Difference between i++ and ++i?

问题

This is My Code:

int main()
{
    int i=2,j=2,c=2,d;

    d= ++i || ++j && c;
    printf("I is %d\n",i);
    printf("J is %d\n",j);
    printf("C is %d\n",c);
    printf("D is %d\n",d);
    return 0;
}

The output of the following code is:

Now my question is, if ++i is 3 then why ++j is 2?

What is the difference between ++i & i++? Also I want to know that how come d is 1 ?

回答1:

You asked:

Now my question is if ++i is 3 then why ++j is 2?

++j is never executed since ++i evaluates to true.

The expression ++i || ++j && c is equivalent to ++i || (++j && c) due to operator precedence.

The run time evaluates the expression ++i || ++j && c from left to right.
The first term, ++i evaluates to 3, with the side effect that the value of i is 3. Since it is non-zero, the run time does not need to evaluate anything after that. Since ++j is never evaluated, the value of j is not changed.

What is the difference between ++i & i++?

The value of the expression ++i is i+1 with the side effect that the value of i is incremented by 1.

The value of the expression i++ is i with the side effect that the value of i is incremented by 1.

int i = 2;
int x = ++i;  // x == 3, i == 3 at the end of execution of the statement
int y = i++;  // y == 3, i == 4 at the end of execution of the statement

I want to know that howcome D is 1

The value of d is set to 1 since the boolean value of the expression is true.

回答2:

++i is pre increment.
i++ is post increment.
Suppose here i = 2

In Post In crement

printf("%d",i++); // it will print output 2 then increment value of i to 3

In Pre Increment

printf("%d",++i); //first it will increment value of i to 3, then print output 3

d= ++i || ++j && c;

Condition will be performed in this manner ++i || (++j && c);
As i=2, after ++i it will become 3.
In OR if first condition is true it will skip second condition. means ++j && c will not be performed.
Result of ++i || ++j && c is 1
so d = 1.

printf("I is %d\n",i);  // i = 3;
printf("J is %d\n",j);  // j = 2;
printf("C is %d\n",c);  // c = 2;
printf("D is %d\n",d);  // d = 1;

Short Circuit evaluation

回答3:

Reason for the evaluation is short-circuiting of the boolean operators && and ||.

For &&, if the left-hand side expression is false, the combined result is false (the right-hand side expression is never evaluated).
For ||, if the left-hand side expression is true, the combined result is true (the right-hand side expression is never evaluated).

This is mostly important when the right-hand expression has side effects, such as altering values(or calling some functions with side-effects), for example in your case:

d= ++i || ++j && c;

Note that the combined conditional expression would increment i by 1 since the conditional expression on the left of || is true, the conditional expression on the right-hand side i.e. ++j && c is never evaluated.

Thus, the total expression will lead eventually to:

d = 1 || ++j && c;

which explains why d=1?

Now if you change the value of i=-1 then the expression on the right of the || will be executed and the j will be incremented as per your expectation

Now for the performance effects of i++ vs ++i check here :)

来源：https://stackoverflow.com/questions/29998697/difference-between-i-and-i

标签

short-circuiting