I've asked a question here, and that led me to an another question.
In R, there's no fundamental distinction between a string and a
character. A "string" is just a character variable that contains one
or more characters.
and
There is a distinction between a scalar character variable, and a
vector. A character vector is a set of strings stored as a single
object.
So I wonder how to allocate a string with c api of R? For example, what do I get from:
result = Rf_allocVector(STRSXP, dst_size);
is it(the result) a scalar character variable or a vector? or could I use other API for allocating string?
Thanks.
We have that as a motivating example in our introductory vignette in the Rcpp package (and this is also published as a paper JSS in 2011):
In the C API you must do allocate a vector of STRSXP:
SEXP ab;
PROTECT(ab = allocVector(STRSXP, 2));
SET_STRING_ELT( ab, 0, mkChar("foo") );
SET_STRING_ELT( ab, 1, mkChar("bar") );
UNPROTECT(1);
which imposes on the programmer knowledge of PROTECT, UNPROTECT,
SEXP, allocVector, SET_STRING_ELT, and mkChar.
Whereas with Rcpp and
using the Rcpp::CharacterVector class, we can express the same code more concisely:
Rcpp::CharacterVector ab(2);
ab[0] = "foo";
ab[1] = "bar";
Related
I want to do something like this (C sample):
char str[] = "Hello\x90\x90\xcc\x00";
How?
The equivalent of this in Fortran would be:
character(*), parameter :: str = "Hello"//char(144)//char(144)//char(204)//char(0)
I made this a named (PARAMETER) constant here, but the expression for initializing would be the same in a normal assignment context. Standard Fortran doesn't allow the use of hex constants (such as Z'90') as an argument to CHAR, though many compilers support that as an extension.
In RcppArmadillo, I need to know how I can convert arma::mat to c-style array double * for use in other functions.
When I run the following functions, the computer crashes:
R part:
nn3 <- function(x){
results=.Call("KNNCV", PACKAGE = "KODAMA", x)
results
}
C++ part:
double KNNCV(arma::mat x) {
double *cvpred = x.memptr();
return cvpred[1];
}
and at the end, I try:
nn3(as.matrix(iris[,-5]))
Can you help me to find the errors, please?
First, there is no such such thing as vector stored in a double*. You can cast to a C-style pointer to double; but without length information that does not buy you much.
By convention, most similar C++ classes give you a .begin() iterator to the beginning of the memory block (which Armadillo happens to guarantee to be contiguous, just like std::vector) so you can try that.
Other than that the (very fine indeed) Armadillo documentation tells you about memptr() which is probably what you want here. Straight copy from the example there:
mat A = randu<mat>(5,5);
const mat B = randu<mat>(5,5);
double* A_mem = A.memptr();
const double* B_mem = B.memptr();
I come from a C# background where System.String is immutable and string concatenation is relatively expensive (as it requires reallocating the string) we know to use the StringBuilder type instead as it preallocates a larger buffer where single characters (Char, a 16-bit value-type) and short strings can be concatenated cheaply without extra allocation.
I'm porting some C# code to Swift which reads from a bit-array ([Bool]) at sub-octet indexes with character lengths less than 8 bits (it's a very space-conscious file format).
My C# code does something like this:
StringBuilder sb = new StringBuilder( expectedCharacterCount );
int idxInBits = 0;
Boolean[] bits = ...;
for(int i = 0; i < someLength; i++) {
Char c = ReadNextCharacter( ref idxInBits, 6 ); // each character is 6 bits in this example
sb.Append( c );
}
In Swift, I assume NSMutableString is the equivalent of .NET's StringBuilder, and I found this QA about appending individual characters ( How to append a character to string in Swift? ) so in Swift I have this:
var buffer: NSMutableString
for i in 0..<charCount {
let charValue: Character = readNextCharacter( ... )
buffer.AppendWithFormat("%c", charValue)
}
return String(buffer)
But I don't know why it goes through a format-string first, that seems inefficient (reparsing the format-string on every iteration) and as my code is running on iOS devices I want to be very conservative with my program's CPU and memory usage.
As I was writing this, I learned my code should really be using UnicodeScalar instead of Character, problem is NSMutableString does not let you append a UnicodeScalar value, you have to use Swift's own mutable String type, so now my code looks like:
var buffer: String
for i in 0..<charCount {
let x: UnicodeScalar = readNextCharacter( ... )
buffer.append(x)
}
return buffer
I thought that String was immutable, but I noticed its append method returns Void.
I still feel uncomfortable doing this because I don't know how Swift's String type is implemented internally, and I don't see how I can preallocate a large buffer to avoid reallocations (assuming Swift's String uses a growing algorithm).
(This answer was written based on documentation and source code valid for Swift 2 and 3: possibly needs updates and amendments once Swift 4 arrives)
Since Swift is now open-source, we can actually have a look at the source code for Swift:s native String
swift/stdlib/public/core/String.swift
From the source above, we have following comment
/// Growth and Capacity
/// ===================
///
/// When a string's contiguous storage fills up, new storage must be
/// allocated and characters must be moved to the new storage.
/// `String` uses an exponential growth strategy that makes `append` a
/// constant time operation *when amortized over many invocations*.
Given the above, you shouldn't need to worry about the performance of appending characters in Swift (be it via append(_: Character), append(_: UniodeScalar) or appendContentsOf(_: String)), as reallocation of the contiguous storage for a certain String instance should not be very frequent w.r.t. number of single characters needed to be appended for this re-allocation to occur.
Also note that NSMutableString is not "purely native" Swift, but belong to the family of bridged Obj-C classes (accessible via Foundation).
A note to your comment
"I thought that String was immutable, but I noticed its append method returns Void."
String is just a (value) type, that may be used by mutable as well as immutable properties
var foo = "foo" // mutable
let bar = "bar" // immutable
/* (both the above inferred to be of type 'String') */
The mutating void-return instance methods append(_: Character) and append(_: UniodeScalar) are accessible to mutable as well as immutable String instances, but naturally using them with the latter will yield a compile time error
let chars : [Character] = ["b","a","r"]
foo.append(chars[0]) // "foob"
bar.append(chars[0]) // error: cannot use mutating member on immutable value ...
From what I understand, strings in Nim are basically a mutable sequence of bytes and that they are copied on assignment.
Given that, I assumed that sizeof would tell me (like len) the number of bytes, but instead it always gives 8 on my 64-bit machine, so it seems to be holding a pointer.
Given that, I have the following questions...
What was the motivation behind copy on assignment? Is it because they're mutable?
Is there ever a time when it isn't copied when assigned? (I assume non-var function parameters don't copy. Anything else?)
Are they optimized such that they only actually get copied if/when they're mutated?
Is there any significant difference between a string and a sequence, or can the answers to the above questions be equally applied to all sequences?
Anything else in general worth noting?
Thank you!
The definition of strings actually is in system.nim, just under another name:
type
TGenericSeq {.compilerproc, pure, inheritable.} = object
len, reserved: int
PGenericSeq {.exportc.} = ptr TGenericSeq
UncheckedCharArray {.unchecked.} = array[0..ArrayDummySize, char]
# len and space without counting the terminating zero:
NimStringDesc {.compilerproc, final.} = object of TGenericSeq
data: UncheckedCharArray
NimString = ptr NimStringDesc
So a string is a raw pointer to an object with a len, reserved and data field. The procs for strings are defined in sysstr.nim.
The semantics of string assignments have been chosen to be the same as for all value types (not ref or ptr) in Nim by default, so you can assume that assignments create a copy. When a copy is unneccessary, the compiler can leave it out, but I'm not sure how much that is happening so far. Passing strings into a proc doesn't copy them. There is no optimization that prevents string copies until they are mutated. Sequences behave in the same way.
You can change the default assignment behaviour of strings and seqs by marking them as shallow, then no copy is done on assignment:
var s = "foo"
shallow s
I'm having an absolute hell of a time trying to figure out how to get a plain, mutable C string (a char*) from a D string (a immutable(char)[]) to that I can pass the character data to legacy C code. toStringz doesn't work, as I get an error saying that I "cannot implicitly convert expression (toStringz(this.fileName())) of type immutable(char)* to char*". Do I need to recreate a new, mutable array of char and copy the characters over?
If you can change the header of the D interface of that legacy C code, and you are sure that legacy C code will not modify the string, you could make it accept a const(char)*, e.g.
char* strncpy(char* dest, const(char)* src, size_t count);
// ^^^^^^^^^^^^
Yeah, it's not pretty, because the result is immutable.
This is why I always return a mutable copy of new arrays in my code. There's no point in making them immutable.
Solutions:
You could just do
char[] buffer = (myString ~ '\0').dup; //Concatenate a null terminator, then dup
then use buffer.ptr as the pointer.
However:
This wastes a string. A better approach might be:
char[] buffer = myString.dup;
buffer ~= '\0'; //Hopefully this doesn't reallocate
and using buffer.ptr afterwards.
Another solution is to use a method like this one:
char* toStringz(in char[] s)
{
string result;
if (s.length > 0 && s[$ - 1] == '\0') //Is it null-terminated?
{ result = s.dup; }
else { result = new char[s.length + 1]; result[0 .. s.length][] = s[]; }
return result.ptr;
}
This one is the most efficient but also the longest.
(Edit: Whoops, I had a typo in the if; fixed it.)
If you want to pass a mutable char* to a C function, you're going to need to allocate a mutable char[]. string isn't going to work, because it's immutable(char)[]. You can't alter immutable variables, so there is no way to pass a string to a function (C or otherwise) which needs to alter its elements.
So, if you have a string, and you need to pass it to a function which takes a char[], then you can use to!(char[]) or dup and get a mutable copy of it. In addition, if you want to pass it to a C function, you're going to need to append a '\0' to it so that it's zero-terminated. The easiest way to do that is just to do ~= '\0' on the char[], but the more efficient way would probably be to do something like this:
auto cstring = new char[](str.length + 1);
cstring[0 .. str.length] = str[];
cstring[$ - 1] = '\0';
In either case, you then pass cstring.ptr to the C function that you're calling.
If you know that the C function that you're calling isn't going to alter the string, then you can either do as KennyTM suggests and alter the C function's signature in D to take a const(char)*, or you can cast the string. e.g.
auto cstring = toStringz(str);
cfunc(cast(char*)cstring.ptr);
Altering the C function's signature would be more correct and less error-prone though.
It sounds like we may be altering std.conv.to to be smart enough to turn strings into zero-terminated strings when cast to char*, const(char)*, etc. So, once that's done, getting a zero-terminated mutable string should be easier, but for the moment, you pretty much just need to copy the string and append a '\0' to it so that it's zero-terminated. But regardless, you're never going to be able to pass a string to a C function which needs to modify it, because a string can't be mutated.
Without any context on which function you're calling it's hard to say what is the right solution.
Typically, if the C function wants to modify or write to the string it probably expects you to provide a buffer and a length. Usually what I do is:
Allocate a buffer:
auto buffer = new char[](256); // your own length instead of 256 here
And call the C function:
CWriteString(buffer.ptr, buffer.length);
You can try the following :
char a[]="abc";
char *p=a;
Now you can pass pointer 'p' to the array in any function.
Hope it works.