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Harryu

(C++) difference between #define and const int

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I was looking at the first few lines of my game today, and I saw that I had stuff like: #define BLAH 10 #define BLAHBLAH 20 const int BLAHBLAHBLAH = 30; And I was wondering what the difference was between using #define, and using const int. I assume there is a fairly important difference that I am missing, but if anybody could tell me that would be good. Thanks.

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#define is a preprocessor directive. Before compiling, the middle symbol is replaced by the right hand symbol(s).

#define String std::string

for example.

const int myInt = 0;

is a better way of doing it because it creates a constant 'variable'. This avoids the use of the preprocessor, which is good for many reasons.

Dave

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defines are preprocessor text replacements and therefore BLAH BLAHBLAH have no type, where as BLAHBLAHBLAH does. C++ is a very type based language so "int const" is a far better method.
[edit] far too slow:)

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What a #define does is that the preprocessor (before compilation) will replace in your code all instances of BLAH with 10. In this cas it causes no problem but you should be using constants for the following reason :

// If you declare
#define BLUH 1+1
// and then you do
int myvariable = BLUH * 2;
// myvariable now equals 3!
// Because he preprocessor got through your code and did :
// int myvariable = BLUH * 2; // There's a BLUH, replace it by 1+1!
// int myvariable = 1+1 * 2; // And now this gives 3

// Of course you could do the following :
#define BLUH (1+1)
// and then it'd work (int myvariable = (1+1) * 2)
// but you'd better get used now to use constants

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Quote:
Original post by OrangyTang
#define-s for constants aren't type safe. Where possible prefer const (or static const if appropriate) over defines.


In which cases would static const be appropriate over a standard const?

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if const int is actual variable (with memory location and size) wouldn't it result in slower code?

for example

const int x=4;
#define y 5

z = 2*x; // slower - load from memory location and mult
z = 2*y; // faster - precompiled calc placed into instruction

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No real compiler will ever perform a load and multiply in that situation. Real compilers will always simply perform a move 8 into the target variable (or maybe not even that, depending on other optimisations). There are occasional reasons to use the preprocessor instead of the language, but run-time performance is never* one of them.

Σnigma

*unless someone can prove me wrong on this

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