I often write code like this:
int result = someMethod(arg1,arg2,...);
assert(result==0)
Let's say assert() is defined something like this:
#ifdef DEBUG
#define assert(e) if(!e) printf("something's wrong");
#else
#define assert(...)
#endif
The first piece of code would give a warning about 'result' being an unused var.
I could do something like:
#ifdef DEBUG
int result = someMethod(arg1,arg2,...);
#else
someMethod(arg1,arg2,...);
#endif
assert(result==0)
But that seems quite non-dry to me ...
What else could I do?
int result = someMethod(arg1,arg2,...);
assert(result==0);
(void)result;
A macro like assertion_code to enable certain code pieces only in debug configuration.
#if defined(NDEBUG)
#define assertion_code(v)
#else
#define assertion_code(v) v
#endif
Now you can write
assertion_code(int result =) expr();
assert(result == 0);
Whitespace and line breaks are irrelevant to the compiler, so...
#ifdef DEBUG
int result =
#endif
someMethod(arg1,arg2,...);
assert(result==0)
There is the explicit way:
static inline void debug(const char *msg)
{
#ifdef DEBUG
printf("%s\n", msg);
#else
(void)msg;
#endif
}
Then:
if (someMethod(arg1,arg2,...) != 0) {
debug(message);
}
As there is a fair chance you won't want to continue as normal when something is wrong, the explicit conditional clause is perhaps going to be useful anyway. Or were you planning to ignore errors when DEBUG is turned off?
Related
Using #ifdef <macro> <statement> #endif allows one to have verbose messages displayed only during development and is quite handy. I wonder if something like the code below is possible, becoming even shorter:
// pseudo-code:
#define IN_DEV
#define DEBUG_ONLY(statement) (#ifdef IN_DEV (statement) #endif)
int main(void)
{
DEBUG_ONLY(printf("hello from debug mode\n");)
//...
}
This would only cost me a very readable one-liner which can be turned on or off. Is something like this / close to this possible?
You could change the meaning of DEBUG_ONLY dependent on if IN_DEV is defined:
// pseudo-code:
#ifdef IN_DEV
#define DEBUG_ONLY(statement) {statement}
#else
#define DEBUG_ONLY(statement) // Nothing
#endif
int main(void)
{
DEBUG_ONLY(printf("hello from debug mode\n");)
//...
}
Running example: Link
It doesn't make much sense to pass whole expressions as parameters to macros. That is both dangerous and unmaintainable. Worse yet, it's taking us down the obfuscation road of "lets invent our own private macro language instead of using readable C that anyone can understand". That's a terrible idea even for debugging purposes.
The more sensible approach would be a function-like printing macro which only prints something in debug build.
#ifdef INDEV
#define DEBUG_PRINT(...) printf(__VA_ARGS__)
#else
#define DEBUG_PRINT(...)
#endif
int main(void)
{
DEBUG_PRINT("hello from debug mode\n");
}
Optionally this macro can be narrowed down to only accept strings and be made more type safe (C17 only):
#ifdef INDEV
#define DEBUG_PRINT(str) _Generic((str), char*: puts(str))
#else
#define DEBUG_PRINT(str) _Generic((str), char*: (void)0)
#endif
That is not possible. You cannot use #if inside a macro definition.
What you can do is this:
#define IN_DEV
#ifdef
#define DEBUG_ONLY(statement) (statement)
#else
#define DEBUG_ONLY(statement)
#endif
int main(void)
{
DEBUG_ONLY(printf("hello from debug mode\n");)
//...
}
This is also switchable with only a single macro IN_DEV.
I'm working on some C code that does lot's of error reporting and logging when a DEBUG flag is set, which sometimes produces unused variable warnings when compiling with the DEBUG flag not set.
#ifdef DEBUG
#define CHECK(expr) foo(expr)
#else
#define CHECK(expr)
#endif /* DEBUG */
int x = bar(a, b, c); /* bar has to be called for both DEBUG begin defined and undefined */
CHECK(x == SOME_VALUE); /* Produces an "unused variable" warning if DEBUG is undefined
Edit: Just a little reminder (not sure if it is of any consequence): the argument for the CHECK macro is an expression, not a single variable.
For this pattern, what is the best way to get rid of the unused variable warning?
What I tried/though of:
#ifdef DEBUG
int x = bar(a, b, c);
#else
bar(a, b, c);
#endif
CHECK(x == SOME_VALUE);
and then, to avoid writing the call to bar (which is more complicated in the actual call) twice:
#ifdef DEBUG
int x =
#endif
bar(a, b, c);
CHECK(x == SOME_VALUE);
However, I feel like this is not exactly a clean and readable solution. Is there a better way? Note that for performance reasons the CHECK(expr) macro should not produce any code if DEBUG is undefined (EDIT: and thus, expr should not be evaluated).
Is there a more elegant way than the one I outlined above?
#ifdef DEBUG
#define CHECK(x) x
#else
#define CHECK(x) ((void)sizeof((void)(x),0))
#endif
I think this addresses all of the possible issues :
sizeof ensures that the expression is not evaluated at all, so its side-effects don't happen. That is to be consistent with the usual behaviour of debug-only constructs, such as assert.
((x), 0) uses the comma operator to swallow the actual type of (x). This is to prevent VLAs from triggering evaluation.
(void) explicitly ignores the result of (x) and sizeof so no "unused value" warning appears.
If I understood your question correctly, you can do something like
#ifdef DEBUG
.
.
#else
#define CHECK(expr) ((void)(expr))
#endif /* DEBUG */
to get rid of the warning.
With this solution there is no need for an intermediate variable.
#define DEBUG
#define DOCHECK(a) printf("DOCHECK %d\n", a)
#ifndef DEBUG
#define CHECK(a, b) a
#else
#define CHECK(a, b) do {int x = a; DOCHECK(x == b);} while (0)
#endif
int bar(int x, int y)
{
return x+y;
}
int main()
{
CHECK(bar(2,3), 2+3);
return 0;
}
#include <stdio.h>
#define LOG_D(x) { printf("D:"); printf(x);}
#define LOG_E(x) { printf("E:"); printf(x);}
void test(void)
{
LOG_D("ALL is well " );
}
I have a very huge code it has different levels of log, like above code.
In the final tested library I just need only one error logs in order to reduce the code size .
so I want something like this
#define ENABLE_DEBUG_LOG 0
#define ENABLE_ERROR_LOG 1
#define LOG_D(x) {#if(ENABLE_DEBUG_LOG==1) printf("D:"); printf(x); #endif}
#define LOG_E(x) {#if(ENABLE_ERROR_LOG==1) printf("E:"); printf(x);#endif}
I added this #if(ENABLE_DEBUG_LOG==1) just for explaining, I need some solution which can compile.
Another option - you can just comment / uncomment ENABLE_DEBUG_LOG and ENABLE_ERROR_LOG to disable / enable corresponding log level.
// #define ENABLE_DEBUG_LOG // disable DEBUG_LOG
#define ENABLE_ERROR_LOG // enable ERROR_LOG
#ifdef ENABLE_DEBUG_LOG
#define LOG_D(x) { printf("D:"); printf(x);}
#else
#define LOG_D(x) // nothing
#endif
#ifdef ENABLE_ERROR_LOG
#define LOG_E(x) { printf("E:"); printf(x);}
#else
#define LOG_E(x) // nothing
#endif
You cannot nest preprocessor directives. But you can make two versions of your macro and define them in exclusive parts of an #if or #ifdef:
#define ENABLE_DEBUG_LOG 0
#if ENABLE_DEBUG_LOG != 0
#define LOG_D(...) printf("D: " __VA_ARGS__)
#else
#define LOG_D(...) // Do nothing
#endif
Here, the disabled version just "eats" the LOG_D macro and doesn't do anything. (Note that undefined macros are treated as the value 0 in #if conditionals.)
You should be able to do something like this:
#if ENABLE_DEBUG_LOG == 1
# define LOG_D(x) { printf("D:"); printf(x);}
#else
# define LOG_D(x)
#end
That way the debug log statements will just disappear if ENABLE_DEBUG_LOG is undefined or has a different value.
Regarding the other answers, it is not good idea to define the macros completely empty when they are not enabled, as this would go wrong when error logging is enabled:
if (some_error)
LOG_E("Oops...");
do_something();
If LOG_E(x) expands to nothing, then do_something() would only be called if some_error is true, which is probably not what you want!
So you could define the "do nothing" variant of LOG_E(x) like this:
#define LOG_E(x) { }
Rather than starting and ending with braces, I tend to use the do { blah; } while (0) construct as it forces you to put a semicolon on the end when you use it. Something like this:
#if ENABLE_ERROR_LOG
#define LOG_E(x) do { printf("E:"); printf(x); } while (0)
#else
#define LOG_E(x) do ; while (0)
#endif
Then,
if (some_error)
LOG_E("Oops")
would result in a syntax error because of the missing semicolon, forcing you to write it as
if (some_error)
LOG_E("Oops");
Another thing you can do is concatenate the "E:" or "D:" tag with the passed in string, although this requires the parameter to be a string literal, rather than a general char *:
#define LOG_E(x) printf("E:" x)
Another thing you can do is to define the macro with a variable number of parameters (a variadic macro) to increase your options:
#define LOG_E(...) printf("E:" __VA_ARGS__)
Then you can do:
if (some_error)
LOG_E("Oops, got error: %d\n", some_error);
Another thing you can do is let the compiler optimize out the call to printf and define it like this:
#define LOG_E(...) do if (ENABLE_ERROR_LOG) printf("E:" __VA_ARGS__); while (0)
A decent compiler will notice that the if condition is constant and either optimize out the call to printf completely (if the constant condition is false), or include it (if the constant condition is true). For some compilers, you might need to suppress warnings about constant conditions in an if statement.
I am not sure if this is what you want, but you could check the #ifdef directive.
#include <stdio.h>
/* #define DEBUG */
#ifdef DEBUG
#define LOG_D(x) { printf("D: %s\n",x); }
#define LOG_E(x) { printf("E: %s\n",x); }
#else
#define LOG_D(x)
#define LOG_E(x)
#endif
int main() {
LOG_D("blah...");
return 0;
}
If you uncomment the #define DEBUG line, the program will print D: blah...
I have the following code, and while compiling it with gcc-4.6 I get warning:
warning: variable ‘status’ set but not used [-Wunused-but-set-variable]
#if defined (_DEBUG_)
#define ASSERT assert
#else /* _DEBUG_ */
#define ASSERT( __exp__ )
#endif
static inline void cl_plock(cl_plock_t * const p_lock)
{
status_t status;
ASSERT(p_lock);
ASSERT(p_lock->state == INITIALIZED);
status = pthread_rwlock_unlock(&p_lock->lock);
ASSERT(status == 0);
}
When _DEBUG_
flag isn't set I get the warning.
Any ideas how can I workaround this warning?
You can change your ASSERT macro to:
#if defined (_DEBUG_)
#define ASSERT assert
#else /* _DEBUG_ */
#define ASSERT( exp ) ((void)(exp))
#endif
If the expression has no sideeffects, then it should still be optimised out, but it should also suppress the warning (if the expression does have side-effects, then you would get different results in debug and non-debug builds, which you don't want either!).
The compiler option to turn off unused variable warnings is -Wno-unused.
To get the same effect on a more granular level you can use diagnostic pragmas like this:
int main()
{
#pragma GCC diagnostic ignored "-Wunused-variable"
int a;
#pragma GCC diagnostic pop
// -Wunused-variable is on again
return 0;
}
This is, of course, not portable but you can use something similar for VS.
You could surround the variable declaration of status with a #ifdef clause.
#ifdef _DEBUG_
status_t status
#endif
EDIT: You have to surround the call also:
#ifdef _DEBUG_
status = pthread_rwlock_unlock(&p_lock->lock);
#else
pthread_rwlock_unlock(&p_lock->lock);
#endif
or you can switch off the error message.
I'm trying to do a debug system but it seems not to work.
What I wanted to accomplish is something like this:
#ifndef DEBUG
#define printd //
#else
#define printd printf
#endif
Is there a way to do that? I have lots of debug messages and I won't like to do:
if (DEBUG)
printf(...)
code
if (DEBUG)
printf(...)
...
No, you can't. Comments are removed from the code before any processing of preprocessing directives begin. For this reason you can't include comment into a macro.
Also, any attempts to "form" a comment later by using any macro trickery are not guaranteed to work. The compiler is not required to recognize "late" comments as comments.
The best way to implement what you want is to use macros with variable arguments in C99 (or, maybe, using the compiler extensions).
A common trick is to do this:
#ifdef DEBUG
#define OUTPUT(x) printf x
#else
#define OUTPUT(x)
#endif
#include <stdio.h>
int main(void)
{
OUTPUT(("%s line %i\n", __FILE__, __LINE__));
return 0;
}
This way you have the whole power of printf() available to you, but you have to put up with the double brackets to make the macro work.
The point of the double brackets is this: you need one set to indicate that it's a macro call, but you can't have an indeterminate number of arguments in a macro in C89. However, by putting the arguments in their own set of brackets they get interpreted as a single argument. When the macro is expanded when DEBUG is defined, the replacement text is the word printf followed by the singl argument, which is actually several items in brackets. The brackets then get interpreted as the brackets needed in the printf function call, so it all works out.
С99 way:
#ifdef DEBUG
#define printd(...) printf(__VA_ARGS__)
#else
#define printd(...)
#endif
Well, this one doesn't require C99 but assumes compiler has optimization turned on for release version:
#ifdef DEBUG
#define printd printf
#else
#define printd if (1) {} else printf
#endif
On some compilers (including MS VS2010) this will work,
#define CMT / ## /
but no grantees for all compilers.
You can put all your debug call in a function, let call it printf_debug and put the DEBUG inside this function.
The compiler will optimize the empty function.
The standard way is to use
#ifndef DEBUG
#define printd(fmt, ...) do { } while(0)
#else
#define printd(fmt, ...) printf(fmt, __VA_ARGS__)
#endif
That way, when you add a semi-colon on the end, it does what you want.
As there is no operation the compiler will compile out the "do...while"
Untested:
Edit: Tested, using it by myself by now :)
#define DEBUG 1
#define printd(fmt,...) if(DEBUG)printf(fmt, __VA_ARGS__)
requires you to not only define DEBUG but also give it a non-zer0 value.
Appendix:
Also works well with std::cout
In C++17 I like to use constexpr for something like this
#ifndef NDEBUG
constexpr bool DEBUG = true;
#else
constexpr bool DEBUG = false;
#endif
Then you can do
if constexpr (DEBUG) /* debug code */
The caveats are that, unlike a preprocessor macro, you are limited in scope. You can neither declare variables in one debug conditional that are accessible from another, nor can they be used at outside function scopes.
You can take advantage of if. For example,
#ifdef debug
#define printd printf
#else
#define printd if (false) printf
#endif
Compiler will remove these unreachable code if you set a optimization flag like -O2. This method also useful for std::cout.
As noted by McKay, you will run into problems if you simply try to replace printd with //. Instead, you could use variadric macros to replace printd with a function that does nothing as in the following.
#ifndef DEBUG
#define printd(...) do_nothing()
#else
#define printd(...) printf(__VA_ARGS__)
#endif
void do_nothing() { ; }
Using a debugger like GDB might help too, but sometimes a quick printf is enough.
I use this construct a lot:
#define DEBUG 1
#if DEBUG
#if PROG1
#define DEBUGSTR(msg...) { printf("P1: "); printf( msg); }
#else
#define DEBUGSTR(msg...) { printf("P2: "); printf( msg); }
#endif
#else
#define DEBUGSTR(msg...) ((void) 0)
#endif
This way I can tell in my console which program is giving which error message... also, I can search easily for my error messages...
Personally, I don't like #defining just part of an expression...
It's been done. I don't recommend it. No time to test but the mechanism is kind of like this:
#define printd_CAT(x) x ## x
#ifndef DEBUG
#define printd printd_CAT(/)
#else
#define printd printf
#endif
This works if your compiler processes // comments in the compiler itself (there's no guarantee like the ANSI guarantee that there are two passes for /* comments).