conditional operator in C question

Question

I just have a quick question about the conditional operator. Still a budding programmer here. I am given x = 1, y = 2, and z = 3.

I want to know, why after this statement:

y += x-- ? z++ : --z;

That y is 5. The values after the statement are x = 0, y = 5, and z = 4. I know the way the conditional operator works is that it is formatted like this: variable = condition ? value if true : value if false.

For the condition, y += x-- , how does y become 5? I can only see 2 (2 += 0) and 3 (2 += 1)(then x-- becomes zero) as possibilities. Any help is much appreciated. :)

Answer 1

When it evaluates the condition (x != 0) x is still 1 (that is not 0). So it picks z++. Which is still 3. 2 + 3 = 5. At the end of the day x has become 0 and z has become 4.

Take a look here for details. It's important to remember a simple thing: when you say x ++ the current value of x is used and then it is incremented. When you say ++x it is first incremented and then used.

Answer 2

Hi,

The reason for this is that post-decrement/increment operator (x++ or x--) does the following:

1. increment or decrement the variable
2. return the original value.

So the return value of x-- is 1, indicating true, so the statement z++ is evaluated, returning the original value 3.

Since y = 2, y += 3 is 5.

Answer 3

Just break it down into a similar if statement:

if (x--)
    y += z++;
else
    y += --z;

In your case, since x is 1, you'll take the "true" side of this if statement. That means you're adding z++ to y, giving 3 + 2, resulting in 5.

Please don't write code like this.

Answer 4

x-- would mean the expression is evaluated at current value of x and after that x is reduced by 1. Same is the case with Z++. This is the other way around for --z, which means this is evaluated at the new value of z.

So at the time of evaluation x is 1, z is 3. and after the evaluation of expression x becomes 0 and z 4; and y = 2 + 3 = 5

Answer 5

Remember that the increment and decrement operators return different things depending on whether they're placed before or after the name of a variable.

In particular, when x-- is evaluated, it decreases x by 1, but returns the unmodified value of x, which in this case is 1. In C, 1 evaluates to true, so the ternary operator will return z++.

And again, because the ++ operator is placed after the variable, the return value of z++ is the unmodified value of z, which is 3.

Thus, this comes down to y += 3, which results in y being 5.

Answer 6

As a budding programmer just know that you should NEVER write anything like this so that way you can forget about it worrying you!

Answer 7

x-- and z++ decrement and increment after they are used.You end up with the following when the ternary operator is evaluated:

y += (1) ? (3) : (--z);

--z never gets called, the conditional evaluates to true and executes the first option in the ternary operator. After use, x is then decremented, and z is incremented.

Answer 8

The Operator ?: has higher precedence than the operator +=. So Your expression is evaluated as

y += (x-- ? z++ : --z);

the value of x-- ? z++ : --z expression is the value of the expression z++ (that is 3) because the value of the expression x-- is 1

Answer 9

It is normal because it first "runs" ternary operator then does decrement as your decrement operator (x--) is postfix, so you got z++ which is 3 so you have 5 in y.

Answer 10

The expression x-- evaluates to the current value of x, which is 1. Thus the result of the conditional expression is z++, which evaluates to 3. 3 gets added to y, for a total of 5.

Answer 11

I think your fundamental issue here is that you're assuming 'y+= x--' is your condition, when in fact your condition is merely 'x--'. There is a return from the conditional operator which makes y += the result of the conditional operation: "x-- ? z++ : --z;" is 5. Other comments have the reason why it actually evaluates to 5.