Memory optimization while using const vs #define in C++. Is it dependent on types?

Asked Aug 18 '15 at 06:09

Active Aug 18 '15 at 06:26

Viewed 82 times

While reading merits and demerits of using a #define macro vs an equivalent const type I encountered contradictory statements.

Preprocessor replaces instance of macro with its values and as many object copies are created as many instances whereas using a const leads to only one object copy.
Compilers do not set aside storage for constant integral types. Also macros do not do any unnecessary memory allocation.

Example quoted for statement 1

#define PI 3.14

const float pi = 3.14;

Example quoted for statement 2

#define COUNT 10

const int count = 10;

Is optimization behavior dependent on types used under context as it looks from quoted examples? Why?

edited Aug 18 '15 at 06:26

asked Aug 18 '15 at 06:09

Tarun Singal

2

http://stackoverflow.com/questions/6442328/what-is-the-difference-betweem-define-and-const – Arun Aug 18 '15 at 06:14
Your first example (PI) is not an *integral* type. – Paul R Aug 18 '15 at 06:17
2

Just don't bother. Just **don't.** This is **not** a good place for "optimization". This is unnecessary bit-whacking, not optimization. – The Paramagnetic Croissant Aug 18 '15 at 06:26
Both sentences are meaningless. Modern compilers just do the right thing, always. – n. m. could be an AI Aug 18 '15 at 06:31

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