CODEWITHSUNDEEP
c++c
Kartik Sethi
Kartik Sethi

Reputation: 3

Using #define for printf, does it effect on speed

I am using 

#define printInt(x)  printf ("%d",x)

In main()

I can use it like this:

int var=10;
printInt (var);

Which is easier to use than typing 

printf ("%d",var);

Will using my own #define for printing an int, float etc make my program slower?

Upvotes: 0

Views: 463

Answers (2)

splotz90
splotz90

Reputation: 331

No, it doesn't affect the speed of your program.

The #define instructions are processed by the preprocessor before your program is compiled.

For example the call 

printInt(var);

is replaced with

printf ("%d",var);

by the preprocessor.

Therefore the compiler can't determine if a #define was used or not. In both cases it leads to the same code (and the same program). Thats the reason why it isn't possible that both programs differ in their speed.

EDIT: If you use a lot of #defines in your program, it is possible that the speed of the proprocessing step decreases. But in most cases this should be no problem.

Upvotes: 3

walwart
walwart

Reputation: 51

No this will not effect the speed. The macro is expanded during pre-processing so that in every instance that you use printInt(myInt) what is actually passed to the compiler will be printf("%d", myInt). So I think the binary output would be identical either way.

Upvotes: 5

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