关于C ++中的i ++和++ i

Question

There are following two programs for finding largest and secondlargest integers:

1.

int main()
{
    int const no_of_elements =10;
    int list[no_of_elements] = { 1, 2, 3, 4, 5, 10, 9, 6, 7, 10 };
    int largest = list[0];
    int second_largest = list[0];
    int pos_largest = 0;
    int pos_second_largest = 0;
    int i;

    for (i = 0; i < no_of_elements; i++)
    {
        if (largest <= list[i])
        {
            second_largest = largest;
            largest = list[i];
            pos_second_largest = pos_largest;
            pos_largest = i;
        }
    }

    cout << "Largest number is : " << largest << "\n";
    cout << "And it is at the position : " << ++pos_largest << endl;
    cout << "Second largest number is : " << second_largest << " and its at the position : " << ++pos_second_largest << endl;
    return 0;
}

2. In for loop, i do pos_largest = ++i
They give different answers:
1. gives largest=10 secondlargest=10 ie as required
but 2. gives largest=10 secondlargest=9

Kindly explain????

Answer 1

i++ is post-increment. This means "give me the value of i , but after that, increment it".

++i is pre-increment. This means "increment i , then give me its new value".

int i = 1;
int a = i++; // a is now 1, i is 2

i = 1; // reset i's value to 1
int b = ++i; // i's value is incremented to 2, b's value is 2

Regarding the for loops, both of these have the same effect

int i;
for (i = 0; i < 10; i++) {
    // ...
}

for (i = 0; i < 10; ++i) {
    // ...
}

The third section in the for loop is run after each iteration (execution) of the loop body.

Answer 2

In your #2, when you hit the first 10 in your list, you assign pos_largest to be 6 (ie,. where 9 is) instead of 5 (where the 10 is) because you've pre-incremented the i before the assignment. Thus, later when you hit the second 10 in your list, it becomes the largest and your second largest is now set to 9 (ie, the value in spot 6).

Answer 3

++i and i++ is i+=1 . The former will increase i first then does operation, the latter does the operation with i , after that increase its value by 1 .

Answer 4

Well your main problem is to understand the difference between ++i (preincrement) and i++ (postincrement). ++i means that it quickly adds the value 1 in the next coming variable or lvalue. And the meaning of the i++ is that it will not add 1 to the previous lvalue or variable untill a sequence point comes. And remember that sequence points are ; { } && || etc. e. if you have a code like this.

int test=6;
cout<<(test++)<<endl;
cout<<test<<endl;

You will be thinking that it will print like this

7
7

But it will not because it is a post increment. It will not add 1 to test untill a sequence point comes and the next sequence point is ; . so it will print like this.

6
7

Opposite is the case for preincrement. It will quickly add 1 to the variable (ie lvalue otherwise it will give error ) and then display 7 without waiting for the sequence point.

Answer 5

If you use i++ it completes the operation before incrementing the variable. If you use ++i it increments before completing the operation. In most cases you want to use i++, though it greatly depends on how you're using it. For example:

for(int i = 0; i < 5; i++){
    cout << "i = "<< i << endl;
}

This will complete the operation, then increment. However, if you use this:

for(int i = 0; i < 5; ++i){
    cout << "i = " << i << endl;
}

it will increment first before printing the value