+ 3
Plz help me understand this piece of code.
int x=3 while(x++<10) {x+=2} court<<x
4 Answers
+ 18
6 //loop1 , 3 <10 , 4+2=6
9 //loop2, 6 <10, 7+2 =9
12 //loop3 , 9 <10 , 10+2 =12
//13 will be the final value of x as 12 is not smaller than 10 , so loop stops but 12++ is 13
/* don't much familiar with c++ syntax , so don't know whether to put ; or not , btw mine explanation is about what will happen in loop & what will be the value of x achieved after each cycle âș*/
+ 16
same i was thinking @John sir
//that ; is missing & added in /**/of my answer , haha c++ is also like java or java is like c++ âș
//btw U write lovely comments with full explanation
+ 8
You have a bunch of compile errors:
1: int x=3; // Added missing ;
2: while(x++<10)
3: {x+=2;} // Added missing ;
4: cout<<x; //Assumed court was cout add missing ;
1 x becomes 3
2 x or 3 is < 10 so execute loop after assigning 4 to x
3 add 2 to x for 6
2 x or 6 is < 10 so execute loop after assigning 7 to x
3 add 2 to x for 9
2 x or 9 is < 10 so execute loop after assigning 10 to x
3 add 2 to x for 12
2 x or 12 is > 10 so skip loop after assigning 13 to x
4 output x or 13
+ 3
The x variable of type integer (whole number) is initialized to 3. Then, a loop is entered. The condition of the loop increases the value of x by one, and compares the previous value of x (before adding the one) to the number 10. If the previous value of x is less than 10, the loop body is executed. In the loop body, the number 2 is added to the value of x.
The last line redirects (<<) the value of x to the standard output stream (cout), and therefore displays it in the console.