How '<<' operator does work with #define in C++

Question

I have two operation, and I am assuming both are doing ShiftLeft bitwise operation.

#define TACH_MAX_OWN_ORDERS 1<<6
int myVal = 1<<6
cout<<"Value after operation|"<<myVal <<"|"<<TACH_MAX_OWN_ORDERS<<endl;

output of TACH_MAX_OWN_ORDERS value always surprise me.

Value after operation|64|16

Do anyone have any clue, how it comes??? Thanks

A macro is just a text substitution. Look at the result of `cout<<1<<6<< endl;`. — molbdnilo, Sep 07 '22 at 10:41
why it is declare in my code and using in this loop ` for(int i=0;i — Pankaj Belwal, Sep 07 '22 at 11:33

Stack Danny · Answer 1 · 2022-09-07T10:36:34.503

Macros replace text as is, so it will result in

cout<<"Value after operation|"<<myVal <<"|"<<1<<6<<endl;

the << won't result in (int)1<<6 but rather ([...] << 1) << 6 where [...] will have std::cout at the deepest level. This means your macro will always result in 16 when used in std::cout, because 1 and 6 are shifted into the out stream ("1" + "6") instead of the actual numerical value 64.

You should put parantheses around the statement to avoid this:

#define TACH_MAX_OWN_ORDERS (1<<6)

or even better, since you should avoid macros, if available try to use compile time constants:

constexpr int TACH_MAX_OWN_ORDERS = 1 << 6;

How '<<' operator does work with #define in C++

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