#define macro doesn't print variable [duplicate]

Question

I have this code:

#define PRINT(fmt, ...) printf("%s:%d: "fmt, __FILE__, __LINE__, ##  __VA_ARGS__)

It works well when I pass some value to it, eg PRINT("Hello, world"). But when I pass variable as an argument, eg. PRINT(somevar) it doesn't print the value of variable int somevar. How should I rewrite the macros to make it print variables as well?

You mean I have better to rewrite it as a conventional routine and use va_list? — cwenz1980
– cwenz1980, Commented Feb 8, 2015 at 5:17
The macro will not work unless the fmt (first) argument to the macro can be concatenated with the "%s:%d: " string, which means it too must be a literal. To print int somevar, the macro invocation must be PRINT("%d\n", somevar);. — Jonathan Leffler
– Jonathan Leffler, Commented Feb 8, 2015 at 5:20
In short, you can't invoke printf as printf(some_int_var) either; a format string is needed. This is no different (and in fact mandatory as a const-literal, as your logic now mandates it as concat-able, and a computed format string is not feasible, fair warning). The only immediate you can get away with sending to this is just like printf, a char* (const or otherwise) itself, and just like printf that would be discouraged for security reasons. — WhozCraig
– WhozCraig, Commented Feb 8, 2015 at 5:28

Rob · Accepted Answer · 2015-02-08 05:23:49Z

3

The problem isn't in the macro. It is in how you use it. Essentially, your usage requires you to use PRINT with a format string, which must be a string literal that describes what to do with the subsequent arguments.

For example, try

PRINT("x = %d y = %d\n", x, y);

where x and y are variables of type int.

answered Feb 8, 2015 at 5:23

Rob

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1 Comment

cwenz1980 Over a year ago

@Jonathan Leffler and Rob. Terrific!! You have solved my problem in no time at all!!

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