I have a C program with a lot of optimizations that can be enabled or disabled with #defines. When I run my program, I would like to know what macros have been defined at compile time.
So I am trying to write a macro function to print the actual value of a macro. Something like this:
SHOW_DEFINE(X){\
if( IS_DEFINED(X) )\
printf("%s is defined and as the value %d\n", #X, (int)X);\
else\
printf("%s is not defined\n", #X);\
}
However I don't know how to make it work and I suspect it is not possible, does anyone has an idea of how to do it?
(Note that this must compile even when the macro is not defined!)
As long as you are willing to put up with the fact that SOMESTRING=SOMESTRING indicates that SOMESTRING has not been defined (view it as the token has not been redefined!?!), then the following should do:
#include <stdio.h>
#define STR(x) #x
#define SHOW_DEFINE(x) printf("%s=%s\n", #x, STR(x))
#define CHARLIE -6
#define FRED 1
#define HARRY FRED
#define NORBERT ON_HOLIDAY
#define WALLY
int main()
{
SHOW_DEFINE(BERT);
SHOW_DEFINE(CHARLIE);
SHOW_DEFINE(FRED);
SHOW_DEFINE(HARRY);
SHOW_DEFINE(NORBERT);
SHOW_DEFINE(WALLY);
return 0;
}
The output is:
BERT=BERT
CHARLIE=-6
FRED=1
HARRY=1
NORBERT=ON_HOLIDAY
WALLY=
Writing a MACRO that expands to another MACRO would require the preprocessor to run twice on it.
That is not done.
You could write a simple file,
// File check-defines.c
int printCompileTimeDefines()
{
#ifdef DEF1
printf ("defined : DEF1\n");
#else // DEF1
printf ("undefined: DEF1\n");
#endif // DEF1
// and so on...
}
Use the same Compile Time define lines on this file as with the other files.
Call the function sometime at the start.
If you have the #DEFINE lines inside a source file rather than the Makefile,
Move them to another Header file and include that header across all source files,
including this check-defines.c.
Hopefully, you have the same set of defines allowed across all your source files.
Otherwise, it would be prudent to recheck the strategy of your defines.
To automate generation of this function,
you could use the M4 macro language (or even an AWK script actually).
M4 becomes your pre-pre-processor.
#ifdef MYMACRO
printf("MYMACRO defined: %d\r\n", MYMACRO);
#endif
I don't think what you are trying to do is possible. You are asking for info at runtime which has been processed before compilation. The string "MYMACRO" means nothing after CPP has expanded it to its value inside your program.
You did not specify the compiler you were using. If this is gcc, gcc -E may help you, as it stops after the preprocessing stage, and prints the preprocessing result. If you diff a gcc -E result with the original file, you may get part of what you want.
Why not simply testing it with the preprocessor ?
#if defined(X)
printf("%s is defined and as the value %d\n", #X, (int)X);
#else
printf("%s is not defined\n", #X);
#endif
One can also embed this in another test not to print it everytime:
#if define(SHOW_DEFINE)
#if defined(X)
printf("%s is defined and as the value %d\n", #X, (int)X);
#else
printf("%s is not defined\n", #X);
#endif
#endif
Wave library from boost can be helpful also to do what you want. If your project is big enough, I think it is worth trying. It is a C++ preprocessor, but they are the same any way :)
I believe what you are trying to do is not possible. If you are able to change the way your #defines work, then I suggest something like this:
#define ON 1
#define OFF 2
#define OPTIMIZE_FOO ON
#define OPTIMIZE_BAR OFF
#define SHOW_DEFINE(val)\
if(val == ON) printf(#val" is ON\n");\
else printf(#val" is OFF\n");
It is not possible indeed, the problem being the "not defined" part. If you'd relied on the macro values to activate/deactivate parts of your code, then you could simply do:
#define SHOW_DEFINE(X) \
do { \
if (X > 0) \
printf("%s %d\n", #X, (int)X); \
else \
printf("%s not defined\n", #X); \
} while(0)
Based on Chrisharris, answer I did this.
Though my answer is not very nice it is quite what I wanted.
#define STR(x) #x
#define SHOW_DEFINE(x) printf("%s %s\n", #x, strcmp(STR(x),#x)!=0?"is defined":"is NOT defined")
Only bug I found is the define must not be #define XXX XXX (with XXX equals to XXX).
This question is very close from Macro which prints an expression and evaluates it (with __STRING). Chrisharris' answer is very close from the answer to the previous question.
you can use an integer variable initialized to 1.
multiply the #define with this variable and print the value of variable.
Related
Is there any idea to pass C string as part of the defined macro like below code?
#define AAA_NUM 10
#define BBB_NUM 20
#define PREFIX_NUM(string) string##_NUM
int main()
{
char *name_a = "AAA";
char *name_b = "AAA";
printf("AAA_NUM: %d\n", PREFIX_NUM(name_a));
printf("BBB_NUM: %d\n", PREFIX_NUM(name_b));
return 0;
}
Expected output
AAA_NUM: 10
BBB_NUM: 20
As mentioned in other posts, you can't use run-time variables in the pre-processor. You could however create enum that way. Though it is usually not a good idea to generate identifiers with macros either, save for special cases like when maintaining an existing code base and you are limited in how much of the existing code you can/want to change. So it should be used as a last resort only.
The least bad way to write such macros would be by using a common design pattern called "X macros". These are used when it is important that code repetition should be reduced to a single place in the project. They tend to make the code look rather alien though... Example:
#define PREFIX_LIST(X) \
/* pre val */ \
X(AAA, 10) \
X(BBB, 20) \
X(CCC, 30) \
enum // used to generate constants like AAA_NUM = 10,
{
#define PREFIX_ENUMS(pre, val) pre##_NUM = (val),
PREFIX_LIST(PREFIX_ENUMS)
};
#include <stdio.h>
int main (void)
{
// one way to print
#define prefix_to_val(pre) pre##_NUM
printf("AAA_NUM: %d\n", prefix_to_val(AAA));
printf("BBB_NUM: %d\n", prefix_to_val(BBB));
// another alternative
#define STR(s) #s
#define print_all_prefixes(pre, val) printf("%s: %d\n", STR(pre##_NUM), val);
PREFIX_LIST(print_all_prefixes)
return 0;
}
A macro is only processed before compilation and not at runtime. Your code example does not work as you can see here.
Good practice (for example MISRA coding rules) recommend to use macros as little as possible since it is error prone.
Preprocessor works at compile time and here name_a and name_b are non constant, and even if they were (i.e. const char *str is a real constant in C++ but not in C), there is a literal substitution and the preprocessor does not know the contents of variables.
This works (notice that the parameter should be expanded by another macro in order to get a valid token):
#include <stdio.h>
#define AAA_NUM 10
#define BBB_NUM 20
#define _PREFIX_NUM(string) string##_NUM
#define PREFIX_NUM(string) _PREFIX_NUM(string)
int main(void)
{
#define name_a AAA
#define name_b BBB
printf("AAA_NUM: %d\n", PREFIX_NUM(name_a));
printf("BBB_NUM: %d\n", PREFIX_NUM(name_b));
return 0;
}
There is no way in C to create runtime symbols and use them. C is a compiled language and all symbols have to be known before the compilation.
The preprocessor (which do changes on the text level before the compilation) does not know anything about the C language.
What does the following statement mean:
#define FAHAD
I am familiar with the statements like:
#define FAHAD 1
But what does the #define statement without a token signify?
Is it that it is similar to a constant definition?
Defining a constant without a value acts as a flag to the preprocessor, and can be used like so:
#define MY_FLAG
#ifdef MY_FLAG
/* If we defined MY_FLAG, we want this to be compiled */
#else
/* We did not define MY_FLAG, we want this to be compiled instead */
#endif
it means that FAHAD is defined, you can later check if it's defined or not with:
#ifdef FAHAD
//do something
#else
//something else
#endif
Or:
#ifndef FAHAD //if not defined
//do something
#endif
A real life example use is to check if a function or a header is available for your platform, usually a build system will define macros to indicate that some functions or headers exist before actually compiling, for example this checks if signal.h is available:
#ifdef HAVE_SIGNAL_H
# include <signal.h>
#endif/*HAVE_SIGNAL_H*/
This checks if some function is available
#ifdef HAVE_SOME_FUNCTION
//use this function
#else
//else use another one
#endif
Any #define results in replacing the original identifier with the replacement tokens. If there are no replacement tokens, the replacement is empty:
#define DEF_A "some stuff"
#define DEF_B 42
#define DEF_C
printf("%s is %d\n", DEF_A, DEF_B DEF_C);
expands to:
printf("%s is %d\n", "some stuff", 42 );
I put a space between 42 and ) to indicate the "nothing" that DEF_C expanded-to, but in terms of the language at least, the output of the preprocessor is merely a stream of tokens. (Actual compilers generally let you see the preprocessor output. Whether there will be any white-space here depends on the actual preprocessor. For GNU cpp, there is one.)
As in the other answers so far, you can use #ifdef to test whether an identifier has been #defined. You can also write:
#if defined(DEF_C)
for instance. These tests are positive (i.e., the identifier is defined) even if the expansion is empty.
#define FAHAD
this will act like a compiler flag, under which some code can be done.
this will instruct the compiler to compile the code present under this compiler option
#ifdef FAHAD
printf();
#else
/* NA */
#endif
I regularly use object-like preprocessor macros as boolean flags in C code to turn on and off sections of code.
For example
#define DEBUG_PRINT 1
And then use it like
#if(DEBUG_PRINT == 1)
printf("%s", "Testing");
#endif
However, it comes a problem if the header file that contains the #define is forgotten to be included in the source code. Since the macro is not declared, the preprocessor treats it as if it equals 0, and the #if statement never runs.
When the header file is forgotten to be included, non-expected, unruly behaviour can occur.
Ideally, I would like to be able to both check that a macro is defined, and check that it equals a certain value, in one line. If it is not defined, the preprocessor throws an error (or warning).
I'm looking for something along the lines of:
#if-def-and-true-else-throw-error(DEBUG_PRINT)
...
#endif
It's like a combination of #ifdef and #if, and if it doesn't exist, uses #error.
I have explored a few avenues, however, preprocessor directives can't be used inside a #define block, and as far as I can tell, there is no preprocessor option to throw errors/warnings if a macro is not defined when used inside a #if statement.
This may not work for the general case (I don't think there's a general solution to what you're asking for), but for your specific example you might consider changing this sequence of code:
#if(DEBUG_PRINT == 1)
printf("%s", "Testing");
#endif
to:
if (DEBUG_PRINT == 1) {
printf("%s", "Testing");
}
It's no more verbose and will fail to compile if DEBUG_PRINT is not defined or if it's defined to be something that cannot be compared with 1.
as far as I can tell, there is no preprocessor option to throw errors/warnings if a macro is not defined when used inside a #if statement.
It can't be an error because the C standard specifies that behavior is legal. From section 6.10.1/3 of ISO C99 standard:
After all replacements due to macro expansion and the defined unary
operator have been performed, all remaining identifiers are replaced with the pp-number
0....
As Jim Balter notes in the comment below, though, some compilers (such as gcc) can issue warnings about it. However, since the behavior of substituting 0 for unrecognized preprocessor tokens is legal (and in many cases desirable), I'd expect that enabling such warnings in practice would generate a significant amount of noise.
There's no way to do exactly what you want. If you want to generate a compilation failure if the macro is not defined, you'll have to do it explicitly
#if !defined DEBUG_PRINT
#error DEBUG_PRINT is not defined.
#endif
for each source file that cares. Alternatively, you could convert your macro to a function-like macro and avoid using #if. For example, you could define a DEBUG_PRINT macro that expands to a printf call for debug builds but expands to nothing for non-debug builds. Any file that neglects to include the header defining the macro then would fail to compile.
Edit:
Regarding desirability, I have seen numerous times where code uses:
#if ENABLE_SOME_CODE
...
#endif
instead of:
#ifdef ENABLE_SOME_CODE
...
#endif
so that #define ENABLE_SOME_CODE 0 disables the code rather than enables it.
Rather than using DEBUG_PRINT directly in your source files, put this in the header file:
#if !defined(DEBUG_PRINT)
#error DEBUG_PRINT is not defined
#endif
#if DEBUG_PRINT
#define PrintDebug([args]) [definition]
#else
#define PrintDebug
#endif
Any source file that uses PrintDebug but doesn't include the header file will fail to compile.
If you need other code than calls to PrintDebug to be compiled based on DEBUG_PRINT, consider using Michael Burr's suggestion of using plain if rather than #if (yes, the optimizer will not generate code within a false constant test).
Edit:
And you can generalize PrintDebug above to include or exclude arbitrary code as long as you don't have commas that look like macro arguments:
#if !defined(IF_DEBUG)
#error IF_DEBUG is not defined
#endif
#if IF_DEBUG
#define IfDebug(code) code
#else
#define IfDebug(code)
#endif
Then you can write stuff like
IfDebug(int count1;) // IfDebug(int count1, count2;) won't work
IfDebug(int count2;)
...
IfDebug(count1++; count2++;)
Yes you can check both:
#if defined DEBUG && DEBUG == 1
# define D(...) printf(__VA_ARGS__)
#else
# define D(...)
#endif
In this example even when #define DEBUG 0 but it is not equal to 1 thus nothing will be printed.
You can do even this:
#if defined DEBUG && DEBUG
# define D(...) printf(__VA_ARGS__)
#else
# define D(...)
#endif
Here if you #define DEBUG 0 and then D(1,2,3) also nothing will be printed
DOC
Simply create a macro DEBUG_PRINT that does the actual printing:
#define DEBUG_PRINT(n, str) \
\
if(n == 1) \
{ \
printf("%s", str); \
} \
else if(n == 2) \
{ \
do_something_else(); \
} \
\
#endif
#include <stdio.h>
int main()
{
DEBUG_PRINT(1, "testing");
}
If the macro isn't defined, then you will get a compiler error because the symbol is not recognized.
#if 0 // 0/1
#define DEBUG_PRINT printf("%s", "Testing")
#else
#define DEBUG_PRINT printf("%s")
#endif
So when "if 0" it'll do nothing and when "if 1" it'll execute the defined macro.
What MACRO can be used to switch off printf statements, rather than removing them all for deployment builds, I just want to switch them off, skip them, ignore them.
EDIT: I personally use gcc, but code is part of a larger project which will be compiled on a Panda board running Ubuntu.
Not exactly what you ask for, but I use this construct in my code for debug output when I do not have a proper logging system handy:
#if 1
#define SPAM(a) printf a
#else
#define SPAM(a) (void)0
#endif
So I can do this all over my code
SPAM(("foo: %d\n", 42));
and then disable all of them by changing 1 to 0 in #if above.
But if you have variadic macro support in all compilers that you write code for, then you may go for other answers and just redefine printf. (That being said, I find it useful to distinct debugging prints from regular ones in code — using a different function name helps readability.)
Note that you also can redirect stdout to the /dev/null, but I assume that you want to get rid from runtime overhead as well.
#ifdef IGNORE_PRINTF
#define printf(fmt, ...) (0)
#endif
See also C #define macro for debug printing which discusses some important issues closely related to this.
Two options, either:
#define printf(...)
(requires C99 variadic macro parameters), you need to put it in some common header file which is never included before stdio.h, if there is one..
Or you can tell the linker to link it to something else, in GCC you would define
int wrap_printf(void) {return 0;}
and link using
--wrap printf
All that said, you should probably not be using printf for printing debug output, but rather a macro or utility function (which in turn can use printf if you'd like) which you have better control over.
Hope that helps.
If you want to avoid the potential warning that Jonathan's answer may give you and if you don't mind an empty call to printf you could also do something like
#define printf(...) printf("")
This works because C macros are not recursive. The expanded printf("") will just be left as such.
Another variant (since you are using gcc) would be something like
inline int ignore_printf(char const*, ...)
__attribute__ ((format (printf, 1, 2)));
inline int ignore_printf(char const*, ...) { return 0; }
#define printf ignore_printf
and in one compilation unit
int ignore_printf(char const*, ...)
I use to prefix the debug printf()s (not all of them) with PDEB.
For the debug builds, I compile with -DPDEB= (nothing)
For the release builds, I compile with -DPDEB="0&&" or -DPDEB="0 && "
That way, the following code (test.c):
#include <stdio.h>
void main(void) {
printf("normal print\n");
PDEB printf("debug print\n");
}
outputs:
either (in release mode):
normal print
either (in debug mode):
normal print
debug print
Ideally, one could aim for turning the PDEB into the "//" (comments mark), except that this is not possible under the standard pre-/processing chain.
Another possibility would be something like freopen("/dev/null", "w", stdout);
This doesn't exactly disable printf though -- it's roughly equivalent to running your program with stdout redirected to /dev/null, like: ./myprog > /dev/null at the shell prompt.
I included #define printf // in common header file. It will suppress all the printf.
Below simple function serves the purpose, I use the same.
int printf(const char *fmt, ...)
{
return (0)
}
Use this macro to enable or disable the printf.
//Uncomment the following line to enable the printf function.
//#define ENABLE_PRINTF
#ifdef ENABLE_PRINTF
#define DEBUG_PRINTF(f,...) printf(f,##__VA_ARGS__)
#else
#define DEBUG_PRINTF(f,...)
#endif
Then call "DEBUG_PRINTF" instead of "printf".
For example:
DEBUG_PRINTF("Hello world: %d", whateverCount);
I have used two macros for this. The first one defines the condition to print. In this simple example we print any time the parameter is not zero. More complex expressions can be used.
The second one determines, based on the first macro, to call or not printf.
If the condition can be determined by the compiler (with the right optimization settings) no code is generated.
If the condition cannot be determined at compile time then will be at run time. One of the advantages of this method is that if printf is not going to happen then the whole printf is not evaluated avoiding many conversions to string that can happen in a complex printf statement.
#define need_to_print(flag) ((flag) != 0))
#define my_printf(debug_level, ...) \
({ \
if(need_to_print(debug_level)) \
printf(__VA_ARGS__); \
})
to use it call my_printf instead of printf and add a parameter at the beginning for the print condition.
my_printf(0, "value = %d\n", vv); //this will not print
my_printf(1, "value = %d\n", vv); //this will print
my_printf(print_debug, "value = %d\n", vv); //this will print if print_debug != 0
the ( ... ) parenthesis surrounding the macro make it a single statement.
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).