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bikola_p

c++ inconsistency...really, look inside.

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Why doesnt the macro equivilent of the Inline function act the same, the code is a replica. However, strange enough i get diff results. Here look.
#include <iostream>
using namespace std;

#define BAND(x) (((x)>5 && (x)<10) ? (x) : 0) //LOOK AT THIS LINE


int main() {
  for(int i = 4; i < 11; i++) {
    int a = i;
    cout << "a = " << a << endl << '\t';
    cout << "BAND(++a)=" << BAND(++a) << endl;
    cout << "\t a = " << a << endl;
  }
  return 0;
}

IN COMPARISON TO 

#include <iostream>
using namespace std;

inline band(int x) {return (x>5 && x<10)? x : 0;} //THIS LINE


int main() {
  for(int i = 4; i < 11; i++) {
    int a = i;
    cout << "a = " << a << endl << '\t';
    cout << "band(++a)=" << band(++a) << endl;
    cout << "\t a = " << a << endl;
  }
  return 0;
}
THE output of both programs are different, after follwoing the programs, the inline function is the correct one, what is wrong with macro's?

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The wrong thing about macros is, that they are only textual replacements.
In you case the line

cout << "BAND(++a)=" << BAND(++a) << endl;

is replaced by

cout << "BAND(++a)=" << (((++x)>5 && (++x)<10) ? (++x) : 0) << endl;


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oh lol, so its something that was around, i thought this discovery would make me famous like the pioneers..nah just playin, thanks for that though...

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I'm pretty sure that the behaviour with the macro version isn't even (pardon the pun) defined.

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Quote:
Original post by Zahlman
I'm pretty sure that the behaviour with the macro version isn't even (pardon the pun) defined.


Invoking the macro on its own with ++a is well defined -- logical operations and the ternary operation imply sequence points and have a well defined order.

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Quote:
Original post by Polymorphic OOP
Quote:
Original post by Zahlman
I'm pretty sure that the behaviour with the macro version isn't even (pardon the pun) defined.


Invoking the macro on its own with ++a is well defined -- logical operations and the ternary operation imply sequence points and have a well defined order.


Can you go into more detail? I never understood what would be defined behavior in cases like these and not.

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C++ has the concept of sequence points. A sequence point is basically a point in the code where everything before it has to have finished evaluating before execution proceeds any further. For example: the end of a statement is a sequence point. So this is well defined:
++a;
++a;

Even though this is not:
++a + ++a;


Another example is a function call: all arguments to the function call must be evaluated before the function is actually called. So if you have foo(++a, ++b), you know that both ++a and ++b were both evaluated before foo() was called.

Logical and and or, the ternary operator and the comma operator also have a sequence point after the first expression. So in A && B, A || B, A , B and A ? B : C the A has to finish evaluating, with all its side effects before B (or C) is evaluated. So ++a && ++a has well defined behaviour even though ++a + ++a</t> doesn't.

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Quote:
Original post by SiCrane
Another example is a function call: all arguments to the function call must be evaluated before the function is actually called. So if you have foo(++a, ++b), you know that both ++a and ++b were both evaluated before foo() was called.


Just to add one more: while you can be sure that they will be evaluated before the function call, you can't be sure about the order. So doing something like foo(a,++a) and relying on a left to right order isn't a good idea (and real fun to debug when differently optimized builds or builds from different compilers behave differently).

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