Need help - C++ beginner

davekw7x · #11 (**permalink**) August 7th, 2004, 03:22 PM

I don't agree: since the behaviour is undefined, there is no such thing as a "wrong" answer any more than there is a "right" answer.

You can play games, guessing how different implementations might handle it, but compilers are not required to give answers consistent with your guesses about specific implementations. (No matter how reasonable your assumptions seem to you.)

The point of the exercise is not to decide which is the more nearly correct answer, but to caution programmers that operators or functions with side effects can give results that mislead you.

Note that there is nothing illegal about the syntax of any of the statements, but they are symantic nonsense, since the C standard is very explicit that order of evaluation of all functions and of certain operators (including ++) is undefined.

Here's a simple rule of thumb. If you see an expression with ++ or -- applied more than once to the same variable, raise the red flag: change it!!!

Regards,

Dave

mehdi62b · #12 (**permalink**) August 7th, 2004, 04:04 PM

yes,okay
>>what values for a,b,c,r after below instruction(r=b=c=0,a=-1)
r=++a||++b||++c
---
answer:
a=0,b=1,c=0,r=1
Do you know why c is zero?because in OR when b gets 1,compiler doesn't consider ++c
(I dont know what could happand to them in Borland)
Cheers.

--------------------------------------------
Mehdi.:)

davekw7x · #13 (**permalink**) August 7th, 2004, 08:59 PM

Now here's one that has an unambiguously unequivocal "right" answer, since the C standard says that expressions connected by the "||" are evaluated left to right, and whenever a non-zero value is found for an term in the expression, no further evaluation takes place. Note that ++ has a higher precedence than ||, so the first thing that is evaluated is ++a. This sets a to a new value 1, and the value of the expression is the new value of a, and since this is non-zero, no further evaluation takes place past the ||.

Borland gives the same (correct) answer, as does Gnu gcc, and all of the other valid C compilers, including Microsoft Visual C++.

What's the answer?

compile and run this:

Code:

#include <stdio.h>

int main()
{

  int a = 0;
  int b = 1;
  int c = 0;
  int r;

  r=++a||++b||++c;

  printf("a = %d, b = %d, c = %d, r = %d\n", a, b, c, r);
  return 0;
}

If you don't get the following results, ask for your money back on your C compiler:

a = 1, b = 1, c = 0, r = 1

Dave

davekw7x · #14 (**permalink**) August 7th, 2004, 09:18 PM

Quote:

quote:Originally posted by davekw7x
Now here's one that has an unambiguously unequivocal "right" answer, since the C standard says that expressions connected by the "||" are evaluated left to right, and whenever a non-zero value is found for an term in the expression, no further evaluation takes place. Note that ++ has a higher precedence than ||, so the first thing that is evaluated is ++a. This sets a to a new value 1, and the value of the expression is the new value of a, and since this is non-zero, no further evaluation takes place past the ||.

Borland gives the same (correct) answer, as does Gnu gcc, and all of the other valid C compilers, including Microsoft Visual C++.

What's the answer?

compile and run this:

Code:

#include <stdio.h>

int main()
{

  int a = 0;
  int b = 0;
  int c = 0;
  int r;

  r=++a||++b||++c;

  printf("a = %d, b = %d, c = %d, r = %d\n", a, b, c, r);
  return 0;
}

If you don't get the following results, ask for your money back on your C compiler:

a = 1, b = 0, c = 0, r = 1

Dave