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1. Solve all: Question 11 of the first picture.

1) Select C, event name.

2）① p*(1/i) ③i=i+1。

Question 11 of the second picture.

1）caption

2) i=i-1 t=mid(s,10-val(ch),1)

Question 11 of the third picture.

1) 3 (explanation: forms are also a class).

2）①i=4 ②x=0 ③ 2d

If you need to explain, ask again!

This question also has to be written by yourself.

As you can see from the title, no matter what value j takes, y is 0, so let's look at it directly.

for k = 1 to 4

y = y + 1

next k

That is, k = 1 to 4, execute four cycles, the value of y in each cycle is increased by 1, because y=0 before the cycle, so y=4 after four cycles

As for the value of x, we have to calculate it slowly.

When j = 1, x = 1 - 1 = 0, after for k = 1 to 4, x = 4;

When j=2, x=4-1=3, after for k = 1 to 4, x=7;

When J=3, x=7-1=6, after for k = 1 to 4, x=10;

When j=4, x=10-1=9, after for k = 1 to 4, x=13;

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.