how i c++ macro act function?

April 15, 2014

let's contend reason need macro: macro(x,y). (let's assume there's good reason can't an inline function.) wish macro heed duty relapse value.

example 1: should work expected.

if (x > y)
  macro(x, y);
do_something();

example 2: should outcome compiler error.

if (x > y)
  macro(x, y);
else
  macro(y - x, x - y);

example 3: should not compile.

do_something();
macro(x, y)
do_something();

the nave proceed macro this:

#define macro(x,y)                       \
cout << "1st arg is:" << (x) << endl;    \
cout << "2nd arg is:" << (y) << endl;    \
cout << "sum is:" << ((x)+(y)) << endl;

this unequivocally bad fortitude fails 3 examples, i shouldn't need explain why.

ignore macro indeed does, that's point.

now, proceed i many mostly macros combined hang curly braces, this:

#define macro(x,y)                         \
{                                          \
  cout << "1st arg is:" << (x) << endl;    \
  cout << "2nd arg is:" << (y) << endl;    \
  cout << "sum is:" << ((x)+(y)) << endl;  \
}

this solves instance 1, since macro matter block. nonetheless instance 2 damaged since put semicolon after macro. creates compiler cruise semicolon matter itself, means else matter doesn't conform any statement! lastly, instance 3 compiles ok, even nonetheless there semicolon, since formula retard doesn't need semicolon.

is there proceed macro pass 3 examples?

note: i am submitting possess answer biased , nonetheless anyone improved fortitude feel giveaway post here, competence some-more votes method. :)

code1