C＃中的运算符优先级和关联性

Question

I've two lines of coding as below

int? i = 1;
int j = i ?? 2 +1;

now "j is 1"

int? i = 1;
int j = (i ?? 2) +1;

now "j is 2"

Could you explain how?

Answer 1

Sure - it's simple a matter of precedence. The (Microsoft) C# specification lists operator precedence in section 7.3.1 (in the C# 4 and C# 5 specifications, anyway; it's 7.2.1 in the C# 3 spec), although that's only really an informative table - it's really governed by the grammar.

Anyway, the null-coalescing operator ( ?? ) has lower precedence than the binary addition operator ( + ) so your first code is equivalent to:

int? i = 1;
int j = i ?? (2 + 1);

As the value of i is non-null, the right hand operand of the null-coalescing operator isn't even evaluated - the result of i ?? (2 + 1) i ?? (2 + 1) is just 1.

Compare that with your second example, where the null-coalescing operator expression again evaluates to 1, but then 1 is added to the result of that. It's effectively:

int tmp = i ?? 2; // tmp is now 1
int j = tmp + 1;  // j is 2

Associativity is irrelevant here, has it only controls the ordering/grouping when an operand occurs between two operators with the same precedence.

Answer 2

Maybe you will understand it better without the use of the ?? operator

Scenario 1:

int? i = 1;
int j;
if (i != null)
{
    //i is not null so this is hit
    j = i;
}
else
    j = 2 + 1;
}

So j = 1

Scenario 2:

int? i = 1;
int j;
if (i != null)
{
    //i is not null so this is hit
    j = i;
}
else
    j = 2;
}

So j = 1

//No matter the result of the above if, 1 is always added.
j = j + 1;

So j = 2

Answer 3

int? i = 1;
int j = i ?? 2 + 1; //If I is null make j = to 3

Now j is 1

int? i = 1;
int j = (i ?? 2) + 1;  //If I is null make j = 2 else make j = to 0 and then add 1 

Now j is 2