The difference between i++ and ++i: tell their differences easily

Recently, a friend asked me a question , The title is like this ：

Assume that all variables are integers , Expression (x=2,y=5,y++,x+y) The value of is ( )

A. 7                     B. 8                        C.9                        D. 2

Many people don't understand a problem , Because a lot of people （ Including teachers ） It's described like this ：

i++ Is the assignment first , And then add up ;++i It's self increasing first , Post assignment

This will produce a A little misunderstanding . In this formula , expression y++, It is in this statement that +1 Well , Or in the next sentence +1？

The answer to this question is 8, choose B.

The reason is , I think it's understandable ：

Programming Expressions usually have return values ,i++ And ++i No exception .

Such as ：

int y = 3 + 2;

You can put the 3 + 2 regard as expression , Calculation result assigned to y.

Empathy , We can understand it as ：

i++ And ++i Can change variables i Self increasing 1, But before that ：

i++ The expression increases by itself , The return value is i 

++i The expression increases by itself , The return value is i+1

These two sentences are very important , If you use this conclusion to look at other topics , There is no difficulty in finding ！

Such as ：

int a = 1;
int b;
b = a++;
// b  by  1

because a++ The return value of a, Assigned to b. all b The value of is at this time a Value .

Such as ：

int a = 1;
int b;
b = ++a;
// b  by  2

because a++ The return value of a+1, Assigned to b. all b The value of is at this time a+1 Value .

The same is true in the output , Such as ：

int a = 1;
printf("%d\n",a++);
//  Output 1

because a++ yes printf The parameters of the function , So the output is  a++ expression The return value of , That is to say a

Such as ：

int a = 1;
printf("%d\n",++a);
//  Output 2

because ++a yes printf The parameters of the function , So the output is ++a expression The return value of , That is to say a+1

Of course , If no one uses their return value , Their effect Exactly the same ：

Such as ：

int a = 1;
a++;

and

int a = 1;
++a;

Now let's look at the problem at the beginning ：

Assume that all variables are integers , Expression (x=2,y=5,y++,x+y) The value of is ( )

A. 7                     B. 8                        C.9                        D. 2

Can be seen as y++ In the expression, but no one uses its return value .

So at the moment y++ And ++y It's exactly the same , It just did y+=1 It's just the operation of !

without doubt , This question 2 + 6 = 8 ！

I still use C Functions of language simulation For a moment i++ and ++i Internal operations ：

/**
 *  amount to i++ Function of 
 * @return i
 */
int iPlusPlus(int * i){
    *i = *i + 1;
    return *i;
}

/**
 *  amount to ++i Function of 
 * @return i+1
 */
int plusPlusI(int * i){
    *i = *i + 1;
    return *i+1;
}

It's not easy to create , Give me a compliment if you like ！ You are also welcome to point out mistakes in the comment area ！