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c++operator-precedence
user6490375
user6490375

Reputation:

Operator precedence C++

What operator precedence takes place in the following C++ statement?

int lower = height[height[l] < height[r] ? l++ : r--];

Assume, height=[0,1,0,2,1,0,1,3,2,1,2,1] and int l = 0, r = height.size()-1;.

As per my understanding, height[l] < height[r]? would be evaluated first, followed by l++ or r-- as the case might be (l++ in this case). Then the value of height[updated l] should be assigned to lower. Thus lower should be 1; but the output says it is 0. How? What would be the correct order of the evaluation of the statements then?

Upvotes: 1

Views: 182

Answers (3)

Logman
Logman

Reputation: 4189

The code is equal to:

int lower = height[height[l] < height[r] ? l : r];
(height[l] < height[r]) ? l+=1 : r-=1;

Upvotes: -1

NineBerry
NineBerry

Reputation: 28499

The l++ uses the post-increment operator. The value of l is increased after the value of the expression is evaluated. So, when l is 0, the value of l++ is 0 and not 1, even though the variable l will have the value 1 afterwards.

Using ++l would show a different behaviour.

The following statement in the question is incorrect:

Then the value of height[updated l] should be assigned to lower

The updated value of l is not used as the array index, but the value of the expression l++, which is the value of l before it is increased.

Upvotes: 3

Fantastic Mr Fox
Fantastic Mr Fox

Reputation: 33864

l++ is the post increment operator, which means it will be incremented after the operation is complete. Hence the output 0, everything else is correct.

Upvotes: 1

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