Botcho_Otkho wrote:

TheAdriCoolManDude wrote:

when [variable v] = (6) :: events hat
//Workaround

forever
if <(variable) = [6]> then
...
end
end

Meh, this is better:

when green flag clicked
forever
wait until <(foo) = (. . .)>
. . .
end

TheAdriCoolManDude wrote:

I don't have a lot of workarounds, but I'm willing to help people if I can.

sonicfan12p wrote:

TheMonsterOfTheDeep wrote:

-snip-

“But that's too long!” I hear you say. Note that this is the absolute complete workaround, that does everything mentioned in the OP (reverse output on backwards indices) and then some (return the same thing as letter (0) of (str) if indices are out of bounds).

For the majority of cases, something much simpler will work, such as:

define letters (a) to (b) of (str) => result
set [i v] to (a)
set [result v] to []
repeat (((b) - (a)) + (1))
set [result v] to (join (result) (letter (i) of (str)))

Note that the only things this block does not support are a) reverse strings, b) bounds checking, and c) strings above the join block limit, which is barely even a problem, as most string manipulation will involve the join block anyways.

Also note that this block will still return a valid value for invalid bounds, simply due to how Scratch bound checks the letter of () block.

-snip-

I would like to point out that this script is broken, it will return letter a of str a total of the number of letters to check + 1. So for the string “Happy Birthday!” trying to get letters 4 - 7 with this script would set result to pppp. To fix it, make it look like this.

define letters (a) to (b) of (str) => result
set [i v] to (a)
set [result v] to []
repeat (((b) - (a)) + (1))
set [result v] to (join (result) (letter (i) of (str)))
change [i v] by (1)
end

Or this.

define letters (a) to (b) of (str) => result
set [i v] to ((a) - (1))
set [result v] to []
repeat (((b) - (a)) + (1))
change [i v] by (1)
set [result v] to (join (result) (letter (i) of (str)))
end

Now letters 4 - 7 of “Happy Birthday!” will return py B.

MathlyCat wrote:

(snip)
Your example is fairly ineffective for organizing multiple variables.

(how much [letter] in [word]?)

define how much [letter] in [word]
set [num v] to (0)
set [result v] to (0)
repeat (length of (word))
change [num v] by (1)
if<(letter (num) of (word)) = (letter)> then
change [result v] by (1)
end
i think it works :D
example below!

when flag clicked
ask [hello, i can count how many letters are there in a word. type in any word] and wait
set [word v] to (answer)
ask [now type any letter] and wait
how much (answer) in (word)
^^ block above is moreblock we defined at top
say (join [This word has ] (join (result) [letters])) for (2) secs

creeper82 wrote:

I'm not sure if I understand what is this topic about and i understand nothing. i think that's my third post on these forums…
Okay, so… here's my block:

(how much [letter] in [word]?)

and the workaround or whatever is it called (sorry if i do something wrong )

define how much [letter] in [word]
set [num v] to (0)
set [result v] to (0)
repeat (length of (word))
change [num v] by (1)
if<(letter (num) of (word)) = (letter)> then
change [result v] by (1)
end
i think it works :D
example below!

when flag clicked
ask [hello, i can count how many letters are there in a word. type in any word] and wait
set [word v] to (answer)
ask [now type any letter] and wait
how much (answer) in (word)
^^ block above is moreblock we defined at top
say (join [This word has ] (join (result) [letters])) for (2) secs

See my profile!
My best 3D pen project!

if <[1] = [1] > then
   forever
   
   end
end
say [This is a comment!]

MathlyCat wrote:

Workarounds
Note: If you have trouble finding one already in the list then make a post asking about a workaround to a certain block. I'll reply with the workaround if there is one!


____________________

day of year :: sensing reporter 

((days since 2000) - ((365) * ((current [year v]) - (2000))))

____________________
____________________
____________________
____________________


____________________

when(value::)changes::events hat
//is
when[timer v]>((timer)-<not<(value::)=(var)>>)
set[var v]to(value::)

SuperHal989 wrote:

<<<> or <>> and <not <<> and <>>>> // XOR (place the first 2 reporters in the last 2 as well

<not<<input 1::custom-arg>=<input 2::custom-arg>>>

when [timer v] > (10)

when [timer v] > ((timer) - <touching [mouse pointer v]?>)
say [you touched me!] for (2) secs

creeper82 wrote:

When <x> becomes true
Have you ever wanted to do certain action, when you hover over the sprite? Now You can! Thanks to this block:

when [timer v] > (10)

We can make this:

when [timer v] > ((timer) - <touching [mouse pointer v]?>)
say [you touched me!] for (2) secs

creeper82 wrote:

You can do more of them in one project but it may become laggy.

Botcho_Otkho wrote:

creeper82 wrote:

When <x> becomes true
Have you ever wanted to do certain action, when you hover over the sprite? Now You can! Thanks to this block:

when [timer v] > (10)

We can make this:

when [timer v] > ((timer) - <touching [mouse pointer v]?>)
say [you touched me!] for (2) secs

I always recommend to test scripts before suggesting them. This won't absolutely work, and doesn't mean anything in coding.

creeper82 wrote:

You can do more of them in one project but it may become laggy.

Really? I don't think at all few scripts will make the project lag.
I don't want to be rude, but that's how it is.

thelostgames wrote:

MathlyCat wrote:

Workarounds
Note: If you have trouble finding one already in the list then make a post asking about a workaround to a certain block. I'll reply with the workaround if there is one!


____________________

day of year :: sensing reporter 

((days since 2000) - ((365) * ((current [year v]) - (2000))))

____________________
____________________
____________________
____________________


____________________

You would also have to include a multiplication by 1/4 to account for leap years

MathlyCat wrote:

-snip-

thelostgames wrote:

MathlyCat wrote:

-snip-

day of year :: sensing reporter 

((days since 2000) - ((365) * ((current [year v]) - (2000))))

You would also have to include a multiplication by 1/4 to account for leap years

Replace 365 with 365.25

Botcho_Otkho wrote:

creeper82 wrote:

When <x> becomes true
Have you ever wanted to do certain action, when you hover over the sprite? Now You can! Thanks to this block:

when [timer v] > (10)

We can make this:

when [timer v] > ((timer) - <touching [mouse pointer v]?>)
say [you touched me!] for (2) secs

I always recommend to test scripts before suggesting them. This won't absolutely work, and doesn't mean anything in coding.
-snip-

if (1) {
alert('True = 1!')
}

point towards x:(0) y:(0)

thelostgames wrote:

MathlyCat wrote:

Workarounds
Note: If you have trouble finding one already in the list then make a post asking about a workaround to a certain block. I'll reply with the workaround if there is one!


____________________

day of year :: sensing reporter 

((days since 2000) - ((365) * ((current [year v]) - (2000))))

____________________
____________________
____________________
____________________


____________________

You would also have to include a multiplication by 1/4 to account for leap years

andyhanwu wrote:

Can you make a workaround for this:

point towards x:(0) y:(0)

point towards [ v]

PythonKitten wrote:

Could someone give me a workaround to if some variable contains another? EG:

set [ variable1] to [HELLO!!!!!! ]
set [variable2] to [HELLO]
if (variable2) contains (variable1) then
    set [variable1] to (pick random (1) to (10))
end

brendanthescratcher wrote:

PythonKitten wrote:

Could someone give me a workaround to if some variable contains another? EG:

set [ variable1] to [HELLO!!!!!! ]
set [variable2] to [HELLO]
if (variable2) contains (variable1) then
    set [variable1] to (pick random (1) to (10))
end

There is now a block that does that.

() + () + () :: operators reporter

letter () to () of [sk7725] :: operators reporter

About Scratch wrote:

young people learn to think creatively, reason systematically, and work collaboratively.

if <> then {
  
}else if <> then {

}else {

}::control cblock

ScratchDiogoh wrote:

ihgfedcba wrote:

Workaround for inverses of the built in functions of a single number (assuming e is the number)

([aabs v] of (e)::operators)

((0) - (e))

([asqrt v] of (e)::operators)

([10 ^ v] of (([log v] of (e)) * (2)))

aln=e^
alog=10^
a10^=log
ae^=ln

In aabs should Use in Workaround:

((0) - ([abs v] of (e)))