In the first program, why does count gives 6 and in the second program, why 7 instead of 13?

Question

//first program
var count = 0;
for(var i=0; i<3; i++) {
count--;
for(var j=0; j<=2; j++) {
count++;
}
}
alert(count);

//second program
function func(a) {
return a++
a+=5;
}
document.write(func(7));

Answer 1

First:
If you think about it, the first loop runs three times, meaning count would be -3. But there's another for loop, which also executes thrice, meaning count will be added 3 times. If you do the math, which isn't that complicated:
-3 + 3 * 3 = 6

Second:
The return keyword stops executing the function, whatever is after it, is ignored. Meaning it will return 7++, which is still 7

Answer 2

Oh, thanks Airree for your answer, it helped

Answer 3

when first loop run inner loop will run 3 times which means when first loop run count-- is -1 (0-1) and inner loop increment count three times( -1 +3 = 2),
first loop run second time count-- is (2-1=1) and inner loop increment count three times (1+3=4).
first loop run third time count--(4-1=3) and inner loop increment count three times which is 3+3=6 and loop end

//second program
a++ mean a==a+1
postincrement (assign value first and add 1 is ignored after that)
if it is ++a preincrement.