. Here's some common macro-usage that hasn't been phased out, yet.
#if defined VERBOSE || defined DEBUG || defined MSG_ALL
// Verbose message-handling code here
#endif
As another example for macro-usage, let's say you have some upcoming hardware to release, or maybe a specific generation of it that has some tricky workarounds that the others don't require. We'll define this macro as GEN_3_HW.
#if defined GEN_3_HW && defined _WIN64
// Windows-only special handling for 64-bit upcoming hardware
#elif defined GEN_3_HW && defined __APPLE__
// Special handling for macs on the new hardware
#elif !defined _WIN32 && !defined __linux__ && !defined __APPLE__ && !defined __ANDROID__ && !defined __unix__ && !defined __arm__
// Greetings, Outlander! ;)
#else
// Generic handling
#endif
Sometimes a local variable is used for the sole purpose of checking it in an assert(), like so -
int Result = Func();
assert( Result == 1 );
When compiling code in a Release build, assert()s are usually disabled, so this code may produce a warning about Result being set but never read.
A possible workaround is -
int Result = Func();
if ( Result == 1 )
{
assert( 0 );
}
But it requires too much typing, isn't easy on the eyes and causes the condition to be always checked (yes, the compiler may optimize the check away, but still).
I'm looking for an alternative way to express this assert() in a way that wouldn't cause the warning, but still be simple to use and avoid changing the semantics of assert().
(disabling the warning using a #pragma in this region of code isn't an option, and lowering warning levels to make it go away isn't an option either...).
We use a macro to specifically indicate when something is unused:
#define _unused(x) ((void)(x))
Then in your example, you'd have:
int Result = Func();
assert( Result == 1 );
_unused( Result ); // make production build happy
That way (a) the production build succeeds, and (b) it is obvious in the code that the variable is unused by design, not that it's just been forgotten about. This is especially helpful when parameters to a function are not used.
I wouldn't be able to give a better answer than this, that addresses that problem, and many more:
Stupid C++ Tricks: Adventures in assert
#ifdef NDEBUG
#define ASSERT(x) do { (void)sizeof(x);} while (0)
#else
#include <assert.h>
#define ASSERT(x) assert(x)
#endif
As of C++17, the variable can be decorated with an attribute.
[[maybe_unused]] int Result = Func();
assert( Result == 1 );
See https://en.cppreference.com/w/cpp/language/attributes/maybe_unused for details.
This is better than the (void)Result trick because you directly decorate the variable declaration, rather than add something as an afterthought.
You could create another macro that allows you to avoid using a temporary variable:
#ifndef NDEBUG
#define Verify(x) assert(x)
#else
#define Verify(x) ((void)(x))
#endif
// asserts that Func()==1 in debug mode, or calls Func() and ignores return
// value in release mode (any braindead compiler can optimize away the comparison
// whose result isn't used, and the cast to void suppresses the warning)
Verify(Func() == 1);
int Result = Func();
assert( Result == 1 );
This situation means that in release mode, you really want:
Func();
But Func is non-void, i.e. it returns a result, i.e. it is a query.
Presumably, besides returning a result, Func modifies something (otherwise, why bother calling it and not using its result?), i.e. it is a command.
By the command-query separation principle (1), Func shouldn't be a command and a query at the same time. In other words, queries shouldn't have side effects, and the "result" of commands should be represented by the available queries on the object's state.
Cloth c;
c.Wash(); // Wash is void
assert(c.IsClean());
Is better than
Cloth c;
bool is_clean = c.Wash(); // Wash returns a bool
assert(is_clean);
The former doesn't give you any warning of your kind, the latter does.
So, in short, my answer is: don't write code like this :)
Update (1): You asked for references about the Command-Query Separation Principle. Wikipedia is rather informative. I read about this design technique in Object Oriented Software Construction, 2nd Editon by Bertrand Meyer.
Update (2): j_random_hacker comments "OTOH, every "command" function f() that previously returned a value must now set some variable last_call_to_f_succeeded or similar". This is only true for functions that don't promise anything in their contract, i.e. functions that might "succeed" or not, or a similar concept. With Design by Contract, a relevant number of functions will have postconditions, so after "Empty()" the object will be "IsEmpty()", and after "Encode()" the message string will be "IsEncoded()", with no need to check. In the same way, and somewhat symetrically, you don't call a special function "IsXFeasible()" before each and every call to a procedure "X()"; because you usually know by design that you're fulfilling X's preconditions at the point of your call.
You could use:
Check( Func() == 1 );
And implement your Check( bool ) function as you want. It may either use assert, or throw a particular exception, write in a log file or to the console, have different implementations in debug and release, or a combination of all.
With C++17 we can do:
[[maybe_unused]] int Result = Func();
though it involves a bit of extra typing compared to a assert substitution. See this answer.
Note: Added this because is the first google hit for "c++ assert unused variable".
You should move the assert inside the function before the return value(s). You know that the return value is not an unreferenced local variable.
Plus it makes more sense to be inside the function anyway, because it creates a self contained unit that has its OWN pre- and post-conditions.
Chances are that if the function is returning a value, you should be doing some kind of error checking in release mode on this return value anyway. So it shouldn't be an unreferenced variable to begin with.
Edit, But in this case the post condition should be X (see comments):
I strongly disagree with this point, one should be able to determine the post condition from the input parameters and if it's a member function, any object state. If a global variable modifies the output of the function, then the function should be restructured.
Most answers suggest using static_cast<void>(expression) trick in Release builds to suppress the warning, but this is actually suboptimal if your intention is to make checks truly Debug-only. The goals of an assertion macro in question are:
Perform checks in Debug mode
Do nothing in Release mode
Emit no warnings in all cases
The problem is that void-cast approach fails to reach the second goal. While there is no warning, the expression that you've passed to your assertion macro will still be evaluated. If you, for example, just do a variable check, that is probably not a big deal. But what if you call some function in your assertion check like ASSERT(fetchSomeData() == data); (which is very common in my experience)? The fetchSomeData() function will still be called. It may be fast and simple or it may be not.
What you really need is not only warning suppression but perhaps more importantly - non-evaluation of the debug-only check expression. This can be achieved with a simple trick that I took from a specialized Assert library:
void myAssertion(bool checkSuccessful)
{
if (!checkSuccessful)
{
// debug break, log or what not
}
}
#define DONT_EVALUATE(expression) \
{ \
true ? static_cast<void>(0) : static_cast<void>((expression)); \
}
#ifdef DEBUG
# define ASSERT(expression) myAssertion((expression))
#else
# define ASSERT(expression) DONT_EVALUATE((expression))
#endif // DEBUG
int main()
{
int a = 0;
ASSERT(a == 1);
ASSERT(performAHeavyVerification());
return 0;
}
All the magic is in the DONT_EVALUATE macro. It is obvious that at least logically the evaluation of your expression is never needed inside of it. To strengthen that, the C++ standard guarantees that only one of the branches of conditional operator will be evaluated. Here is the quote:
5.16 Conditional operator [expr.cond]
logical-or-expression ? expression : assignment-expression
Conditional expressions group right-to-left. The first expression is
contextually converted to bool. It is evaluated and if it is true, the
result of the conditional expression is the value of the second
expression, otherwise that of the third expression. Only one of these
expressions is evaluated.
I have tested this approach in GCC 4.9.0, clang 3.8.0, VS2013 Update 4, VS2015 Update 4 with the most harsh warning levels. In all cases there are no warnings and the checking expression is never evaluated in Release build (in fact the whole thing is completely optimized away). Bare in mind though that with this approach you will get in trouble really fast if you put expressions that have side effects inside the assertion macro, though this is a very bad practice in the first place.
Also, I would expect that static analyzers may warn about "result of an expression is always constant" (or something like that) with this approach. I've tested for this with clang, VS2013, VS2015 static analysis tools and got no warnings of that kind.
The simplest thing is to only declare/assign those variables if the asserts will exist. The NDEBUG macro is specifically defined if asserts won't be effected (done that way round just because -DNDEBUG is a convenient way to disable debugging, I think), so this tweaked copy of #Jardel's answer should work (cf. comment by #AdamPeterson on that answer):
#ifndef NDEBUG
int Result =
#endif
Func();
assert(Result == 1);
or, if that doesn't suit your tastes, all sorts of variants are possible, e.g. this:
#ifndef NDEBUG
int Result = Func();
assert(Result == 1);
#else
Func();
#endif
In general with this stuff, be careful that there's never a possibility for different translation units to be build with different NDEBUG macro states -- especially re. asserts or other conditional content in public header files. The danger is that you, or users of your library might accidentally instantiate a different definition of an inline function from the one used inside the compiled part of the library, quietly violating the one definition rule and making the runtime behaviour undefined.
This is a bad use of assert, IMHO. Assert is not meant as an error reporting tool, it's meant to assert preconditions. If Result is not used elsewhere, it's not a precondition.
Certainly you use a macro to control your assert definition, such as "_ASSERT". So, you can do this:
#ifdef _ASSERT
int Result =
#endif /*_ASSERT */
Func();
assert(Result == 1);
int Result = Func();
assert( Result == 1 );
Result;
This will make the compiler stop complaining about Result not being used.
But you should think about using a version of assert that does something useful at run-time, like log descriptive errors to a file that can be retrieved from the production environment.
I'd use the following:
#ifdef _DEBUG
#define ASSERT(FUNC, CHECK) assert(FUNC == CHECK)
#else
#define ASSERT(FUNC, CHECK)
#endif
...
ASSERT(Func(), 1);
This way, for release build, the compiler don't even need to produce any code for assert.
If this code is inside a function, then act on and return the result:
bool bigPicture() {
//Check the results
bool success = 1 != Func();
assert(success == NO, "Bad times");
//Success is given, so...
actOnIt();
//and
return success;
}
// Value is always computed. We also call assert(value) if assertions are
// enabled. Value is discarded either way. You do not get a warning either
// way. This is useful when (a) a function has a side effect (b) the function
// returns true on success, and (c) failure seems unlikely, but we still want
// to check sometimes.
template < class T >
void assertTrue(T const &value)
{
assert(value);
}
template < class T >
void assertFalse(T const &value)
{
assert(!value);
}
I haven't succeeded in using [[maybe_unused]] but
you can use the unused attribute
int Result __attribute__((__unused__)) = Func();
gcc Variable-Attributes
I've inherited a sizeable codebase where someone, somehow, has written several conditionals like so:
enum
{
FOO_TYPE_A,
FOO_TYPE_B,
FOO_TYPE_C,
FOO_TYPE_D
};
void bar(int fooType)
{
if (fooType == FOO_TYPE_A || FOO_TYPE_B) // <-- This will always be true, since FOO_TYPE_B is nonzero!
{
// Do something intended for only type A or B
}
// Do things general to A,B,C,D
}
where the condition check should clearly be:
if (fooType == FOO_TYPE_A || fooType == FOO_TYPE_B)
Is there a warning in gcc I can turn on to find them all, similar to MSDN's C4127?
Specifically, I'm using the Android NDK r9d.
If not, why not? It seems like a useful catch-all for unintentional assignment, unsigned > 0 as well as the above foolishness.
EDIT: Made the code more verbose to illustrate the problem.
I do not see a warning that corresponds to MSDN C4127. GCC does have a warning that is somewhat similar in intent, but not to address your problem: -Wtype-limits
Warn if a comparison is always true or always false due to the
limited range of the data type, but do not warn for constant
expressions. For example, warn if an unsigned variable is compared
against zero with < or >=. This warning is also enabled by
-Wextra.
As you can see, GCC explicitly states it does not warn about constant expressions. The motivation for this may be due to the common use of constant expressions to leverage the compiler's dead code elimination so that macros (or portions of it) can be optimized away by using a compile time constant. It would be used as an alternative to conditional compilation (#if defined() and #if X == Y), as the macro reads more naturally like a regular function. As a hypothetical example:
#define VERIFY(E) \
do { \
if (NO_VERIFY) break; \
if (!(E) && (VERIFY_LOG_LEVEL >= log_level() || VERIFY_LOG_ALWAYS)) { \
log("validation error for: " #E); \
} \
} while (0)
I think the problem is that variable is a define or a const. CONSTANT_2 is a non-zero constant also make this problem.
I'm writing system-level code for an embedded system without memory protection (on an ARM Cortex-M1, compiling with gcc 4.3) and need to read/write directly to a memory-mapped register. So far, my code looks like this:
#define UART0 0x4000C000
#define UART0CTL (UART0 + 0x30)
volatile unsigned int *p;
p = UART0CTL;
*p &= ~1;
Is there any shorter way (shorter in code, I mean) that does not use a pointer? I looking for a way to write the actual assignment code as short as this (it would be okay if I had to use more #defines):
*(UART0CTL) &= ~1;
Anything I tried so far ended up with gcc complaining that it could not assign something to the lvalue...
#define UART0CTL ((volatile unsigned int *) (UART0 + 0x30))
:-P
Edited to add: Oh, in response to all the comments about how the question is tagged C++ as well as C, here's a C++ solution. :-P
inline unsigned volatile& uart0ctl() {
return *reinterpret_cast<unsigned volatile*>(UART0 + 0x30);
}
This can be stuck straight in a header file, just like the C-style macro, but you have to use function call syntax to invoke it.
I'd like to be a nitpick: are we talking C or C++ ?
If C, I defer to Chris' answer willingly (and I'd like the C++ tag to be removed).
If C++, I advise against the use of those nasty C-Casts and #define altogether.
The idiomatic C++ way is to use a global variable:
volatile unsigned int& UART0 = *((volatile unsigned int*)0x4000C000);
volatile unsigned int& UART0CTL = *(&UART0 + 0x0C);
I declare a typed global variable, which will obey scope rules (unlike macros).
It can be used easily (no need to use *()) and is thus even shorter!
UART0CTL &= ~1; // no need to dereference, it's already a reference
If you want it to be pointer, then it would be:
volatile unsigned int* const UART0 = 0x4000C000; // Note the const to prevent rebinding
But what is the point of using a const pointer that cannot be null ? This is semantically why references were created for.
You can go one further than Chris's answer if you want to make the hardware registers look like plain old variables:
#define UART0 0x4000C000
#define UART0CTL (*((volatile unsigned int *) (UART0 + 0x30)))
UART0CTL &= ~1;
It's a matter of taste which might be preferable. I've worked in situations where the team wanted the registers to look like variables, and I've worked on code where the added dereference was considered 'hiding too much' so the macro for a register would be left as a pointer that had to be dereferenced explicitly (as in Chris' answer).
#define UART0 ((volatile unsigned int*)0x4000C000)
#define UART0CTL (UART0 + 0x0C)
I like to specify the actual control bits in a struct, then assign that to the control address. Something like:
typedef struct uart_ctl_t {
unsigned other_bits : 31;
unsigned disable : 1;
};
uart_ctl_t *uart_ctl = 0x4000C030;
uart_ctl->disable = 1;
(Apologies if the syntax isn't quite right, I haven't actually coded in C for quite awhile...)
Another option which I kinda like for embedded applications is to use the linker to define sections for your hardward devices and map your variable to those sections. This has the advantage that if you are targeting multiple devices, even from the same vendor such as TI, you will typically have to alter the linker files on a device by device basis. i.e. Different devices in the same family have different amounts of internal direct mapped memory, and board to board you might have different amounts of ram as well and hardware at different locations. Here's an example from the GCC documentation:
Normally, the compiler places the objects it generates in sections
like data and bss. Sometimes, however, you need additional sections,
or you need certain particular variables to appear in special
sections, for example to map to special hardware. The section
attribute specifies that a variable (or function) lives in a
particular section. For example, this small program uses several
specific section names:
struct duart a __attribute__ ((section ("DUART_A"))) = { 0 };
struct duart b __attribute__ ((section ("DUART_B"))) = { 0 };
char stack[10000] __attribute__ ((section ("STACK"))) = { 0 };
int init_data __attribute__ ((section ("INITDATA")));
main()
{
/* Initialize stack pointer */
init_sp (stack + sizeof (stack));
/* Initialize initialized data */
memcpy (&init_data, &data, &edata - &data);
/* Turn on the serial ports */
init_duart (&a);
init_duart (&b);
}
Use the section attribute with global variables and not local variables, as shown in the example.
You may use the section attribute with initialized or uninitialized
global variables but the linker requires each object be defined once,
with the exception that uninitialized variables tentatively go in the
common (or bss) section and can be multiply “defined”. Using the
section attribute will change what section the variable goes into and
may cause the linker to issue an error if an uninitialized variable
has multiple definitions. You can force a variable to be initialized
with the -fno-common flag or the nocommon attribute.
In a coding style question about infinite loops, some people mentioned they prefer the for(;;) style because the while(true) style gives warning messages on MSVC about a conditional expression being constant.
This surprised me greatly, since the use of constant values in conditional expressions is a useful way of avoiding #ifdef hell. For instance, you can have in your header:
#ifdef CONFIG_FOO
extern int foo_enabled;
#else
#define foo_enabled 0
#endif
And the code can simply use a conditional and trust the compiler to elide the dead code when CONFIG_FOO isn't defined:
if (foo_enabled) {
...
}
Instead of having to test for CONFIG_FOO every time foo_enabled is used:
#ifdef CONFIG_FOO
if (foo_enabled) {
...
}
#endif
This design pattern is used all the time in the Linux kernel (for instance, include/linux/cpumask.h defines several macros to 1 or 0 when SMP is disabled and to a function call when SMP is enabled).
What is the reason for that MSVC warning? Additionally, is there a better way to avoid #ifdef hell without having to disable that warning? Or is it an overly broad warning which should not be enabled in general?
A warning doesn't automatically mean that code is bad, just suspicious-looking.
Personally I start from a position of enabling all the warnings I can, then turn off any that prove more annoying than useful. That one that fires anytime you cast anything to a bool is usually the first to go.
I think the reason for the warning is that you might inadvertently have a more complex expression that evaluates to a constant without realizing it. Suppose you have a declaration like this in a header:
const int x = 0;
then later on, far from the declaration of x, you have a condition like:
if (x != 0) ...
You might not notice that it's a constant expression.
I believe it's to catch things like
if( x=0 )
when you meant
if( x==0 )
A simple way to avoid the warning would be:
#ifdef CONFIG_FOO
extern int foo_enabled;
#else
extern int foo_enabled = 0;
#endif