Why the code inside the #ifdef INITIALISATION and #endif is not executed?
int main(void)
{
uint8_t DLEVEL = 10;
#if DLEVEL > 5
#define INITIALISATION
#endif
while (1)
{
#ifdef INITIALISATION
Display(); // This line is never being executing :: please help for a solution
#endif
}
}
edited...
Thanks for the reply
Actually the below mentioned logic of the code is used as a part of my code memory optimization.
Now the code memory size is overflowed, so i need to execute the one time initializations when the SYM_DLEVEL value is 10 and then the value of SYM_DLEVEL is changed from 10 to 2, then the initial sections i need to be automatically commented or disabled and then only the Display_2() function needed to be enable and need to execute.
Is it possible?
#define SYM_DLEVEL 10
int main(void)
{
#if SYM_DLEVEL > 5
Display_1(); // need to execute this line once(before the value changes from 10 t0 2)
#endif
#define SYM_DLEVEL 2 // after this line execution i need to automatically disable the above section and automatically enable the below section
#if SYM_DLEVEL < 5
#define INITIALISATION // need to execute only when value changes from 10 to 2
#endif
while (1)
{
#ifdef INITIALISATION
Display_2();
#endif
}
}
It doesn't execute code because the preprocessor runs in the compiler, while the local variable DLEVEL is assigned during execution.
The preprocessor simply looks for a preprocessor symbol DLEVEL that evidently is undefined.
Undefined symbols are equated to 0, so the symbol INITIALISATION isn't defined.
To make it work define a preprocessor symbol, i.e. SYM_DLEVEL, to be used in the conditional preprocessing, and eventually assign it to the runtime variable. I.e.
#define SYM_DLEVEL 10
int main(void)
{
uint8_t DLEVEL = SYM_DLEVEL;
#if SYM_DLEVEL > 5
#define INITIALISATION
#endif
while (1)
{
#ifdef INITIALISATION
Display(); // This line is now is executed
#endif
}
}
Remember that preprocessing is more or less a text processor (which actuates on base piece of text, tokens, as outlined in comments below), and is executed before the compilation of the code. It doesn't know about runtime variables and their assignments, it only knows symbols defined with a preprocessor directive (#define).
Don't confuse actions as dead code removal or other optimization made by compiler with what the preprocessor can do. The golden rule is: "preprocessing only understand preprocessing objects, C code is a plain text without any special meaning for it".
Related
What is the difference between these two codes:
#define check 0
int main(void)
{
#if check
this is a normal line
#endif
return 0;
}
and this one:
int main(void)
{
/*
this is a normal line
*/
return 0;
}
Are they exactly the same ?
No they are not the same.
#define check 0
int main(void) {
#if check
this is a normal line
#endif
return 0;
}
This segment, when compiled, is completely dependent on the value of the preprocessed check value, currently at 0. Since the sole condition for compiling the code between your preprocessing if-endif is the 0|1 value of check, at runtime, they will be the same. However, if you were to change check to 1, it would compile the conditional line as well.
At compile-time, the first segment processes the #define check 0, where as the second, obviously, never has to and simply ignores the comment.
Yes, except that the first code won't compile for #define check ;-).
I access registers by concatenated defines unsing the function GETREG
#define NUMBER 1 //changes
#define REG1 register.N1
#define REG2 register.N2
#define REG8 register.N8
#define GETREG_(N) REG ## N
#define GETREG(N) GETREG_(N)
Sometimes the registers for that NUMBER are not defined so i want to make sure the macro correctly expands before inserting it in the code so i tried to do:
#define NUMBER 5
#ifdef GETREG(NUMBER)
GETREG(NUMBER) = 0
#endif
However this always seems to evaluate as true and the compiler prints
Warning:extra tokens at end of #ifdef directive
Background Story:
In my projects I create libraries to interface with a HAL to abstract the hardware level. Often from one project to another the codebasis in the HAL stays exactly the same but just the location of pins changes. For that reason i would like to use macro-expansion to access the pins. The following macro does the job for me at adressing the analog features of the pin:
#define ANSEL_(Pin) _ANS ## Pin
#define ANSEL(...) ANSEL_(__VA_ARGS__)
that way i can turn on or off analog features by:
#define PIN_RX A0
ANSEL(PIN_RX)= 0;
and other registers by similar macros. The problem i am facing now is that some pins for example do not have analog features (e.g. Pin A5). Because of that i would like to test if the define in the library esists. I was tryin to do this by:
#ifdef ANSEL(PIN_RX)
ANSEL(PIN_RX)= 0;
#endif
However this simple approach is not working.
The microcontroller (PIC33) lib:
Edit: Shipped from the manufactorer
#define ANSELA ANSELA
extern volatile unsigned int ANSELA __attribute__((__sfr__));
typedef struct tagANSELABITS {
unsigned ANSA0:1;
unsigned ANSA1:1;
unsigned ANSA2:1;
unsigned ANSA3:1;
unsigned ANSA4:1;
unsigned :4;
unsigned ANSA9:1;
} ANSELABITS;
extern volatile ANSELABITS ANSELAbits __attribute__((__sfr__));
/* ANSELA */
#define _ANSA0 ANSELAbits.ANSA0
#define _ANSA1 ANSELAbits.ANSA1
#define _ANSA2 ANSELAbits.ANSA2
#define _ANSA3 ANSELAbits.ANSA3
#define _ANSA4 ANSELAbits.ANSA4
#define _ANSA9 ANSELAbits.ANSA9
You have the wrong expectation for the behavior of the #ifdef directive. That directive is equivalent to #if defined, where defined is a preprocessing operator in that context. Although the condition of an overall #if directive is evaluated only after the line is fully macro-expanded, the defined operator operates on the next preprocessing token, which must have the form of an identifier, before macro expansion:
Prior to evaluation, macro invocations in the list of preprocessing tokens that will become the controlling constant expression are replaced (except for those macro names modified by the defined unary operator) [...]
[C2011, 6.10.1/4]
Thus, your
#ifdef GETREG(NUMBER)
tests whether GETREG is a defined macro, and the (NUMBER) constitutes a sequence of unexpected extra tokens.
The defined operator therefore cannot serve your purpose. It is conceivable that there is a way to use a more general #if directive to achieve your aim, but I'm not seeing it at the moment. Instead, I'm inclined to suggest letting the compiler find such errors for you, instead of expecting the preprocessor to do it. If compile time is too late, then perhaps you need to look to a build configuration system such as the Autotools or CMake could help you create.
You can do more or less what you want if you are prepared to use two #defines instead of one for each pin:
#define HAS_ANS_A0 1
#define ANS_A0 ANSELAbits.ANSA0
#define HAS_ANS_A1 1
#define ANS_A1 ANSELAbits.ANSA1
#define HAS_ANS_A5 1
#define ANS_A5 ANSELAbits.ANSA5
#define HAS_ANSEL_(Pin) HAS_ANS_ ## Pin
#define HAS_ANSEL(...) HAS_ANSEL_(__VA_ARGS__)
#define ANSEL_(Pin) ANS_ ## Pin
#define ANSEL(...) ANSEL_(__VA_ARGS__)
#if HAS_ANSEL(PIN_RX)
ANSEL(PIN_RX)= 0;
#endif
This works because in an #if, an undefined identifier token (that is, an identifier token which has not been #defined) has the value 0. It is not an error, or even a warning.
Note: I changed the symbols starting _ANS to conform with ยง7.1.3 (Reserved Names), paragraph 1:
All identifiers that begin with an underscore and either an uppercase letter or another underscore are always reserved for any use.
How does an #if Preprocessor stage statement which contain variable of compile time get resolved at preprocessor stage itself?
Below is the code which runs without any error:
#include<stdio.h>
void main()
{
int num=10; /* compile time */
#if((num%2)==0) /* #if is preprocessor stage but has num of compile time why not error here? */
printf("\nNumber is Even");
#else
printf("\nNumber is Odd");
#endif
}
For the evaluation in #if, the preprocessor replaces all identifiers that are already defined macros by their appropriate expansion. All identifiers that remain after that have the value 0.
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.
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.