question about ? and : in c++

Question

Why this statement :

int a = 7, b = 8, c = 0;
c = b>a?a>b?a++:b++:a++?b++:a--;
cout << c;

is not equal to :

int a = 7, b = 8, c = 0;
c = (b>a?(a>b?a++:b++):a++)?b++:a--;
cout << c;

and is equal to :

int a = 7, b = 8, c = 0;
c = b>a?(a>b?a++:b++):(a++?b++:a--);
cout << c;

Please give me some reason. Why ?

Answer 1

Because ? : ? : is right-to-left associative. It's defined like that in the language.

Answer 2

I believe @sth has provided the correct answer, however, I think @Skilldrick got it right on the comments - why the hell would you ever write something like that.

As well as the precedence issue, you really need to be careful when incrementing the same variables in a single statement. There may or may not be sequence points in the statement, and therefore the order of evaluation of the increments might not be guaranteed. You could end up with different results with different compilers or even different optimization settings on the same compiler.

Answer 3

The operators && , || , and ?: perform flow control within expressions. ?: behaves like an if-else statement.

c = b>a?a>b?a++:b++:a++?b++:a--;

if ( b>a )
    if ( a>b )
        a ++;
    else
        b ++;
else if ( a ++ )
    b ++;
else
    a --;

b>a? (
    a>b ?
        a ++
    :
        b ++
) : ( a ++ ?
    b ++
:
    a --
)

The associativity is necessary to have behavior like if … else if … else .

Sometimes I use an expression similar to yours for lexicographic sequence comparision:

operator< ()( arr &l, arr &r ) {
    return l[0] < r[0]? true
         : r[0] < l[0]? false
         : l[1] < r[1]? true
         : r[1] < l[1]? false
         : l[2] < r[2];
}