Ok, i've always kind of known that computers treat strings as a series of numbers under the covers, but i never really looked at the details of how it works. What sort of magic is going on in the average compiler/processor when we do, for instance, the following?
string myString = "foo";
myString += "bar";
print(myString) //replace with printing function of your choice
The answer is completely dependent on the language in question. But C is usually a good language to kind of see how things happen behind the scenes.
In C:
In C strings are array of char with a 0 at the end:
char str[1024];
strcpy(str, "hello ");
strcpy(str, "world!");
Behind the scenes str[0] == 'h' (which has an int value), str[1] == 'e', ...
str[11] == '!', str[12] == '\0';
A char is simply a number which can contain one of 256 values. Each character has a numeric value.
In C++:
strings are supported in the same way as C but you also have a string type which is part of STL.
string literals are part of static storage and cannot be changed directly unless you want undefined behavior.
It's implementation dependent how the string type actually works behind the scenes, but the string objects themselves are mutable.
In C#:
strings are immutable. Which means you can't directly change a string once it's created. When you do += what happen is a new string gets created and your string now references that new string.
The implementation varies between language and compiler of course, but typically for C it's something like the following. Note that strings are essentially syntactical sugar for char arrays (char[]) in C.
1.
string myString = "foo";
Allocate 3 bytes of memory for the array and set the value of the 1st byte to 'f' (its ASCII code rather), the 2nd byte to 'o', the 2rd byte to 'o'.
2.
foo += "bar";
Read existing string (char array) from memory pointed to by foo.
Allocate 6 bytes of memory, fill the first 3 bytes with the read contents of foo, and the next 3 bytes with b, a, and r.
3.
print(foo)
Read the string foo now points to from memory, and print it to the screen.
This is a pretty rough overview, but hopefully should give you the general idea.
Side note: In some languages/compuilers, char != byte - for example, C#, where strings are stored in Unicode format by default, and notably the length of the string is also stored in memory. C++ typically uses null-terminated strings, which solves the problem in another way, though it means determining its length is O(n) rather than O(1).
Its very language dependent. However, in most cases strings are immutable, so doing that is going to allocate a new string and release the old one's memory.
I'm assuming a typo in your sample and that there is only one variable called either foo or myString, not two variables?
I'd say that it'll depend a lot on what compiler you're using. In .Net strings are immutable so when you add "bar" you're not actually adding it but rather creating a new string containing "foobar" and telling it to put that in your variable.
In other languages it will work differently.
Related
I want to check if string A is just a reordered version of string B. For example, "abc" = "bca" = "cab"...
There are other solutions here: https://www.geeksforgeeks.org/check-if-two-strings-are-permutation-of-each-other/
However, I was thinking a hash function would be an easy way of doing this, but the typical hash function takes order into consideration. Are there any hash functions that do not care about character order?
Are there any hash functions that do not care about character order?
I don't know of real-world hash functions that have this property, no. Because this is not a problem they are designed to solve.
However, in this specific case, you can make your own "hash" function (a very very bad one) that will indeed ignore order: just sum ASCII codes of characters. This works due to the commutative property of addition (a + b == b + a)
def isAnagram(self,a,b):
sum_a = 0
sum_b = 0
for c in a:
sum_a += ord(c)
for c in b:
sum_b += ord(c)
return sum_a == sum_b
To reiterate, this is absolutely a hack, that only happens to work because input strings are limited in content in the judge system (only have lowercase ASCII characters and do not contain spaces). It will not work (reliably) on arbitrary strings.
For a fast check you could use a kind af hash-funkction
Candidates are:
xor all characters of a String
add all characters of a String
multiply all characters of a String (be careful might lead to overflow for large Strings)
If the hash-value is equal, it could still be a collision of two not 'equal' strings. So you still need to make a dedicated compare. (e.g. sort the characters of each string before comparing them).
I have a variable which contains a single char. I want to convert this char to upper case. However, the to_uppercase function returns a rustc_unicode::char::ToUppercase struct instead of a char.
Explanation
ToUppercase is an Iterator, that may yield more than one char. This is necessary, because some Unicode characters consist of multiple "Unicode Scalar Values" (which a Rust char represents).
A nice example are the so called ligatures. Try this for example (on playground):
let fi_upper: Vec<_> = 'fi'.to_uppercase().collect();
println!("{:?}", fi_upper); // prints: ['F', 'I']
The 'fi' ligature is a single character whose uppercase version consists of two letters/characters.
Solution
There are multiple possibilities how to deal with that:
Work on &str: if your data is actually in string form, use str::to_uppercase which returns a String which is easier to work with.
Use ASCII methods: if you are sure that your data is ASCII only and/or you don't care about unicode symbols you can use std::ascii::AsciiExt::to_ascii_uppercase which returns just a char. But it only changes the letters 'a' to 'z' and ignores all other characters!
Deal with it manually: Collect into a String or Vec like in the example above.
ToUppercase is an iterator, because the uppercase version of the character may be composed of several codepoints, as delnan pointed in the comments. You can convert that to a Vector of characters:
c.to_uppercase().collect::<Vec<_>>();
Then, you should collect those characters into a string, as ker pointed.
How does one detect if a string is binary safe or not in Go?
A function like:
IsBinarySafe(str) //returns true if its safe and false if its not.
Any comment after this are just things I have thought or attempted to solve this:
I assumed that there must exist a library that already does this but had a tough time finding it. If there isn't one, how do you implement this?
I was thinking of some solution but wasn't really convinced they were good solutions.
One of them was to iterate over the bytes, and have a hash map of all the illegal byte sequences.
I also thought of maybe writing a regex with all the illegal strings but wasn't sure if that was a good solution.
I also was not sure if a sequence of bytes from other languages counted as binary safe. Say the typical golang example:
世界
Would:
IsBinarySafe(世界) //true or false?
Would it return true or false? I was assuming that all binary safe string should only use 1 byte. So iterating over it in the following way:
const nihongo = "日本語abc日本語"
for i, w := 0, 0; i < len(nihongo); i += w {
runeValue, width := utf8.DecodeRuneInString(nihongo[i:])
fmt.Printf("%#U starts at byte position %d\n", runeValue, i)
w = width
}
and returning false whenever the width was great than 1. These are just some ideas I had just in case there wasn't a library for something like this already but I wasn't sure.
Binary safety has nothing to do with how wide a character is, it's mainly to check for non-printable characters more or less, like null bytes and such.
From Wikipedia:
Binary-safe is a computer programming term mainly used in connection
with string manipulating functions. A binary-safe function is
essentially one that treats its input as a raw stream of data without
any specific format. It should thus work with all 256 possible values
that a character can take (assuming 8-bit characters).
I'm not sure what your goal is, almost all languages handle utf8/16 just fine now, however for your specific question there's a rather simple solution:
// checks if s is ascii and printable, aka doesn't include tab, backspace, etc.
func IsAsciiPrintable(s string) bool {
for _, r := range s {
if r > unicode.MaxASCII || !unicode.IsPrint(r) {
return false
}
}
return true
}
func main() {
fmt.Printf("len([]rune(s)) = %d, len([]byte(s)) = %d\n", len([]rune(s)), len([]byte(s)))
fmt.Println(IsAsciiPrintable(s), IsAsciiPrintable("test"))
}
playground
From unicode.IsPrint:
IsPrint reports whether the rune is defined as printable by Go. Such
characters include letters, marks, numbers, punctuation, symbols, and
the ASCII space character, from categories L, M, N, P, S and the ASCII
space character. This categorization is the same as IsGraphic except
that the only spacing character is ASCII space, U+0020.
I am getting input from the user, however when I try to compare it later on to a string literal it does not work. That is just a test though.
I would like to set it up so that when a blank line is entered (just hitting the enter/return key) the program exits. I don't understand why the strings are not comparing because when I print it, it comes out identical.
in := bufio.NewReader(os.Stdin);
input, err := in.ReadBytes('\n');
if err != nil {
fmt.Println("Error: ", err)
}
if string(input) == "example" {
os.Exit(0)
}
string vs []byte
string definition:
string is the set of all strings of 8-bit bytes, conventionally but not necessarily representing UTF-8-encoded text. A string may be empty, but not nil. Values of string type are immutable.
byte definition:
byte is an alias for uint8 and is equivalent to uint8 in all ways. It is used, by convention, to distinguish byte values from 8-bit unsigned integer values.
What does it mean?
[]byte is a byte slice. slice can be empty.
string elements are unicode characters, which can have more then 1 byte.
string elements keep a meaning of data (encoding), []bytes not.
equality operator is defined for string type but not for slice type.
As you see they are two different types with different properties.
There is a great blog post explaining different string related types [1]
Regards the issue you have in your code snippet.
Bear in mind that in.ReadBytes(char) returns a byte slice with char inclusively. So in your code input ends with '\n'. If you want your code to work in desired way then try this:
if string(input) == "example\n" { // or "example\r\n" when on windows
os.Exit(0)
}
Also make sure that your terminal code page is the same as your .go source file. Be aware about different end-line styles (Windows uses "\r\n"), Standard go compiler uses utf8 internally.
[1] Comparison of Go data types for string processing.
Is there any way to replace a character at position N in a string in Lua.
This is what I've come up with so far:
function replace_char(pos, str, r)
return str:sub(pos, pos - 1) .. r .. str:sub(pos + 1, str:len())
end
str = replace_char(2, "aaaaaa", "X")
print(str)
I can't use gsub either as that would replace every capture, not just the capture at position N.
Strings in Lua are immutable. That means, that any solution that replaces text in a string must end up constructing a new string with the desired content. For the specific case of replacing a single character with some other content, you will need to split the original string into a prefix part and a postfix part, and concatenate them back together around the new content.
This variation on your code:
function replace_char(pos, str, r)
return str:sub(1, pos-1) .. r .. str:sub(pos+1)
end
is the most direct translation to straightforward Lua. It is probably fast enough for most purposes. I've fixed the bug that the prefix should be the first pos-1 chars, and taken advantage of the fact that if the last argument to string.sub is missing it is assumed to be -1 which is equivalent to the end of the string.
But do note that it creates a number of temporary strings that will hang around in the string store until garbage collection eats them. The temporaries for the prefix and postfix can't be avoided in any solution. But this also has to create a temporary for the first .. operator to be consumed by the second.
It is possible that one of two alternate approaches could be faster. The first is the solution offered by Paŭlo Ebermann, but with one small tweak:
function replace_char2(pos, str, r)
return ("%s%s%s"):format(str:sub(1,pos-1), r, str:sub(pos+1))
end
This uses string.format to do the assembly of the result in the hopes that it can guess the final buffer size without needing extra temporary objects.
But do beware that string.format is likely to have issues with any \0 characters in any string that it passes through its %s format. Specifically, since it is implemented in terms of standard C's sprintf() function, it would be reasonable to expect it to terminate the substituted string at the first occurrence of \0. (Noted by user Delusional Logic in a comment.)
A third alternative that comes to mind is this:
function replace_char3(pos, str, r)
return table.concat{str:sub(1,pos-1), r, str:sub(pos+1)}
end
table.concat efficiently concatenates a list of strings into a final result. It has an optional second argument which is text to insert between the strings, which defaults to "" which suits our purpose here.
My guess is that unless your strings are huge and you do this substitution frequently, you won't see any practical performance differences between these methods. However, I've been surprised before, so profile your application to verify there is a bottleneck, and benchmark potential solutions carefully.
You should use pos inside your function instead of literal 1 and 3, but apart from this it looks good. Since Lua strings are immutable you can't really do much better than this.
Maybe
"%s%s%s":format(str:sub(1,pos-1), r, str:sub(pos+1, str:len())
is more efficient than the .. operator, but I doubt it - if it turns out to be a bottleneck, measure it (and then decide to implement this replacement function in C).
With luajit, you can use the FFI library to cast the string to a list of unsigned charts:
local ffi = require 'ffi'
txt = 'test'
ptr = ffi.cast('uint8_t*', txt)
ptr[1] = string.byte('o')