In TI-BASIC, the + operation is overloaded for string concatenation (in this, if nothing else, TI-BASIC joins the rest of the world).
However, any attempt to concatenate involving an empty string raises a Dimension Mismatch error:
"Fizz"+"Buzz"
FizzBuzz
"Fizz"+""
Error
""+"Buzz"
Error
""+""
Error
Why does this occur, and is there an elegant workaround? I've been using a starting space and truncating the string when necessary (doesn't always work well) or using a loop to add characters one at a time (slow).
The best way depends on what you are doing.
If you have a string (in this case, Str1) that you need to concatenate with another (Str2), and you don't know if it is empty, then this is a good general-case solution:
Str2
If length(Str1
Str1+Str2
If you need to loop and add a stuff to the string each time, then this is your best solution:
Before the loop:
" →Str1
In the loop:
Str1+<stuff_that_isn't_an_empty_string>→Str1
After the loop:
sub(Str1,2,length(Str1)-1→Str1
There are other situations, too, and if you have a specific situation, then you should post a simplified version of the relevant code.
Hope this helps!
It is very unfortunate that TI-Basic doesn't support empty strings. If you are starting with an empty string and adding chars, you have to do something like this:
"?
For(I,1,3
Prompt Str1
Ans+Str1
End
sub(Ans,2,length(Ans)-1
Another useful trick is that if you have a string that you are eventually going to evaluate using expr(, you can do "("+Str1+")"→Str1 and then freely do search and replace on the string. This is a necessary workaround since you can't search and replace any text involving the first or last character in a string.
Related
It's a weird problem
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120"
And two strings below:
s1="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\[Content_Types].xml"
s2="D:\\Users\\UserKnown\\PycharmProjects\\ProjectKnown\\PT\\collections\\120\\_rels\.rels"
When I use the command below:
s1.strip(to_be_stripped)
s2.strip(to_be_stripped)
I get these outputs:
'[Content_Types].x'
'_rels\\.'
If I use lstrip(), they will be:
'[Content_Types].xml'
'_rels\\.rels'
Which is the right outputs.
However, if we replace all Project Known with zeus_pipeline:
to_be_stripped="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120"
And:
s2="D:\\Users\\UserKnown\\PycharmProjects\\zeus_pipeline\\PT\\collections\\120\\_rels\.rels"
s2.lstrip(to_be_stripped)will be '.rels'
If I use / instead of \\, nothing goes wrong. I am wondering why this problem happens.
strip isn't meant to remove full strings exactly. Rather, you give it a string, and every character in that string is removed from the start and of the string to be stripped.
In your case, the variable to_be_stripped contains the characters m and l, so those are stripped from the end of s1. However, it doesn't contain the character x, so the stripping stops there and no characters beyond that are removed.
Check out this question. The accepted answer is probably more extensive than you need - I like another user's suggestion of using replace instead of strip. This would look like:
s1.replace(to_be_stripped, "")
I'm trying to get a substring from an initial string in Smalltalk. I'm wondering if there's a way to do it. For example in Java, the method aStringObject.substring(index), allows you to trim a String object using an index (or its position in the array). I've been looking in the browser for something that works in a similar way, but couldn't find it. So far every trimming method uses a character or string to do the separation.
As an example of what I'm looking for:
initialString:='Hello'.
finalString:=initialString substring: 1
The value of finalString should be 'ello'.
In Smalltalk a String is a type of SequencableCollection so you can use the copying protocol messages as well.
For example you could use:
copyFrom: start to: stop
allButFirst (will not copy the first character)
allButFirst: n (more generally answer a copy of the receiver containing all but the first n elements.
I want to remove all the characters from a string expect whatever character is between a certain set of characters. So for example I have the input of Grade:2/2014-2015 and I want the output of just the grade, 2.
I'm thinking that I need to use the FIND function to grab whatever is between the : and the / , this also needs to work with double characters such 10 however I believe that it would work so long as the defining values with the FIND function are correct.
Unfortunately I am totally lost on this when using the FIND function however if there is another function that would work better I could probably figure it out myself if I knew what function.
It's not particularly elegant but =MID(A1,FIND(":",A1)+1,FIND("/",A1) - FIND(":",A1) - 1) would work.
MID takes start and length,FIND returns the index of a given character.
Edit:
As pointed out, "Grade:" is fixed length so the following would work just as well:
=MID(A1,7,FIND("/",A1) - 7)
You could use LEFT() to remove "Grade:"
And then use and then use LEFTB() to remove the year.
Look at this link here. This is the way I would go about it.
=SUBSTITUTE(SUBSTITUTE(C4, "Grade:", ""), "/2014-2015", "")
where C4 is the name of your cell.
I'm currently teaching myself Lua for iOS game development, since I've heard lots of very good things about it. I'm really impressed by the level of documentation there is for the language, which makes learning it that much easier.
My problem is that I've found a Lua concept that nobody seems to have a "beginner's" explanation for: nested brackets for quotes. For example, I was taught that long strings with escaped single and double quotes like the following:
string_1 = "This is an \"escaped\" word and \"here\'s\" another."
could also be written without the overall surrounding quotes. Instead one would simply replace them with double brackets, like the following:
string_2 = [[This is an "escaped" word and "here's" another.]]
Those both make complete sense to me. But I can also write the string_2 line with "nested brackets," which include equal signs between both sets of the double brackets, as follows:
string_3 = [===[This is an "escaped" word and "here's" another.]===]
My question is simple. What is the point of the syntax used in string_3? It gives the same result as string_1 and string_2 when given as an an input for print(), so I don't understand why nested brackets even exist. Can somebody please help a noob (me) gain some perspective?
It would be used if your string contains a substring that is equal to the delimiter. For example, the following would be invalid:
string_2 = [[This is an "escaped" word, the characters ]].]]
Therefore, in order for it to work as expected, you would need to use a different string delimiter, like in the following:
string_3 = [===[This is an "escaped" word, the characters ]].]===]
I think it's safe to say that not a lot of string literals contain the substring ]], in which case there may never be a reason to use the above syntax.
It helps to, well, nest them:
print [==[malucart[[bbbb]]]bbbb]==]
Will print:
malucart[[bbbb]]]bbbb
But if that's not useful enough, you can use them to put whole programs in a string:
loadstring([===[print "o m g"]===])()
Will print:
o m g
I personally use them for my static/dynamic library implementation. In the case you don't know if the program has a closing bracket with the same amount of =s, you should determine it with something like this:
local c = 0
while contains(prog, "]" .. string.rep("=", c) .. "]") do
c = c + 1
end
-- do stuff
In MATLAB, ... is used to continue a line to the next line. But if I want to continue a long string within quotation, what can I do? ... will be treated as a part of the string itself.
Using [] is not a perfect solution since in most cases I use sprintf/fprintf to parse a long string like sql query. Using [] would be cumbersome. thanks.
If you put the string in brackets, you can build it in several pieces:
s = ['abc' 'def' ...
'ghi'];
You can then split that statement into several lines between the strings.
answer=['You can divide strings '...
,'by adding a comma '...
,'(as you probably know one year later).'];
You can use strcat or horzcat, which gives you somewhat more options than [], including the ability to mix in variables along with the hardcoded values.