I searched the site but did not find the answer I was looking for so here is a really quick question.
I am trying to do something like that :
#ifdef _WIN32 || _WIN64
#include <conio.h>
#endif
How can I do such a thing? I know that _WIN32 is defined for both 32 and 64 bit windows so I would be okay with either for windows detection. I am more interested in whether I can use logical operators like that with preprocessor directives, and if yes how, since the above does not work.
Compiling with gcc I get :
warning: extra tokens at end of #ifdef directive , and it basically just takes the first MACRO and ignores the rest.
Try:
#if defined(_WIN32) || defined(_WIN64)
// do stuff
#endif
The defined macro tests whether or not a name is defined and lets you apply logical operators to the result.
You must use #if and special operator defined
I think it should be possible this way:
#if defined block1 || defined block2 /*or any other boolean operator*/
/*Code*/
#endif
More information here
Use defined:
#if defined(A) || defined(B)
#include <whatever.h>
#endif
Related
I'm trying to use #ifndef as below.
#ifndef MACRO1 || #ifndef MACRO2
....
#endif
I already tried:
#ifndef (MACRO1 || MACRO2)
..
#endif
But for both cases am getting below error
error: extra tokens at end of #ifndef directive
Use the #if preprocessor directive instead:
#if !defined(MACRO1) || !defined(MACRO2)
#ifdef and #ifndef are special abbreviations for #if defined(...) and #if !defined(...). However, they can only be used for a single macro and do not allow for logical operations. So, if checking for multiple macros, use #if with the defined() operator instead. Being a regular operatior, this can be combined with logical operations, as the for !defined() already does.
You can use following code
#if !defined(MACRO1) || !defined(MACRO2)
#endif
You can use the defined operator in the #if directive to use
expressions that evaluate to 0 or 1 within a preprocessor line.
You can use logical operators in preprocessor directives, but in order to check something defined, use the defined directive:
#if !defined MACRO1 || !defined MACRO2
....
#endif
I have a pre-processor directive as such,
//#define SPEC_CONTROL // Is not defined, but can be if need be
#define SPEC_A_CONTROL // Is defined
#ifdef SPEC_CONTROL || SPEC_A_CONTROL
; // do something
#else
; // do something else
#endif
Is it correct syntax to use the || in the manner I did?
I can assure yout that at least
#if defined(SPEC_CONTROL) || defined(SPEC_A_CONTROL)
works on Windows and several UNIX platforms. Im not sure about
#ifdef SPEC_CONTROL || SPEC_A_CONTROL
You can use boolean logic in a C preprocessor directive by using the if defined directive.
See this post for more information:
C Preprocessor testing definedness of multiple macros
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.
I had envisaged one of these in the project preferences
TESTING = HOST
TESTING = TARGET
TESTING not defined at all
My problem is with the latter.
It seems that instead of
#if TESTING==HOST
#error "HOST defined" // add temporarilly for testing porpoises
#endif
I need to code
#ifdef TESTING
#if TESTING==HOST
#error "HOST defined" // add temporarilly for testing porpoises
#endif
#endif
I am convinced that this is not-standard behaviour, since if TESTING is not defined then it certainly doesn't equal HOST, and I do not need that extra #ifdef TESTING with the GCC compiler.
However, when I use the Atmel AVR Studio (which I think is based on MS Visual Studio), it is necessary to add that initial #ifdef TESTING in a few dozen places :-(
It looks like I have no choice, but I just wondered if any C standard acually requires this.
#if TESTING==HOST
If TESTING is not defined, then
it is equivalent to:
#if 0==HOST
From the C Standard:
(C99, 6.10.1p4) "After all replacements due to macro expansion and the defined unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0""
And note that you can do this:
#ifndef TESTING
...
#elif TESTING == HOST
...
#elif TESTING == TARGET
...
#else
#error "Unexpected value of TESTING."
#endif
Also:
#if defined(TESTING) && TESTING == HOST
...
#endif
If you want to collapse the tests. The parenthesis are optional (#if defined TESTING is valid) but I think it's clearer to include them, especially when you start adding additional logic.
The original C preprocessors required explicit #ifdef validation before using a symbol. It is a relatively recent innovation (perhaps driven by scripting languages like Javascript) to assume that undefined symbols have a default value.
Why don't you always insure the symbol is defined?:
#ifndef TESTING
#define TESTING (default value)
#endif
#if TESTING==HOST
...
#elif TESTING==TARGET
...
#else
...
#endif
Alternative, maybe force a selection?:
#ifndef TESTING
#error You must define a value for TESTING
#endif
I had the same problem. It used to be the compiler would error if the #if parameter was not defined which is what I wanted. I learned the hard way this is no longer the case. The solution I came up with is as follows:
#define BUILDOPT 3
#if 3/BUILDOPT == 1
<compile>
#endif
If BUILDOPT is not defined you get a compile error for divide by 0. If BUILDOPT != 3 && > 0 the #if fails.
My final implementation is to set BUILDOPT to 1 to enable the code, > 1 to disable. Then the #if becomes #if 1/BUILDOPT==1.
Is there a way to do check for numerical equality in macros?
I want to do something like
#define choice 3
#if choice == 3
....
#endif
#if choice == 4
...
#endif
Does C macros have support for things like this?
Another way to write your code uses chained #elif directives:
#if choice == 3
...
#elif choice == 4
...
#else
#error Unsupported choice setting
#endif
Note that if choice is not #defined, the compiler (preprocessor) treats it as having the value 0.
Indeed that should work. See http://gcc.gnu.org/onlinedocs/cpp/If.html#If
That reference is accurate, but written in "standards format": abstractly without examples.
As far as i know that should work. What compiler are you using ?
PS : Just for information, the defines names are usually written in caps !
#define CHOICE 3