I am using a #ifdef block to conditionally include a block of code based on -Dflag passed to the compiler.
Is there a way to implement this with an if?
Something along the lines of:
if (defined(flag)) {
}
You use preprocessor to produce a different flag, which could be tested with a run-time if statement, like this:
#ifdef flag
#define flag_defined 1
#else
#define flag_defined 0
#endif
Now you can do this:
if (flag_defined) ...
If you define a macro so that it evaluates to either zero or nonzero, then you can use it in the condition of the if statement. Since it's a compile-time constant, in case it's zero, the dead code eliminator pass in the compiler will remove the unreachable code anyway.
If you are willing to give flag a value (0 or 1) instead of defining it or not, you can do
cc -Dflag=1 file.c
with
if (flag) { ... }
without messing up your code with ugly #ifdefs. You are also protected against forgetting to define flag at all: in this case your compiler treats flag as an identifier. Since you never declared it, you'll get a compile time error about an undeclared or undefined symbol flag.
No, you can't use a C if statement to check if a preprocessor token is defined.
You can use one of these forms though
#ifdef flag
...
#endif
#if defined(flag)
...
#endif
You can however check if the token evaluates to a true/false C expression,
if you use -Dflag=1 you can do
if (flag) {
...
}
And if you want to turn it off, you can define the macro as -Dflag=0
Following this you can define a new preprocessor token that you can check with a C if statement. e.g.
#ifdef flag
#define FLAG_SET 1
#else
#define FLAG_SET 0
#endif
if(FLAG_SET) { //the flag macro was defined
...
}
If all you want to do is is check whether the flag is defined, there's no point to all of this, just use #ifdef. With a C if statement, the most C compilers will optimize away the code, there's nothing you could change at runtime to get any benefit of not using the preprocessor.
#ifdef FLAG
#define FLAG_VALUE 1
#else
#define FLAG_VALUE 0
#endif
int flag = FLAG_VALUE;
...
if(flag)
{
....
}
Alternatively just use -DFLAG_VALUE=1 or -DFLAG_VALUE=0 as appropriate and use FLAG_VALUE directly.
If you plan on being able to change the FLAG_VALUE at runtime you will need a variable. If you change in non-obvious ways (e.g. via a debugger or through other loader trickery) make sure to declare it volatile, otherwise the compiler might be able to do dead-code elimination and remove huge chunks of code.
Also, if you don't plan on changing the value of flag after initializing, then you should probably make it const.
Related
I want to deprecate a macro in such a way that it will print a nice warning even if used inside of a #if statement.
This answer is very nearly what I want, but it throws an error when the macro is accessed from within a #if statement.
#include <stdio.h>
#define DEPRECATED_CONSTANT _Pragma ("GCC warning \"Deprecated constant!\"") 0
#define DEPRECATED_FUNCTION(...) _Pragma ("GCC warning \"Deprecated function!\"") printf(__VA_ARGS__)
int main() {
// Prints a warning (good)
int n = DEPRECATED_CONSTANT;
// Prints a warning (good)
DEPRECATED_FUNCTION("%d\n", n);
// Throws an error (bad)
#if DEPRECATED_CONSTANT
return 1;
#else
return 2;
#endif
}
The error is:
error: missing binary operator before token "("
Bonus points if you can find me a cross-platform compatible solution!
EDIT
I'm trying to handle a breaking change in a library gracefully - I want users to have a nice, clear warning (or error) whenever they use an old macro, so it will be clear that they need to migrate their code to using the new macro. These pragma solutions only work if the value of that constant is used in code, not if the value is accessed in a preprocessor directive.
According to the answers provided below, it seems like there's not a solution to this problem (except possibly when using clang?). Thanks, everyone.
I want to deprecate a macro in such a way that it will print a nice
warning even if used inside of a #if statement.
I was going to suggest the comma operator, but that doesn't seem to work because the _Pragma macro probably yields no real code. Also, gcc, at least, explicitly says you can't do what you suggested with _Pragma():
https://gcc.gnu.org/onlinedocs/cpp/Pragmas.html
The standard is unclear on where a _Pragma operator can appear. The
preprocessor does not accept it within a preprocessing conditional
directive like ‘#if’. To be safe, you are probably best keeping it out
of directives other than ‘#define’, and putting it on a line of its
own.
PS - clang 8.1.0 didn't error on your program and gave the warnings you want ...
As #jschultz410 mentions, what you are trying to do is explicitly forbidden in gcc (see https://gcc.gnu.org/onlinedocs/cpp/Pragmas.html).
Nested macros seem appropriate for such a use case:
#include <stdio.h>
#define DEPRECATED_CONSTANT_VALUE 0
#define DEPRECATED_CONSTANT _Pragma ("GCC warning \"Deprecated constant!\"") DEPRECATED_CONSTANT_VALUE
#define DEPRECATED_FUNCTION(...) _Pragma ("GCC warning \"Deprecated function!\"") printf(__VA_ARGS__)
int main() {
// Prints a warning (good)
int n = DEPRECATED_CONSTANT;
// Prints a warning (good)
DEPRECATED_FUNCTION("%d\n", n);
// Throws an error (bad)
#if DEPRECATED_CONSTANT_VALUE
return 1;
#else
return 2;
#endif
}
Yes, this is kind of gross, but in the land of preprocessor logic we're already giving up on any kind of design elegance in the first place. At least this way the macro interface is maintained in non-preprocessor code. (Yes, this would not print the preprocessor warning in the #if statement, but unfortunately that's not possible with gcc).
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
How does the following code work?
#define ENABLE_DEBUG 1
#if ENABLE_DEBUG
#define LOG_MSG printf
#else
#define LOG_MSG(...)
#endif
Depending on the value of ENABLE_DEBUG, LOG_MSG is either defined to be an alias for printf() or it is defined as a no-op macro. It is implied that you can change the value to 0 to disable debugging. This is a common technique for making it easy to switch between debugging builds which display lots of output and release builds which are quiet.
#define LOG_MSG printf
This makes it an alias for printf().
#define LOG_MSG(...) /* empty */
And this defines it as an empty macro. Notice that here it has a set of parentheses, which means the macro takes parameters. It has nothing afterwards which means it expands to absolutely nothing. And the ... indicates that this macro can take a varying number of arguments. This syntax is a C99 extension so it may not be available on older C compilers.
LOG_MSG("file not found\n");
The result is that a LOG_MSG() call will either print a message or do nothing depending on whether logging is enabled.
// If ENABLE_DEBUG is non-zero, a debugging printout:
printf("file not found\n");
// If ENABLE_DEBUG is zero, an empty statement:
;
For what it's worth, whoever authored this macro could've done a better job by replacing the first definition with one using the ... syntax (which he/she is clearly familiar with), printing to stderr instead of stdout:
#define LOG_MSG(...) fprintf(stderr, __VA_ARGS__)
This uses the preprocessor to change code before compilation.
If ENABLE_DEBUG is defined as 1, whenever the preprocessor sees
LOG_MSG("something happened");
It will replace it with
printf("something happened");
If it is defined as 0, or not defined it will replace it with nothing (as the other answer that has just been published says).
In C/C++
What happens to code placed between an #if 0/#endif block?
#if 0
//Code goes here
#endif
Does the code simply get skipped and therefore does not get executed?
Not only does it not get executed, it doesn't even get compiled.
#if is a preprocessor command, which gets evaluated before the actual compilation step. The code inside that block doesn't appear in the compiled binary.
It's often used for temporarily removing segments of code with the intention of turning them back on later.
It's identical to commenting out the block, except with one important difference: Nesting is not a problem. Consider this code:
foo();
bar(x, y); /* x must not be NULL */
baz();
If I want to comment it out, I might try:
/*
foo();
bar(x, y); /* x must not be NULL */
baz();
*/
Bzzt. Syntax error! Why? Because block comments do not nest, and so (as you can see from SO's syntax highlighting) the */ after the word "NULL" terminates the comment, making the baz call not commented out, and the */ after baz a syntax error. On the other hand:
#if 0
foo();
bar(x, y); /* x must not be NULL */
baz();
#endif
Works to comment out the entire thing. And the #if 0s will nest with each other, like so:
#if 0
pre_foo();
#if 0
foo();
bar(x, y); /* x must not be NULL */
baz();
#endif
quux();
#endif
Although of course this can get a bit confusing and become a maintenance headache if not commented properly.
It permanently comments out that code so the compiler will never compile it.
The coder can later change the #ifdef to have that code compile in the program if he wants to.
It's exactly like the code doesn't exist.
I'd like to add on for the #else case:
#if 0
/* Code here will NOT be complied. */
#else
/* Code will be compiled. */
#endif
#if 1
/* Code will be complied. */
#else
/* Code will NOT be compiled. */
#endif
When the preprocessor sees #if it checks whether the next token has a non-zero value. If it does, it keeps the code around for the compiler. If it doesn't, it gets rid of that code so the compiler never sees it.
If someone says #if 0 they are effectively commenting out the code so it will never be compiled. You can think of this the same as if they had put /* ... */ around it. It's not quite the same, but it has the same effect.
If you want to understand what happened in detail, you can often look. Many compilers will allow you to see the files after the preprocessor has run. For example, on Visual C++ the switch /P command will execute the preprocessor and put the results in a .i file.
Lines beginning with a # are preprocessor directives. #if 0 [...] #endif blocks do not make it to the compiler and will generate no machine code.
You can demonstrate what happens with the preprocessor with a source file ifdef.cxx:
#if 0
This code will not be compiled
#else
int i = 0;
#endif
Running gcc -E ifdef.cxx will show you what gets compiled.
You may choose to use this mechanism to prevent a block of code being compiled during the development cycle, but you would probably not want to check it in to your source control as it just adds cruft to your code and reduces readability. If it's a historical piece of code that has been commented out, then it should be removed: source control contains the history, right?
Also, the answer may be the same for both C and C++ but there is no language called C/C++ and it's not a good habit to refer to such a language.
Not quite
int main(void)
{
#if 0
the apostrophe ' causes a warning
#endif
return 0;
}
It shows "t.c:4:19: warning: missing terminating ' character"
with gcc 4.2.4
It is a cheap way to comment out, but I suspect that it could have debugging potential. For example, let's suppose you have a build that output values to a file. You might not want that in a final version so you can use the #if 0... #endif.
Also, I suspect a better way of doing it for debug purpose would be to do:
#ifdef DEBUG
// output to file
#endif
You can do something like that and it might make more sense and all you have to do is define DEBUG to see the results.
I have the following code
#define PROC_ADD
void main(void)
{
while(1)
{
#ifdef PROC_ADD
// Do this code here then undefined it to run the code in the else
// processing work
#undef PROC_ADD
#else
// now that PROC_ADD has been undefined run this code
// processing work
#endif
}
}
However, it will run the code. But it won't run the code in the else after the PROC_ADD has been undefined.
I think the reason could be that you can only define and undefine at compile time, and not at run-time. However, I am not really sure.
What you are doing is the build time equivalent of:
int x = 1;
int main()
{
if (x)
{
...
x = 0;
}
else
{
...
}
}
ifdef, etc. happen at build time, but for your example, that's not an issue. Once you evaluate the if (either the runtime or build-time form), the decision about which branch to take it made. Changing something after the decision has been made does not change that decision.
The ifdef condition is evaluated when the preprocessor gets to it. When you undef PROC_ADD inside the ifdef'd code, the preprocessor has already decided which section of code to include and which to ignore.
Furthermore, yes: ifdef, undef, etc are processed at pre-processing time -- the compiler never even sees these so-called directives. This of course means run-time code never sees these directives either.
Edit: The preprocessor works by taking a single pass through the text file. The preprocessor does not even care that your text file happens to contain C code! It has zero knowledge that your ifdefs and elses and whatnot happen to be inside a while loop.
#defines only work during preprocessing. So
#define PROC_ADD
void main(void)
{
#ifdef PROC_ADD
// Do this code here then undefined it to run the code in the else
// processing work
#undef PROC_ADD
#else
// now that PROC_ADD has been undefined run this code
// processing work
#endif
}
will be processed the following way: since PROC_ADDR is defined the preprocessor will completely exclude the #else branch and then execute #undef, so the #else branch code never survives preprocessing and never reaches the compiler.
In just about every programming language or syntax, once execution has entered one branch of a conditional (in this case, the conditional being #ifdef, even if the condition changes during execution of the branch, other branches will never be executed.
I'm sure you wouldn't expect this to print "Hello", would you?
if (i == 1)
i = 0;
else
printf("Hello\n");
Basically what you're saying is that the code under the else branch should always execute, then just take it out of a branch, and put it directly in the code.
Both the compiler and the execution only make one pass through conditionals, once a match has been found they look no further.
So you want the code represented by the second comment section to always run? Why not just do
#ifdef PROC_ADD
// Do the stuff to be done if PROC_ADD is defined
#undef PROC_ADD
#endif
// Do the stuff to always be done
edit
OK - if you want run-time behaviour changes you must use run-time constructs (such as a variable to serve as a flag). As we are all saying ;), pre-processor directives are evaluated once only, at compile time.
Think about it this way: else portion of the following code is not executed even though x has been set to false in the if section.
The condition is checked in the if(x) line itself - once it enters that block, it doesn't recalculate each of the subsequent else sections - compiler has already made a decision on that.
bool x = true;
if(x)
{
//do something
x = false;
}
else
{
//else code
}