I++ and ++i details

i++ And ++i Detailed explanation

i++ And ++i Not with assignment number （=） When connected, the effect is equivalent to i=i+1

Examples are as follows ：

//++i
public class demo
{
    
    public static void main(String[] args) {
    
        int i=0;
        ++i;
        System.out.println(i);
    }

}
//i++
public class demo
{
    
    public static void main(String[] args) {
    
        int i=0;
        i++;
        System.out.println(i);
    }

}

The output of both code blocks is 1

i++ And assignment number （=） At the same time, the performance takes precedence i And then add one

public class demo{
    
    public static void main(String[] args){
    
        int q;
        int i=0;
        q=++i;
        System.out.println(q);
    }
}

++i And assignment number （=） When it appears at the same time, add 1 Then take

public class demo{
    
    public static void main(String[] args){
    
        int q;
        int i=0;
        q=++i;
        System.out.println(q);
    }
}

Now we will i++ And ++i And assignment number （=） Put them together and discuss their order

public class demo
{
    
    public static void main(String[] args) {
    
        int q;
        int i=0;
        q=(++i)+(i++);
        System.out.println(q);
        System.out.println(i);
    }

}

We put q=(++i)+(i++) Take it out and see

Before this step, our q Unassigned ,i The value of is 0,

When running here （） The highest priority is to execute the things in brackets first. The first bracket indicates ++i take i+1 Assign to i, here i Turned into 1 Participate outside the brackets + Method operation is 1, Look at the second bracket i++ It means first i Value is assigned to i Give again i+1 here i by 2, But one thing to note ,i++ It is added after the assignment, so the value of the right bracket when participating in the addition operation outside the two brackets is （ Current value -1） namely 1. therefore q,i The value of will end up being 2

Now let's raise the difficulty , Add 3 individual

public class demo
{
    
    public static void main(String[] args) {
    
        int q;
        int i=0;
        q=(++i)+(i++)+(++i);
        System.out.println(q);
        System.out.println(i);
    }

}

Carry it here alone q=(++i)+(i++)+(++i) This statement shows

Above this statement q No value ,i The value is 0

To this sentence, run from left to right in parentheses ：

• In the first bracket 0+1,i The value of becomes 1,1 Participate in external operations
• In the second bracket 1+1,i The value of becomes 2,1 Participate in external operations
• In the third bracket 2+1,i The value of becomes 3,3 Participate in external operations

q To be an assignment 1+1+3 by 5,i Final value for 3

The above statement is equivalent to the following statement ：

public class demo
{
    
    public static void main(String[] args) {
    
        int q;
        int i=0;
        int a=++i;
        int b=i++;
        int c=++i;
        q=a+b+c;
        System.out.println(q);
        System.out.println(i);
    }

}