VB questions, this kind of questions will not be encountered every time, I hope to explain in as muc

You have two calls to sun1 here, static i as integer is the first one.

II. The third ......When the sub procedure is called, the i value does not change to the initial value of 0 and continues to run.

First, call(x,x,z),x is passed to a,a=1, in the same way b=1,c=3,i=1+3+0=4,a=3*3+4=13,b=10,c=24. Because call(x,x,z), a=b, i.e. there is an implicit execution statement a=b=10, so the last first line prints 10 10 24.

After call(x,y,y), because it is passed by address, x=10, y=2, i.e. a=10, b=2, c=2. Because static, i is still equal to 4 and the initial value 0 is not assigned, i = 10 + 2 + 4 = 16, a = 22, b = 20, c = 58 (similarly, implicit b = c = 58), so the last second line prints 22 58 58. Because of the value passing, x=22, y=58, z=24, that is, the third line prints 22 58 24

Emphasis is placed on static variables. (Static variables, the value of the variable will not be cleared after the process is executed, and the value of the last saved variable will not be re-initialized to 0 when the next call is made).

After remembering this, you need to record the value of the static variable separately when doing this problem, and update it once when you execute a line.

As my title:

The first time sun1(1,1,3) is executed, i=0, and i=a+c+i=1+3+0=4 is executed, and i=4 is recorded on paper

The second time sun1(1,2,2) is executed, i=4, and i=a+c+i=1+2+4=7 is executed, i=7 is recorded on paper

The first time sun1(x,x,z) is called, x=1; x=1;z=3 brings a, b, c, i.e. a=1; b=1;c=3;

Function starts calculation:

static i initialization, i=0;

i=a+c+i=1+3+0=4;

a=3*c+i=3*3+4=13;Return to the argument x, i.e., x=13;sun1（13,x,z）

b=2*c+i=2*3+4=10;Return to the argument x, i.e., x=10;sun1(10,10,z) ps: Since x uses the same address to store the value, the value of y should be the same, here the value of x is assigned to 10, so the two x parameters become 10

c=a+b+i=10+10+4=24；Return to the argument x, i.e., z=24;sun1(10,10,24)

print outputs a, b, and c are outputs"10 10 24";

the second call sun1(x,y,y), x=10; y=2;y=2 brings a, b, c, i.e. a=10, b=2, c=2

Since static i is a static variable, the system allocates a memory storage address for i, which stores the value assigned to i by the last call sun1(), that is, i=4;

Function starts calculation:

i=a+c+i=10+2+4=16;

a=3*c+i=3*2+16=22;Return to the argument x, i.e., x=22;sun1（22,y,y）

b=2*c+i=2*2+16=20;Return to the real argument y, i.e., y=20;sun1(22,20,y)

c=a+b+i=22+20+16=58；Return to the real argument y, i.e., y=58;sun1(22,58,58) ps: Since y uses the value stored in the same address, the value of y should be the same, and the value of y is assigned to 58, so the two y parameters become 58

print outputs a, b, and c are outputs"22 58 58";

This kind of topic is not explained to solve the problem, but you need to start from scratch and study hard!

The explanations of so many friends are nothing more than the contents of the book!

Take a look at the basic books of VB and you will understand, simple explanations are useless.