I hate to ask this question, but I've never seen a macro quite like this following piece of code after 3 or 4 years of programming.
#define CREATE_FIXED_FIELD(Label, LabelFrame, ValueFrame, NextFrame, GetValue, SetValue, Digits, Fraction, Min, Max) \
{Label, LabelFrame, EditValue, ValueFrame, NextFrame, FixedDisplay, FixedBeginEdit, FixedFinishEdit, FixedKeyPressed, FixedHighlight, .fixed = {GetValue, SetValue, Digits, Fraction, Min, Max}}
Now I know you can use the define to create constants, but I never knew you could use them as a function. It's actually the second line with the curly braces that is throwing me off. What does it do exactly? Are the variables in the second line being set to the values in the first line? I would think it would be used as the function implementation, but that doesn't look to be the case. And lastly, what is the .fixed variable? I know it is a struct of some sort, but I have never seen a period used like that before in C. Is this define construct essentially a setter function?
It's a c99 designated initializer.
You can use it to initialize structure members (or array elements) in any order.
struct bla {
int a;
int b;
} x = { .a = 42, .b = 0};
You can specify only some members and any order. Omitted members are initialized as if they are static objects.
This macro is intended as a convenience when initializing a certain kind of structure. Assuming the structure definition is something like this:
typedef struct FixedField {
char *label,
Frame *labelFrame,
// ...
Fixed fixed,
// ... Possibly more here ...
} FixedField;
Then you could use that macro to initialize a FixedField like this:
FixedField ff = CREATE_FIXED_FIELD(myLabel, myLabelFrame, /* ... */);
Whether it's better to use macros like this than proper functions depends on the details of the system. This way will consume more static program memory as it expands before compiling, but it will prevent pushing a new stack frame every time you initialize an object like this.
I find macros more difficult to debug than functions, but your mileage may vary.
Which one is better to use among the below statements in C?
static const int var = 5;
or
#define var 5
or
enum { var = 5 };
It depends on what you need the value for. You (and everyone else so far) omitted the third alternative:
static const int var = 5;
#define var 5
enum { var = 5 };
Ignoring issues about the choice of name, then:
If you need to pass a pointer around, you must use (1).
Since (2) is apparently an option, you don't need to pass pointers around.
Both (1) and (3) have a symbol in the debugger's symbol table - that makes debugging easier. It is more likely that (2) will not have a symbol, leaving you wondering what it is.
(1) cannot be used as a dimension for arrays at global scope; both (2) and (3) can.
(1) cannot be used as a dimension for static arrays at function scope; both (2) and (3) can.
Under C99, all of these can be used for local arrays. Technically, using (1) would imply the use of a VLA (variable-length array), though the dimension referenced by 'var' would of course be fixed at size 5.
(1) cannot be used in places like switch statements; both (2) and (3) can.
(1) cannot be used to initialize static variables; both (2) and (3) can.
(2) can change code that you didn't want changed because it is used by the preprocessor; both (1) and (3) will not have unexpected side-effects like that.
You can detect whether (2) has been set in the preprocessor; neither (1) nor (3) allows that.
So, in most contexts, prefer the 'enum' over the alternatives. Otherwise, the first and last bullet points are likely to be the controlling factors — and you have to think harder if you need to satisfy both at once.
If you were asking about C++, then you'd use option (1) — the static const — every time.
Generally speaking:
static const
Because it respects scope and is type-safe.
The only caveat I could see: if you want the variable to be possibly defined on the command line. There is still an alternative:
#ifdef VAR // Very bad name, not long enough, too general, etc..
static int const var = VAR;
#else
static int const var = 5; // default value
#endif
Whenever possible, instead of macros / ellipsis, use a type-safe alternative.
If you really NEED to go with a macro (for example, you want __FILE__ or __LINE__), then you'd better name your macro VERY carefully: in its naming convention Boost recommends all upper-case, beginning by the name of the project (here BOOST_), while perusing the library you will notice this is (generally) followed by the name of the particular area (library) then with a meaningful name.
It generally makes for lengthy names :)
In C, specifically? In C the correct answer is: use #define (or, if appropriate, enum)
While it is beneficial to have the scoping and typing properties of a const object, in reality const objects in C (as opposed to C++) are not true constants and therefore are usually useless in most practical cases.
So, in C the choice should be determined by how you plan to use your constant. For example, you can't use a const int object as a case label (while a macro will work). You can't use a const int object as a bit-field width (while a macro will work). In C89/90 you can't use a const object to specify an array size (while a macro will work). Even in C99 you can't use a const object to specify an array size when you need a non-VLA array.
If this is important for you then it will determine your choice. Most of the time, you'll have no choice but to use #define in C. And don't forget another alternative, that produces true constants in C - enum.
In C++ const objects are true constants, so in C++ it is almost always better to prefer the const variant (no need for explicit static in C++ though).
The difference between static const and #define is that the former uses the memory and the later does not use the memory for storage. Secondly, you cannot pass the address of an #define whereas you can pass the address of a static const. Actually it is depending on what circumstance we are under, we need to select one among these two. Both are at their best under different circumstances. Please don't assume that one is better than the other... :-)
If that would have been the case, Dennis Ritchie would have kept the best one alone... hahaha... :-)
In C #define is much more popular. You can use those values for declaring array sizes for example:
#define MAXLEN 5
void foo(void) {
int bar[MAXLEN];
}
ANSI C doesn't allow you to use static consts in this context as far as I know. In C++ you should avoid macros in these cases. You can write
const int maxlen = 5;
void foo() {
int bar[maxlen];
}
and even leave out static because internal linkage is implied by const already [in C++ only].
Another drawback of const in C is that you can't use the value in initializing another const.
static int const NUMBER_OF_FINGERS_PER_HAND = 5;
static int const NUMBER_OF_HANDS = 2;
// initializer element is not constant, this does not work.
static int const NUMBER_OF_FINGERS = NUMBER_OF_FINGERS_PER_HAND
* NUMBER_OF_HANDS;
Even this does not work with a const since the compiler does not see it as a constant:
static uint8_t const ARRAY_SIZE = 16;
static int8_t const lookup_table[ARRAY_SIZE] = {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}; // ARRAY_SIZE not a constant!
I'd be happy to use typed const in these cases, otherwise...
If you can get away with it, static const has a lot of advantages. It obeys the normal scope principles, is visible in a debugger, and generally obeys the rules that variables obey.
However, at least in the original C standard, it isn't actually a constant. If you use #define var 5, you can write int foo[var]; as a declaration, but you can't do that (except as a compiler extension" with static const int var = 5;. This is not the case in C++, where the static const version can be used anywhere the #define version can, and I believe this is also the case with C99.
However, never name a #define constant with a lowercase name. It will override any possible use of that name until the end of the translation unit. Macro constants should be in what is effectively their own namespace, which is traditionally all capital letters, perhaps with a prefix.
#define var 5 will cause you trouble if you have things like mystruct.var.
For example,
struct mystruct {
int var;
};
#define var 5
int main() {
struct mystruct foo;
foo.var = 1;
return 0;
}
The preprocessor will replace it and the code won't compile. For this reason, traditional coding style suggest all constant #defines uses capital letters to avoid conflict.
It is ALWAYS preferable to use const, instead of #define. That's because const is treated by the compiler and #define by the preprocessor. It is like #define itself is not part of the code (roughly speaking).
Example:
#define PI 3.1416
The symbolic name PI may never be seen by compilers; it may be removed by the preprocessor before the source code even gets to a compiler. As a result, the name PI may not get entered into the symbol table. This can be confusing if you get an error during compilation involving the use of the constant, because the error message may refer to 3.1416, not PI. If PI were defined in a header file you didn’t write, you’d have no idea where that 3.1416 came from.
This problem can also crop up in a symbolic debugger, because, again, the name you’re programming with may not be in the symbol table.
Solution:
const double PI = 3.1416; //or static const...
I wrote quick test program to demonstrate one difference:
#include <stdio.h>
enum {ENUM_DEFINED=16};
enum {ENUM_DEFINED=32};
#define DEFINED_DEFINED 16
#define DEFINED_DEFINED 32
int main(int argc, char *argv[]) {
printf("%d, %d\n", DEFINED_DEFINED, ENUM_DEFINED);
return(0);
}
This compiles with these errors and warnings:
main.c:6:7: error: redefinition of enumerator 'ENUM_DEFINED'
enum {ENUM_DEFINED=32};
^
main.c:5:7: note: previous definition is here
enum {ENUM_DEFINED=16};
^
main.c:9:9: warning: 'DEFINED_DEFINED' macro redefined [-Wmacro-redefined]
#define DEFINED_DEFINED 32
^
main.c:8:9: note: previous definition is here
#define DEFINED_DEFINED 16
^
Note that enum gives an error when define gives a warning.
The definition
const int const_value = 5;
does not always define a constant value. Some compilers (for example tcc 0.9.26) just allocate memory identified with the name "const_value". Using the identifier "const_value" you can not modify this memory. But you still could modify the memory using another identifier:
const int const_value = 5;
int *mutable_value = (int*) &const_value;
*mutable_value = 3;
printf("%i", const_value); // The output may be 5 or 3, depending on the compiler.
This means the definition
#define CONST_VALUE 5
is the only way to define a constant value which can not be modified by any means.
Although the question was about integers, it's worth noting that #define and enums are useless if you need a constant structure or string. These are both usually passed to functions as pointers. (With strings it's required; with structures it's much more efficient.)
As for integers, if you're in an embedded environment with very limited memory, you might need to worry about where the constant is stored and how accesses to it are compiled. The compiler might add two consts at run time, but add two #defines at compile time. A #define constant may be converted into one or more MOV [immediate] instructions, which means the constant is effectively stored in program memory. A const constant will be stored in the .const section in data memory. In systems with a Harvard architecture, there could be differences in performance and memory usage, although they'd likely be small. They might matter for hard-core optimization of inner loops.
Don't think there's an answer for "which is always best" but, as Matthieu said
static const
is type safe. My biggest pet peeve with #define, though, is when debugging in Visual Studio you cannot watch the variable. It gives an error that the symbol cannot be found.
Incidentally, an alternative to #define, which provides proper scoping but behaves like a "real" constant, is "enum". For example:
enum {number_ten = 10;}
In many cases, it's useful to define enumerated types and create variables of those types; if that is done, debuggers may be able to display variables according to their enumeration name.
One important caveat with doing that, however: in C++, enumerated types have limited compatibility with integers. For example, by default, one cannot perform arithmetic upon them. I find that to be a curious default behavior for enums; while it would have been nice to have a "strict enum" type, given the desire to have C++ generally compatible with C, I would think the default behavior of an "enum" type should be interchangeable with integers.
A simple difference:
At pre-processing time, the constant is replaced with its value.
So you could not apply the dereference operator to a define, but you can apply the dereference operator to a variable.
As you would suppose, define is faster that static const.
For example, having:
#define mymax 100
you can not do printf("address of constant is %p",&mymax);.
But having
const int mymax_var=100
you can do printf("address of constant is %p",&mymax_var);.
To be more clear, the define is replaced by its value at the pre-processing stage, so we do not have any variable stored in the program. We have just the code from the text segment of the program where the define was used.
However, for static const we have a variable that is allocated somewhere. For gcc, static const are allocated in the text segment of the program.
Above, I wanted to tell about the reference operator so replace dereference with reference.
We looked at the produced assembler code on the MBF16X... Both variants result in the same code for arithmetic operations (ADD Immediate, for example).
So const int is preferred for the type check while #define is old style. Maybe it is compiler-specific. So check your produced assembler code.
I am not sure if I am right but in my opinion calling #defined value is much faster than calling any other normally declared variable (or const value).
It's because when program is running and it needs to use some normally declared variable it needs to jump to exact place in memory to get that variable.
In opposite when it use #defined value, the program don't need to jump to any allocated memory, it just takes the value. If #define myValue 7 and the program calling myValue, it behaves exactly the same as when it just calls 7.
I would like to create a C macro returning the scalar minimum for any type of static array in input. For example:
float A[100];
int B[10][10];
// [...]
float minA = MACRO_MIN(A);
int minB = MACRO_MIN(B);
How can I do so?
It can be probably be done with GCC extensions, but not in standard C. Other compilers might have suitable extensions, too. It will of course make the code fantastically hard to port. I would advise against it, since it's quite hard to achieve it will be "unexpected" and probably act as a source of confusion (or, worse, bugs) down the line.
You're going to have to declare a temporary variable to hold the max/min seen "so far" when iterating over the array, and the type of that variable is hard to formulate without extensions.
Also returning the value of the temporary is hard, but possible with GCC extensions.
To make the above more concrete, here's a sketch of what I imagine. I did not test-compile this, so it's very likely to have errors in it:
#define ARRAY_MAX(a) ({ typeof(a) tmp = a[0];\
for(size_t i = 1; i < sizeof a / sizeof tmp; ++i)\
{\
if(a[i] > tmp)\
tmp = a[i];\
}\
tmp;\
})
The above uses:
({ and }) is the GCC Statement Expressions extension, allowing the macro to have a local variable which is used as the "return value".
typeof is used to compute the proper type.
Note assumption that the array is not of zero size. This should not be a very limiting assumption.
The use of sizeof is of course standard.
As I wrote the above, I realize there might be issues with multi-dimensional arrays that I hadn't realized until trying. I'm not going to polish it further, though. Note that it starts out with "probably".
I'm trying to figure out there best way to define a global array with a constant size and I've come to the following options, all with their own flaws.
// 1:
#define ASIZE 10
int array[ASIZE];
// 2:
enum {ASIZE = 10};
int array[ASIZE];
// 3:
#define ASIZE_DEF 10
static const int ASIZE = ASIZE_DEF;
int array[ASIZE_DEF];
The problem with the first two is that I can't get the value of ASIZE from GDB. I guess the third option is best because I can still dump the value of the const, but it also leaks in another macro. I can undef the macro after defining the array and const but if the #define and the const are in a separate file from the array declaration, then it gets a bit hairy.
Is there a better way?
Doing something for the sake of the debugger is wrong. Incidentally, gdb knows about this if you compile your code right.
Some languages, such as C and C++, provide a way to define and invoke
“preprocessor macros” which expand into strings of tokens. gdb can
evaluate expressions containing macro invocations, show the result of
macro expansion, and show a macro's definition, including where it was
defined.
Version 3.1 and later of gcc, the gnu C compiler, provides macro
information if you specify the options -gdwarf-2 and -g3; the former
option requests debugging information in the Dwarf 2 format, and the
latter requests “extra information”.
You are dealing with a GDB issue, not a C issue. You can also do #4, which is arguably better than #3.
enum {ASIZE = 10};
static const int ASIZE_FOR_GDB = ASIZE;
int array[ASIZE];
My understanding is that you are defining a constant, using it later to size one or more arrays, and also want that constant to be a symbol, preferably without a messy namespace. (If it were a matter of exporting the size of a single array, I would instead suggest sizeof(array) / sizeof(*array) as missingno did.)
static const int ASIZE = 10;
#define ASIZE 10
int array[ASIZE];
Here, there is a variable with the desired value which will be in the object file, but the preprocessor macro shadows it with the value itself, so the array definition also succeeds.
However, you may find the need to duplicate the value expression ugly. Wouldn't it be nice if we could define the variable in terms of the macro?
static const int ASIZE =
#define ASIZE 10
ASIZE;
int array[ASIZE];
I'm not sure whether this is actually a better idea maintainability-wise than the above, but it works (and I couldn't get gcc to be offended by it) and it contains no duplication other than of the identifier. And it's amusing.
Since you know array is a global array (and not just a pointer) you can find its length using
sizeof(array) / sizeof(*array)
without needing to set an extra variable for that.
I was just curious to know if it is possible to have a pointer referring to #define constant. If yes, how to do it?
The #define directive is a directive to the preprocessor, meaning that it is invoked by the preprocessor before anything is even compiled.
Therefore, if you type:
#define NUMBER 100
And then later you type:
int x = NUMBER;
What your compiler actually sees is simply:
int x = 100;
It's basically as if you had opened up your source code in a word processor and did a find/replace to replace each occurrence of "NUMBER" with "100". So your compiler has no idea about the existence of NUMBER. Only the pre-compilation preprocessor knows what NUMBER means.
So, if you try to take the address of NUMBER, the compiler will think you are trying to take the address of an integer literal constant, which is not valid.
No, because #define is for text replacement, so it's not a variable you can get a pointer to -- what you're seeing is actually replaced by the definition of the #define before the code is passed to the compiler, so there's nothing to take the address of. If you need the address of a constant, define a const variable instead (C++).
It's generally considered good practice to use constants instead of macros, because of the fact that they actually represent variables, with their own scoping rules and data types. Macros are global and typeless, and in a large program can easily confuse the reader (since the reader isn't seeing what's actually there).
#define defines a macro. A macro just causes one sequence of tokens to be replaced by a different sequence of tokens. Pointers and macros are totally distinct things.
If by "#define constant" you mean a macro that expands to a numeric value, the answer is still no, because anywhere the macro is used it is just replaced with that value. There's no way to get a pointer, for example, to the number 42.
No ,It's Not possible in C/C++
You can use the #define directive to give a meaningful name to a constant in your program
We can able to use in two forms.
Please : See this link
http://msdn.microsoft.com/en-us/library/teas0593%28VS.80%29.aspx
The #define directive can contain an object-like definition or a function-like definition.
Iam sorry iam unable to provide one more wink ... Please see the IBM links..since below i pasted linke link
u can get full info from above 2 links
There is a way to overcome this issue:
#define ROW 2
void foo()
{
int tmpInt = ROW;
int *rowPointer = &tmpInt;
// ...
}
Or if you know it's type you can even do that:
void getDefinePointer(int * pointer)
{
*pointer = ROW;
}
And use it:
int rowPointer = NULL;
getDefinePointer(&rowPointer2);
printf("ROW==%d\n", rowPointer2);
and you have a pointer to #define constant.