preprocessor directive in while condition not working - c++

when trying to use a variable defined in a preprocessor directive:
#define TIME_CONST 20;
in a while condition:
while(i<TIME_CONST){...}
I get an error complaining about parantheses...
when i use:
while(i<20)
everything works fine.
what am I doing wrong?
Thank you in advance!

#define TIME_CONST 20;
Remove ; from the end.
Because of ;, the following:
while(i<TIME_CONST){...}
becomes this:
while(i< 20;){...}
which is wrong, isn't it?
In C++ (and C), macros are text-replacement mechanism, so any usage of TIME_CONST will be replaced by 20; because that is how it has been defined.
An advice: avoid macro as much as possible . You've better alternative in C++. Use const (or constexpr ) to define your really constant object:
const int TIME_CONST = 20;
Now ; is fine, infact required by the language.

Related

What is the benefit of using #define to declare a constant? [duplicate]

This question already has answers here:
"static const" vs "#define" vs "enum"
(17 answers)
Closed 7 years ago.
I have seen a lot of programs using #define at the beginning. Why shouldn't I declare a constant global variable instead ?
(This is a C++ answer. In C, there is a major advantage to using macros, which is that they are pretty much the only way you can get a true constant-expression.)
What is the benefit of using #define to declare a constant?
There isn't one.
I have seen a lot of programs using #define at the beginning.
Yes, there is a lot of bad code out there. Some of it is legacy, and some of it is due to incompetence.
Why shouldn't I declare a constant global variable instead ?
You should.
A const object is not only immutable, but has a type and is far easier to debug, track and diagnose, since it actually exists at compilation time (and, crucially, has a name in a debug build).
Furthermore, if you abide by the one-definition rule, you don't have to worry about causing an almighty palaver when you change the definition of a macro and forget to re-compile literally your entire project, and any code that is a dependent of that project.
And, yes, it's ironic that const objects are still called "variables"; of course, in practice, they are not variable in the slightest.
What is the benefit of using #define to declare a constant?
Declaring a constant with #define is a superior alternative to using literals and magic numbers (that is, code is much better off with a value defined as #define NumDaysInWeek (7) than simply using 7), but not a superior alternative to defining proper constants.
You should declare a constant instead of #define-ing it, for the following reasons:
#define performs a token/textual replacement in the source code, not a semantic replacement.
This screws up namespace use (#defined variables are replaced with values and not containing a fully qualified name).
That is, given:
namespace x {
#define abc 1
}
x::abc is an error, because the compiler actually tries to compile x::1 (which is invalid).
abc on the other hand will always be seen as 1, forbidding you from redefining/reusing the identifier abc in any other local context or namespace.
#define inserts it's parameters textually, instead of as variables:
#define max(a, b) a > b ? a : b;
int a = 10, b = 5;
int c = max(a++, b); // (a++ > b ? a++ : b); // c = 12
#define has absolutely no semantic information:
#define pi 3.14 // this is either double or float, depending on context
/*static*/ const double pi = 3.14; // this is always double
#define makes you (the developer) see different code than the compiler
This may not be a big thing, but the errors created this way are obscure, unexpected and waste a lot of time (you could look at an error, where the code looks perfectly fine to you, and curse the compiler for half a day, only to discover later, that one of the symbols in your expression actually means something completely different).
If you get with a debugger to code using one of the declarations of pi above, the first one will cause the debugger to tell you that pi is an invalid symbol.
Edit (valid example for a local static const variable):
const result& some_class::some_function(const int key) const
{
if(map.count(key)) // map is a std::map<int,result> member of some_class
return map.at(key); // return a (const result&) to existing element
static const result empty_value{ /* ... */ }; // "static" is required here
return empty_value; // return a (const result&) to empty element
}
This shows a case when you have a const value, but it's storage needs to outlast the function, because you are returning a const reference (and the value doesn't exist in the data of some_class). It's a relatively rare case, but valid.
According to the "father" of C++, Stroustroup, defining constants using macros should be avoided.
The biggest Problems when using macros as constants include
Macros override all occurrences in the code. e.g. also variable definitions. This may result in compile Errors or undefined behavior.
Macros make the code very difficult to read and understand because the complexity of a macro can be hidden in a Header not clearly visible to the programmer

Using constants instead of macros

Is it possible to replace macros with constants, if macro is defined with another macro like this:
#define START_OFFSET 0
#define ADDRESS_OFFSET (START + START_SIZE)
#define SIZE_OFFSET (ADDRESS_OFFSET + ADDRESS_SIZE)
and so on
I'm not entirely sure, what will happen if I use global constants and initialize them with constants. Can that be considered safe?
The reason for using constants is for the possibility to wrap them into namespace.
Btw, I'm using macros like these only for working with messages which are stored in byte arrays.
Is structure serialization is better option?
I'm not entirely sure, what will happen, if I use global constants and initialize them with constants. Is it safe?
Yes, that's fine.
const int i = 4;
const int j = 6;
const int k = i + j; // legal
Is structure serialization is better option?
It depends on what you want to accomplish. Right now, that question is slightly broad. There are no golden hammers in C++.

c++ #define and concatenate case (I am using gcc)

I have
#define NAME(value) my ## value ## value
when I do NAME(1), it is my1value, which is good!
But I want to pass variable into NAME(),
such as
for(int i=0;i<10;i++)
{
NAME(i);
...
}
But unfortunately, it becomes myivalue, but I want my0value, my1value etc.
What should I change?
Thanks
Macros are substituted at compile-time (actually, they are substituted by the preprocessr even before "real" compilation begins), and it's pure text processing.
You cannot expect your macro to compute its expansion based on the value of a variable.
Macros only do text-replacement before compiling. It's basically the same as doing a 'Find&Replace' over your code. Loops are executed at runtime - so this doesn't make any sense.
But to answer your question anyway .... you could use a (compile-time) macro 'loop':
#include <iostream>
#define NAME(v) my##v##value
#define LOOP10 NAME(0), NAME(1), NAME(2), NAME(3), NAME(4), NAME(5), NAME(6), NAME(7), NAME(8), NAME(9)
int main()
{
int LOOP10; // example usage
}
boost supports preprocessor loops, as far as I know.
## is a pre-processor directive
## concatenates what's before the ## with what's after it in the #define statement
As u said "#define NAME(value) my ## value ## value" this becomes an error.it means "myii" not myivalue

Should I avoid using #define in C++? Why, and what alternatives can I use?

For some kinds of programs I need to use a constant high value to indicate some properties of some variables. I mean let color[i] = 1000000; if the i node in a tree is unexplored. But I quite often miswrite the number of 0s at the end, so I just wondered whether is it better to do it this way:
#define UNEXPLORED 1000000;
color[i] = UNEXPLORED;
I remember that somewhere I have read that it's much better to avoid using #define. Is it right? How would you tackle this problem?
For simple constants, you can use either const or the new constexpr:
constexpr unsigned int UNEXPLORED = 1000000;
In a case like this, it's no difference between using const and constexpr. However, "variables" marked constexpr are evaluated at compile-time and not at run-time, and may be used in places that otherwise only accepts literals.
For example use constants.
const unsigned int UNEXPLORED = 1000000;
or enums
enum { UNEXPLORED = 1000000 };
In the use of constants the two answers above are correct, however #define is not limited to that use alone. Another example of the use of #define is macros.
Macros
Macros are preprocessor-utilised pieces of code, and they work exactly like other #define declarations in that regard. The preprocessor will literally swap out the occurrence of your defined symbol with the code of the macro. An example:
#define HELLO_MAC do{ std::cout << "Hello World" << std::endl; }while(false)
int main(int argc, char** argv)
{
HELLO_MAC;
}
That will literally swap out the HELLO_MAC symbol with the code I declared. If it were a constant it would do the exact same thing. So you can think of #defines for constants as a particular kind of macro.
With macros you can also pass parameters, and it is especially useful I find for enforcing logging/exception policies over code.
For example
#define THROW_EXCEPT( ex_type, ex_msg ) /
do{ throw ex_type( buildExString( (ex_msg), __LINE__, __FILE__ ) ); }while(false)
...
// somewhere else
THROW_EXCEPT( std::runtime_error, "Unsupported operation in current state" );
That code allows me to ensure that everyone logs with the line of the file that threw the exception.
Templates are often a better choice instead of macros, but I cannot use template functions for this example because I need to use the __LINE__ and __FILE__ functions from the place of the throw, not from the location of the template function.
Where should you not use macros? Anywhere you can use something else. Macros, like any #define are preprocessed, so the compiler does not see them at all. This means that there is never any symbols created for HELLO_MAC or THROW_EXCEPT, and so they cannot be seen in a debugger. They can also be confusing if you get compile errors, especially if they are long macros.

Why #define is bad? [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
When are C++ macros beneficial?
Why is #define bad and what is the proper substitute?
Someone has told me that #define is bad. Well, I honestly don't not understand why its bad. If its bad, then what other way can I do this then?
#include <iostream>
#define stop() cin.ignore(numeric_limits<streamsize>::max(), '\n');
#define is not inherently bad. However, there are usually better ways of doing what you want. Consider an inline function:
inline void stop() {
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}
(Really, you don't even need inline for a function like that. Just a plain ordinary function would work just fine.)
It's bad because it's indiscriminate. Anywhere you have stop() in your code will get replaced.
The way you solve it is by putting that code into its own method.
In C++, using #define is not forcibly bad, although alternatives should be preferred. There are some context, such as include guards in which there is no other portable/standard alternative.
It should be avoided because the C preprocessor operates (as the name suggests) before the compiler. It performs simple textual replacement, without regard to other definitions. This means the result input to the compiler sometimes doesn't make sense. Consider:
// in some header file.
#define FOO 5
// in some source file.
int main ()
{
// pre-compiles to: "int 5 = 2;"
// the compiler will vomit a weird compiler error.
int FOO = 2;
}
This example may seem trivial, but real examples exist. Some Windows SDK headers define:
#define min(a,b) ((a<b)?(a):(b))
And then code like:
#include <Windows.h>
#include <algorithm>
int main ()
{
// pre-compiles to: "int i = std::((1<2)?(1):(2));"
// the compiler will vomit a weird compiler error.
int i = std::min(1, 2);
}
When there are alternatives, use them. In the posted example, you can easily write:
void stop() {
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}
For constants, use real C++ constants:
// instead of
#define FOO 5
// prefer
static const int FOO = 5;
This will guarantee that your compiler sees the same thing you do and benefit you with name overrides in nested scopes (a local FOO variable will override the meaning of global FOO) as expected.
It's not necessarily bad, it's just that most things people have used it for in the past can be done in a much better way.
For example, that snippet you provide (and other code macros) could be an inline function, something like (untested):
static inline void stop (void) {
cin.ignore(numeric_limits<streamsize>::max(), '\n');
}
In addition, there are all the other things that code macros force you to do "macro gymnastics" for, such as if you wanted to call the very badly written:
#define f(x) x * x * x + x
with:
int y = f (a + 1); // a + 1 * a + 1 * a + 1 + a + 1 (4a+2, not a^3+a)
int z = f (a++); // a++ * a++ * a++ + a++
The first of those will totally surprise you with its results due to the precedence of operators, and the second will give you undefined behaviour. Inline functions do not suffer these problems.
The other major thing that macros are used for is for providing enumerated values such as:
#define ERR_OK 0
#define ERR_ARG 1
: :
#define ERR_MEM 99
and these are better done with enumerations.
The main problem with macros is that the substitution is done early in the translation phase, and information is often lost because of this. For example, a debugger generally doesn't know about ERR_ARG since it would have been substituted long before the part of the translation process that creates debugging information.
But, having maligned them enough, they're still useful for defining simple variables which can be used for conditional compilation. That's pretty much all I use them for in C++ nowadays.
#define by itself is not bad, but it does have some bad properties to it. I'll list a few things that I know of:
"Functions" do not act as expected.
The following code seems reasonable:
#define getmax(a,b) (a > b ? a : b)
...but what happens if I call it as such?:
int a = 5;
int b = 2;
int c = getmax(++a,b); // c equals 7.
No, that is not a typo. c will be equal to 7. If you don't believe me, try it. That alone should be enough to scare you.
The preprocessor is inherently global
Whenever you use a #define to define a function (such as stop()), it acts across ALL included files after being discovered.
What this means is that you can actually change libraries that you did not write. As long as they use the function stop() in the header file, you could change the behavior of code you didn't write and didn't modify.
Debugging is more difficult.
The preprocessor does symbolic replacement before the code ever makes it to the compiler. Thus if you have the following code:
#define NUM_CUSTOMERS 10
#define PRICE_PER_CUSTOMER 1.10
...
double something = NUM_CUSTOMERS * PRICE_PER_CUSTOMER;
if there is an error on that line, then you will NOT see the convenient variable names in the error message, but rather will see something like this:
double something = 10 * 1.10;
So that makes it more difficult to find things in code. In this example, it doesn't seem that bad, but if you really get into the habit of doing it, then you can run into some real headaches.