#include <iostream>
using namespace std;
#ifndef R
#define N1 10
#ifndef R
#define N2 11
#endif
#endif
int main(){
cout << N2 << endl;
}
From my understanding if R is not defined then N1 will be 10.
And second case where again R is defined again it should not be able to define N2 as 11, because R is already defined. I am just confused about it, And one Can please help to understand the point I have missed.
Let's walk through the code:
#ifndef R
R hasn't been defined yet, so step into the branch:
#define N1 10
N1 becomes 10
#ifndef R
R still hasn't been defined yet, so step into nested branch
#define N2 11
N2 is assigned to 11
#endif
Ends the nested branch
#endif
Ends the primary branch
So in the end R was never defined, so we assign N2 successfully.
You need to explicitly #define R or use -DR in the compiler line to define R, it won't get defined for you within a ifndef R branch.
The second #ifndef is redundant the way you currently have it organized. I'll write this out in a clearer pseudo code
IF R IS NOT DEFINED
SET N1 = 10
IF R IS NOT DEFINED
SET N2 = 11
END IF
END IF
This can be more simply
IF R IS NOT DEFINED
SET N1 = 10
SET N2 = 11
END IF
or in C/C++ preprocessor notation
#ifndef R
#define N1 10
#define N2 11
#endif
Related
I have the defines with integer values in Cpp code. I need to call them in loop. How do I do that?
// Defines:
#define A0 0
#define A1 1
#define A2 2
// ...
#define A50 50
// Now based on loop I need to call these defines
for (int i = 0; i <= 50; i++){
function_name(A<i>, value);
}
#define macros are evaluated only by the preprocessor before the compiler is invoked. As such, preprocessor macros do not exist at runtime, and cannot be referred to by names that are built up using variables whose values are only known at runtime.
For what you are attempting, you will have to use an array, eg:
#define A0 0
#define A1 1
#define A2 2
...
#define A50 50
const int A[51] = {A0, A1, A2, ..., A50};
...
for (int i = 0; i <= 50; i++){
function_name(A[i], value);
}
#define N 1
#define A(N) #N
#define S_A A(N)
#define B_(N) #N
#define B(N) B_(N)
#define S_B B(N)
#include <stdio.h>
int main(void) {
puts(S_A);
puts(S_B);
}
outputs N and 1 instead of two 1s.
Why is the extra indirection making a difference?
It works different because the macro expansion of N only happens on the "use" of N. So S_A exands to A(N), which is expanded to #N. In S_B it is expanded to B(N), which is expanded to _B(1), and then #1. Why? Well, how would you do something like:
#define COMBINE(A, B) A##B
#define foo 1
#define bar 2
int COMBINE(foo, bar) = 34;
so that it generates int foobar = 34;, and not the illegal int 12 = 34;.
Did you intend to use a ## operator for token pasting?
puts(S_A); //Gets expanded into puts(A(N)) --> puts(#N)
and
puts(S_B); //Gets expanded into puts(B(N)) --> puts(_B(1)) --> puts(#1)
Without affecting your macros much, if you intend to print 1,1 both the times, you could Use the macros this way:
#define N "1" //instead of 1
#define A(N) N //instead of #N
#define B_(N) N //instead of #N
So I have a .h file defined as this
#ifndef CONSTANTS_H
#define CONSTANTS_H
#include "math.h"
//CALA = CA/LA, CRLR = CR/LR
#define CALA 0.2
#define CRLR 25.0
//self-propulsion and friction
#define ALPHA .15
#define BETA .05
//Evolution simulation size
#define STUDYSIZE 100
#define STUDYLENGTH 1
.....
#define INITIAL_CONDITION true
#endif
Is there a way to import the constants that I defined here and turn them into matlab variables.
Try this -
%%// Read in data
imp_data = importdata(CPP_H_FILE,'\n');
%%// Remove the leading and trailing whitespaces
imp_data = cellstr(strtrim(char(imp_data)));
%%// Split into strings
split1 = regexp(imp_data,'\s','Split');
%%// Process to evaluate the variables into MATLAB
for k = 1:size(split1,1)
if strcmp(char(split1{k,1}(:,1)),'#define') && numel(split1{k,1})>2
val_str = strtok(char(split1{k,1}(:,end)), '/');
evalstr = strcat(char(split1{k,1}(:,2)),'=',val_str);
evalc(evalstr);
end
end
I have the source code of an application written in C++ and I just want to comment something using:
#ifdef 0
...
#endif
And I get this error
error: macro names must be identifiers
Why is this happening?
The #ifdef directive is used to check if a preprocessor symbol is defined. The standard (C11 6.4.2 Identifiers) mandates that identifiers must not start with a digit:
identifier:
identifier-nondigit
identifier identifier-nondigit
identifier digit
identifier-nondigit:
nondigit
universal-character-name
other implementation-defined characters>
nondigit: one of
_ a b c d e f g h i j k l m
n o p q r s t u v w x y z
A B C D E F G H I J K L M
N O P Q R S T U V W X Y Z
digit: one of
0 1 2 3 4 5 6 7 8 9
The correct form for using the pre-processor to block out code is:
#if 0
: : :
#endif
You can also use:
#ifdef NO_CHANCE_THAT_THIS_SYMBOL_WILL_EVER_EXIST
: : :
#endif
but you need to be confident that the symbols will not be inadvertently set by code other than your own. In other words, don't use something like NOTUSED or DONOTCOMPILE which others may also use. To be safe, the #if option should be preferred.
Use the following to evaluate an expression (constant 0 evaluates to false).
#if 0
...
#endif
This error can also occur if you are not following the marco rules
Like
#define 1K 1024 // Macro rules must be identifiers error occurs
Reason: Macro Should begin with a letter, not a number
Change to
#define ONE_KILOBYTE 1024 // This resolves
#ifdef 0
...
#endif
#ifdef expect a macro rather than expression
when using constant or expression
#if 0
...
#endif
or
#if !defined(PP_CHECK) || defined(PP_CHECK_OTHER)
..
#endif
if #ifdef is used the it reports this error
#ifdef !defined(PP_CHECK) || defined(PP_CHECK_OTHER)
..
#endif
Where #ifdef expect a macro rather than macro expresssion
Note that you can also hit this error if you accidentally type:
#define <stdio.h>
...instead of...
#include <stdio.>
Why won't this work?
0. #define CONCAT(x, y) x ## y
1.
2. #define VAR_LINE(x) \
3. int CONCAT(_anonymous, __LINE__) = x
4.
5. #define VAR_LINE2(x) \
6. int _anonymous ## x = 1
7.
8. int main()
9. {
10. VAR_LINE(1);
11. VAR_LINE(1);
12. VAR_LINE(1);
13. VAR_LINE2(__LINE__);
14. }
The result from the above macro expansion
int _anonymous__LINE__ = 1;
int _anonymous__LINE__ = 1;
int _anonymous__LINE__ = 1;
int _anonymous13 = 1;
It would be convenient if I didn't have to write that __LINE__ macro as an argument.
I'm thinking the problem is pretty clear. I want to be able to generate anonymous variables so that this macro doesn't fail with redefinition error when declaring several variables within the same scope. My idea was to use the predefined __LINE__ macro because no variable will ever be declared on the same line like this. But the macro expansion troubles me, can you help?
Update: Correct answer
Thanks to Luc Touraille. However, there was a tiny problem with the suggested solution. There has to be whitespace between the operands and the ## operator (apparently the standard says otherwise but the the PS3 flavoured GCC would not expand the macro properly if there were no whitespace between the operator and operands).
#define _CONCAT(x,y) x ## y
#define CONCAT(x,y) _CONCAT(x,y)
The VAR_LINE macro now yields:
int _anonymous10 = 1;
int _anonymous11 = 1;
int _anonymous12 = 1;
This has been verified to work under Win32 (Visual Studio 2008), XBOX360 (Xenon) and PS3.
You need to add a level of indirection so that __LINE__ will be expanded:
#define _CONCAT_(x,y) x ## y
#define CONCAT(x,y) _CONCAT_(x,y)
#define VAR_LINE(x) int CONCAT(_anonymous, __LINE__) = x