C中的关联性和序列点

Question

Since the associativity of '?' is from right to left,any 2 consecutive '?' operators must be treated as such,Right?

Now,

int x=-1;
int y=x?x++?x:-1:1;

I expect this to be executed as:

int y = x ? (x++?x:-1) : 1;

Now since its being executed from right to left,when encountering the first '?' in the statement,x's value is 0 and the expression is as

int y= x? 0 : 1;

hence i expected y to be 1,but it shows Zero on my dev-cpp.Where am i wrong?

Answer 1

You have the order of evaluation wrong. In a ? b : c a ? b : c , a is always evaluated first, then either b or c is evaluated.

I've marked up your example so that I can identify subexpressions:

            c
int y=x?x++?x:-1:1;
      a bbbbbbbb

(a) is evaluated, yielding -1, so (b) is evaluated. There, x++ is evaluated, yielding -1 again, so (c) is evaluated. At this point, x is 0.

Or, with more verbose, clearer code, it's as if you said:

int x = -1;
int y;
if (x != 0)
{
    int new_x = x + 1;
    if (x != 0)
    {
        y = new_x;
    }
    else
    {
        y = -1;
    }
}
else
{
    y = 1;
}

Answer 2

Operations:

Assign y to value = 
    if(x): --> x = -1, so true as it is non-zero
    {
      if(x): --> x = -1 ,so true as x will increment later due to post increment
       x= x +1; --> increment x, so x = 0 . This is the value assigned. So y = 0;
     else:
       -1
    }
    else: 
    {
      1
    }

Hope this helps!

Answer 3

The answer to your question is that in C/C++ int y = x ? (x++?x:-1) : 1; int y = x ? (x++?x:-1) : 1; we will hit two sequence points at ? . Any update operations to variable with in a sequence point will be effective after that sequence is over. So lets look at our example in hand.

First sequence point is first ? from left.

x=-1; (Actual Value)
x=-1; (Value used in expression)
y=-1?(x++?x:-1):1;

Second sequence point is second ? from left. As mentioned above the update operations are effective after sequence so even though x++ is there the value used in this sequence is -1 and updated value will be used in following.

x=0; (Actual Value, bcoz of x++)
x=-1; (Value used in expression)
y=-1?x:-1;

Now it will be

x=0; (Actual Value)
x=0; (Value used in expression)
y=x;
y=0;

Hope this make sense now.