Explainig How this MACRO works ?? [printing VAR name in c++]

Question

I'm using this MACRO for debugging in C++ . Printing the variable name and value. Can anyone explain its working ?

#define watch(x) cout << (#x) << " is " << (x) << endl;
int t = 90;
watch(t);

// output is

t is 90

Run the code through the preprocessor to see how the macro is expanded. — Some programmer dude, Aug 01 '19 at 07:16
I also recommend e.g. [this reference about the `#` and `##` macro operators](https://en.cppreference.com/w/cpp/preprocessor/replace#.23_and_.23.23_operators). — Some programmer dude, Aug 01 '19 at 07:20

score 3 · Answer 1 · answered Aug 01 '19 at 07:19

3

"stringizing" operator (#) converts macro parameters to string literals without expanding the parameter definition. It is used only with macros that take arguments.

answered Aug 01 '19 at 07:19

GOVIND DIXIT

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Explainig How this MACRO works ?? [printing VAR name in c++]

1 Answers1