VB concept questions, please explain them in detail, and answer them well will give you extra points

The key to this problem is to save a small value to an integer type, and the result is +1 for the integer part of the decimal value

So when n=4, i=2, fun1 returns false, outputs 4 hs, and when n=5, i=2 brings fun1 into the loop, and fun1 returns ture with output 5 ss

When n=7, i=3 brings in fun1 loop, and fun1 returns ture output 7 ss

It's easy to see the loop.

print "10">"2"

In VB, print is used to display the value of the expression on the screen.

The expression here is"10">"2", note that the expression is a string-type relational expression that compares two strings"10"with"2"When comparing the strings, the ASCII characters in the VB are compared according to the ASCII code value of the characters, and the first string is"10", the first character is"1"with the 1st character in the 2nd string"2"Compare,"1"The ASCII code value is 49, and"2"The ASCII code value is 50, 49 and 50, so the string"10"to be less than a string"2", so the expression is"10">"2", so it is false, and the value that is false in vb is represented by false, so the result is false

Answer: false

1. The function p1 has two parameters, the first is the default addressing (i.e., the change of the form parameter will affect the real argument), and the second is the value finding (that is, the change of the form parameter will not affect the real argument).

Therefore, when p1(a,b) is executed for the first time, the change of y corresponding to b will not affect the value of b; When P1(b,a) is executed for the second time, the change of y corresponding to A will not affect the value of A.

2. The value of z in the comnand click() event is not the same as the variable z in the p1 function, and the assignment of z will not affect the value of z in the event.

3. When running p1 for the first time, the value of the dynamic variable z is assigned 0, and the value of a corresponds to x, x=x+z=1+0=1, [y=x-z=1-0=1, z=10-y=10-1=9]. Skin buried.

The abz printed after the four runs is , 2.

3. When running p1 for the second time, the value of the dynamic variable z is 9, and the value of b corresponds to x, x=x+z=3+9=12, [y=x-z=12-9=3, z=10-y=12-3=9.

The ABZs printed after the run were 3, 12, 2.