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c++pointerscomputer-scienceoperator-precedence
illla
illla

Reputation: 105

C++ Precedence and Associativity

This piece of code:

int scores[] {1,2,3,4};
int *score_ptr {scores};  
//let's say that initial value of score_ptr is 1000
std::cout<<*score_ptr++;

produces the output:

1

As * and ++ have same precedence and then associativity is from right to left shouldn't we apply ++ operator first that is to increase the pointer first and then *(dereference) it?

So accordingly score_ptr will be increased to 1004 and then dereferencing it will give the second element of scores which is 2.

How and why does this give me output of 1 instead of 2?

Upvotes: 2

Views: 336

Answers (2)

anastaciu
anastaciu

Reputation: 23802

The increment will happen first, it has higher precedence, it's equivalent to *(score_ptr++), but it's a post-increment, this means it will only happen after the dereferenced pointer is used, i.e. the expression reaches ;.

If you use

std::cout << *++score_ptr;

Then you have a pre-increment, here it will happen beforehand, the pointer will be incremented before the value is used and the output will be 2. Equivalent to *(++score_ptr).

Note that it's allways a good idea to use parentheses, it will make the code clearer and will avoid missinterpretations.

Upvotes: 1

songyuanyao
songyuanyao

Reputation: 172934

As * and ++ have same precedence

No, postfix operator++ has higher precedence than operator*; then *score_ptr++ is equivalent to *(score_ptr++). Note that the postfix operator++ would increment the operand and return the original value, then *(score_ptr++) would give the value 1.

The result is prvalue copy of the original value of the operand.

On the other hand prefix operator++ returns incremented value. If you change the code to *++score_ptr (which is equivalent to *(++score_ptr)) then the result would be 2 (which might be what you expected).

Upvotes: 7

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