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If-Then question

Started by thejoshinator Feb 21, 2005 at 11:10 AM 9 replies 800+ views

Original Post

Feb 21, 2005 at 11:10 AM

Say I have something like

if (!Thisfunc() || !Thatfunc())
                // insert code here

If Thisfunc() returns 0, will the expression still evaluate Thatfunc() or will it skip to the next line?

Scythen

Feb 21, 2005 at 11:14 AM

Its an 'OR' operation so in order to figure out if one 'OR' the other is true they both may need to be evaluated.

If !Thisfunc() is true, !Thatfunc() will not be evaluated because its already known that one of the arguments to the 'OR' is true.

If !Thisfunc() is false, !Thatfunc() will be evalueated.

Agony

Feb 21, 2005 at 11:15 AM

You could write a program to test it, you know. Wouldn't be too hard.

int Func1(){  printf("Func1\n");  return 0;}int Func2(){  printf("Func2\n");  return 1;}int main(){  if (!Func1() || Func2())    printf("if body\n");  system("pause");  return 0;}

Wan

Feb 21, 2005 at 11:15 AM

Only one of them has to be true.
"!0" is true, so it's skips after the first statement.

thejoshinator

Feb 21, 2005 at 11:16 AM

Ok, so which is it?

Trap

Feb 21, 2005 at 11:18 AM

Not all programming languages do short-cut-evaluation by default. C/C++/Java do it. Eiffel does not.

thejoshinator

Feb 21, 2005 at 11:20 AM

Alright, I think i got it. Thanks!

Wan

Feb 21, 2005 at 11:22 AM

Quote:
Original post by Trap
Not all programming languages do short-cut-evaluation by default. C/C++/Java do it. Eiffel does not.

And VB doesn't either.

Anyway:

int a = 1;int b = 1;int c = 0;// this won't give a "division by zero" error (first statement is true)if ((a / 1) == 1 || (b / c) == 1){    // do stuff}// this however fails (first statement is false, continues to to the second statement)if ((a / 1) == 2 || (b / c) == 1){    // do stuff}

okonomiyaki

Feb 21, 2005 at 11:54 AM

This is generally considered a bad thing to do also. I'm talking about using function return values in short circuited operations, not just taking advantage of the short circuit operation. Not horrible, but makes it a little harder to read.

Zipster

Feb 21, 2005 at 1:08 PM

You could always do a little something like this:

if(!ThisFunc() | !Thatfunc()){   // Something}

4.7.4 of the standard states that true is converted to 1 while false is converted to 0 (here for the function return values), and 4.12.1 states that non-zero values are converted to true and zero values are converted to false (here for the conditional), so this type of bitwise logic and opposed to the boolean logic should avoid the short-circuit evaluation and should work according to the standard. However you should probably stick a comment in there somewhere, just so your peers don't rip your heart out Aztec style for using bitwise logic in a conditional, hehe [smile]

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