Ask me about the VB 0 placeholder

Write your own functions. Tell you a workaround. I won't write it. Write ideas.

Make sure that the number you enter is within 4 digits. For example, you enter 123Then add 10,000 to whatever value you enter.

This makes 123 10123, and then converts 10123 to a string. Use the right function to take the right 4 bits to get 0123. This saves you from having to judge how many digits.

It's easy and fast.

This can be formatted with the format function, and the example program is as follows:

A textbox and a commandbutton on the form, add the following ** under the click event of the commandbutton.

dim a as integer 'Declaring shaping variables.

a = cint( 'Take what you entered in the text box and convert it to an integer, and assign a value to variable a

msgbox "The result after formatting is:" & cstr(format(a,"0000")),vbokonly 'The formatted content is displayed through a dialog box.

There's a format function to try.

"prec] specifies the precision, which works best for floating-point numbers:

format('this is %.2f',[';

Output this is

format('this is %.7f',[';

Lost this is

For integer numbers, if the number of bits of the prec is small, it has no effect.

Conversely, if the number of bits is larger than the integer value, it will be filled with 0 in front of the integer value.

format('this is %.7d',[1234]);

The output is: this is 0001234].

If the premise is that the number must be less than or equal to 4 digits, you can use :

private sub command1_click()= format(, "0000")

end sub

If it really doesn't work, format afterward.

a = right("0000000000" & a, 4)

So force it to come.

Controls, cracked, registration-free.

Please refer to it.

b is a scientific notation that represents a long integer number, which occupies 4 bytes.

c is a string, not a number. Occupies 7 bytes.

So choose C.

private sub form_load()

const lowerbound as integer = 1 'define lowerbound=1

const upperbound as integer = 100 'defines lowerbound=100

dim x as integer, y as integer, z() as boolean

redim z(lowerbound to upperbound) 'defines the array z(1 to 100).

for x = lowerbound to upperbound ’x=1 to100

for y = x to upperbound step x ‘y=x to100 step x

z(y) =not z(y) 'z(y) is negated.

nextnext

for x = lowerbound to upperbound

if z(x) then

x 'outputs the value of z(x).

end if

nextend sub

The key to this problem is to see how many negations z(y) has done.

For example, when x=1, z(1) does a negation operation, so z(1)=1, and z(2), z(3) ......All of them have done a reversal operation, and all of them are 1.

x=2.

When x2=2, z(2) does another negation operation, so z(2)=0, and z(4), z(6) of multiples of the same 2 ......It is also reversed once, which is 0

When x=3, z(3) does another negation operation, so z(3)=0, the same multiple of 3 z(6), z(9) ......They are also negated once, and z(6) has been negated for the third time to be 0

Finally, to sum up, this is a process of factoring prime factors, 1 is only divisible by 1 so z(1)=12 is divisible by 1 and 2, so z(2)=0, 3 is divisible by 1 and 3, so z(3)=0, 4 is divisible by 1,2,4, so z(4)=1,......That is, the one that can be divisible by the even number is 0, and the odd number is divided by 1, I hope you can see it clearly, and if you don't understand it, continue to ask me, I haven't done this kind of math problem for a long time.

VB can't implement such a feature because it's not an interpretive language and doesn't have the ability to add dynamically like you say.

No, it can only be added manually.

private sub command1_click()=str(val( *val( *val( *val( 2end sub

Changed a little.

Double-click the command1 button and enter the ** above in the ** box.

2 types, one winsock and one string, of course, the error.

Just remove the brackets.

First of all, you initialize i=0, so that your judgment statement while i<0 cannot judge success at all, so you just skip the while statement, in fact, there is no loop.

Isn't your condition to loop when i is less than 0? Because the initial value of i is 0, there is no loop at all, and a is naturally the initial value, that is, 0 after being initialized.

i,a The initial value is 0

The loop condition is while i<0

It didn't go into the loop at all, and of course a was 0.

i=0 after i<0 is not established, there is no running in the loop, it is a=0...