I am attempting to overload the == operator then test it with the code in the main function. It gives me an error and says the z in my if statement must be a bool type or converted into one. I am just wondering where I am going wrong here and how to go about setting that part up. Here is the code snippet. I declared double real & double imaginary as private variables also.
Complex Complex::operator==(const Complex &operand2) const
{
if (real == operand2.real, imaginary == operand2.imaginary)
return true;
else
return false;
}
int main()
{
Complex x(1, 2);
Complex y(2, 3);
Complex z, w, v;
z = x + y;
w = x – y;
if (z == w)
cout << " z = w" << endl;
else
cout << " z != w" << endl;
return 0;
}
The code will be something like this :
bool Complex::operator==(const Complex &operand2) const
{
return (real == operand2.real && imaginary == operand2.imaginary) ;
}
The return type should be bool as the result is always true or false.
Since both the real and imaginary parts need to be equal, you use && (AND) operation to join the two conditions.
Also notice that any operation involving the == operator will return a bool value (either true or false) and hence instead of a if condition, you can directly return the result.
Your return a bool from your operator ,so what else do you expect
If you defined operator return value as complex you cannot return boolean.
Complex Complex::operator==/*...*/
This actually returns the Complex type which is not problem, unless you need bool value.
bool Complex::operator==
So return type here is what you want. For further information read this: http://en.cppreference.com/w/cpp/language/operators
Related
I have a tree with the most frequent letters at the top of the tree and I am trying to find the path to a node so I can turn a message into binary. In this program if I go left, I add 0 to path and, if I go right, I add 1 to the path until I find the node. But I have to go straight to the desired node which is not possible. The only thing I could think of is removing the last character or path if a node has no children, but it does not work if a node has grandchildren. Can someone help me on how to approach this? Thanks!
// global variables
String path;
int mark;
// encodes a message to binary
String encoder(char data) {
path = "";
mark = 0;
findPath(root, data);
return path;
}
// finds the path to a node
void findPath(TNode node, char data) {
if(node.data == data) {
mark = 1;
return;
}
if(mark==0 && node.left != null) {
path += 0;
findPath(node.left, data);
}
if(mark==0 && node.right != null) {
path += 1;
findPath(node.right, data);
}
if(mark==0 && node.left == null || node.right == null) {
path = path.substring(0, path.length() - 1);
}
}
This is not how Huffman coding works. Not at all. You assign a code to every symbol, and symbols can have different lengths.
An algorithm to determine the code: Take two symbols with the lowest frequency, say an and b. Combine them into one symbol x with higher frequency obviously, and when we are finished the code for a is the code for x plus a zero, and the code for b is the code for x plus a one. Then you look again for the two symbols with lowest frequency and combine them and so on. When you are down to two symbols you give them codes 0 and 1 and find all the other codes for symbols.
Example: a, b, c, d with frequencies 3, 47, 2 and 109. Combine a and c to x with frequency 5. Combine x and b to y with frequency 52. Then a = code 0, y = code 1. x = code 10 and b = code 11. Then a = code 100 and c = code 101.
You would not encode messages using the tree. You should instead traverse the entire tree once recursively, generating all of the codes for all of the symbols, and make a table of the symbols and their associated codes. Then you use that table to encode your messages.
i have replaced my code with x, y, and z.
At the moment my code runs like this but i have a lot of lines to check x. is there a way of instead of using this
if x != y && x != z
using something like this instead
if x != y,z
sorry everyone I should have explained it better. Here is a piece of the updated code using contains which was very helpful thank you!
if contactCreation.checkAllowed(newContact.contactDetails){
if{contactCreation.blockedCreation.contains(newContact.contactDetails){
watchList.addAttempt(newContact.contactDetails)
}
}else{
newAccount.showReasons()
}
I guess my question was more is there a way to check if x is y and if x is z but not if x is y + z. basically I wanted to call
if x != y && x != z && x != d && != e
without having to get x every-time in one if statement.
so something like this
if x != y && != z && != d && != e
thanks for all the help!
A slightly shorter version is to add an overload to the pattern matching operator (~=):
func ~=<T: Equatable>(pattern: [T], value: T) -> Bool {
return pattern.contains(value)
}
if !([y,z] ~= x) {
}
If that expression looks a little unwieldy and you write it a lot, you can define your own operator:
infix operator !!= : ComparisonPrecedence
func !!=<T: Equatable>(lhs: T, rhs: [T]) -> Bool {
return !rhs.contains(lhs)
}
if x !!= [y,z] {
}
I wouldn't suggest this with what you've shown for your code, but you could use tuples:
let testCase:(String,String) = ("hello","world")
if x != testCase {
x = ("did not", "match")
}
You can rewrite this condition using an array aggregate and contains method, like this:
if ![y, z].contains(x) {
...
}
This does not help much when x is a simple variable, but it lets you remove some code duplication when x is a complex expression.
double a = 10.123420834;
double b = 100.123412321;
if (a > b) {
// do something here
}
I am trying to compare the two values, the code above doesn't seems to work. Any idea?
The code is correct.
Note that your snippet is equivalent to
float a = 10.123420834;
float b = 100.123412321;
if (a > b) {
// do something here
}
since Objective C uses double by default unless the number is followed by an f.
Also note that a < b, so the if statement will always evaluate to FALSE. Hence you may want to do
double a = 10.123420834;
double b = 100.123412321;
if (a > b) {
// do something here
} else {
// do something else here
}
to test this properly.
double a = 10.123420834
double b = 100.123412321
You need to have a semicolon at the end of each of those lines.
The code in your example is correct. Your problem must be in the "do something here", or elsewhere.
I'm working on a verification-tool for some VHDL-Code in MATLAB/Octave. Therefore I need data types which generate "real" overflows:
intmax('int32') + 1
ans = -2147483648
Later on, it would be helpful if I can define the bit width of a variable, but that is not so important right now.
When I build a C-like example, where a variable gets increased until it's smaller than zero, it spins forever and ever:
test = int32(2^30);
while (test > 0)
test = test + int32(1);
end
Another approach I tried was a custom "overflow"-routine which was called every time after a number is changed. This approach was painfully slow, not practicable and not working in all cases at all. Any suggestions?
In MATLAB, one option you have is to overload the methods that handle arithmetic operations for integer data types, creating your own custom overflow behavior that will result in a "wrap-around" of the integer value. As stated in the documentation:
You can define or overload your own
methods for int* (as you can for any
object) by placing the appropriately
named method in an #int* folder within
a folder on your path. Type help
datatypes for the names of the methods
you can overload.
This page of the documentation lists the equivalent methods for the arithmetic operators. The binary addition operation A+B is actually handled by the function plus(A,B). Therefore, you can create a folder called #int32 (placed in another folder on your MATLAB path) and put a function plus.m in there that will be used instead of the built-in method for int32 data types.
Here's an example of how you could design your overloaded plus function in order to create the overflow/underflow behavior you want:
function C = plus(A,B)
%# NOTE: This code sample is designed to work for scalar values of
%# the inputs. If one or more of the inputs is non-scalar,
%# the code below will need to be vectorized to accommodate,
%# and error checking of the input sizes will be needed.
if (A > 0) && (B > (intmax-A)) %# An overflow condition
C = builtin('plus',intmin,...
B-(intmax-A)-1); %# Wraps around to negative
elseif (A < 0) && (B < (intmin-A)) %# An underflow condition
C = builtin('plus',intmax,...
B-(intmin-A-1)); %# Wraps around to positive
else
C = builtin('plus',A,B); %# No problems; call the built-in plus.m
end
end
Notice that I call the built-in plus method (using the BUILTIN function) to perform addition of int32 values that I know will not suffer overflow/underflow problems. If I were to instead perform the integer addition using the operation A+B it would result in a recursive call to my overloaded plus method, which could lead to additional computational overhead or (in the worst-case scenario where the last line was C = A+B;) infinite recursion.
Here's a test, showing the wrap-around overflow behavior in action:
>> A = int32(2147483642); %# A value close to INTMAX
>> for i = 1:10, A = A+1; disp(A); end
2147483643
2147483644
2147483645
2147483646
2147483647 %# INTMAX
-2147483648 %# INTMIN
-2147483647
-2147483646
-2147483645
-2147483644
If you want to get C style numeric operations, you can use a MEX function to call the C operators directly, and by definition they'll work like C data types.
This method is a lot more work than gnovice's overrides, but it should integrate better into a large codebase and is safer than altering the definition for built-in types, so I think it should be mentioned for completeness.
Here's a MEX file which performs the C "+" operation on a Matlab array. Make one of these for each operator you want C-style behavior on.
/* c_plus.c - MEX function: C-style (not Matlab-style) "+" operation */
#include "mex.h"
#include "matrix.h"
#include <stdio.h>
void mexFunction(
int nlhs, mxArray *plhs[],
int nrhs, const mxArray *prhs[]
)
{
mxArray *out;
/* In production code, input/output type and bounds checks would go here. */
const mxArray *a = prhs[0];
const mxArray *b = prhs[1];
int i, n;
int *a_int32, *b_int32, *out_int32;
short *a_int16, *b_int16, *out_int16;
mxClassID datatype = mxGetClassID(a);
int n_a = mxGetNumberOfElements(a);
int n_b = mxGetNumberOfElements(b);
int a_is_scalar = n_a == 1;
int b_is_scalar = n_b == 1;
n = n_a >= n_b ? n_a : n_b;
out = mxCreateNumericArray(mxGetNumberOfDimensions(a), mxGetDimensions(a),
datatype, mxIsComplex(a));
switch (datatype) {
case mxINT32_CLASS:
a_int32 = (int*) mxGetData(a);
b_int32 = (int*) mxGetData(b);
out_int32 = (int*) mxGetData(out);
for (i=0; i<n; i++) {
if (a_is_scalar) {
out_int32[i] = a_int32[i] + b_int32[i];
} else if (b_is_scalar) {
out_int32[i] = a_int32[i] + b_int32[0];
} else {
out_int32[i] = a_int32[i] + b_int32[i];
}
}
break;
case mxINT16_CLASS:
a_int16 = (short*) mxGetData(a);
b_int16 = (short*) mxGetData(b);
out_int16 = (short*) mxGetData(out);
for (i=0; i<n; i++) {
if (a_is_scalar) {
out_int16[i] = a_int16[0] + b_int16[i];
} else if (b_is_scalar) {
out_int16[i] = a_int16[i] + b_int16[0];
} else {
out_int16[i] = a_int16[i] + b_int16[i];
}
}
break;
/* Yes, you'd have to add a separate case for every numeric mxClassID... */
/* In C++ you could do it with a template. */
default:
mexErrMsgTxt("Unsupported array type");
break;
}
plhs[0] = out;
}
Then you have to figure out how to invoke it from your Matlab code. If you're writing all the code, you could just call "c_plus(a, b)" instead of "a + b" everywhere. Alternately, you could create your own numeric wrapper class, e.g. #cnumeric, that holds a Matlab numeric array in its field and defines plus() and other operations that invoke the approprate C style MEX function.
classdef cnumeric
properties
x % the underlying Matlab numeric array
end
methods
function obj = cnumeric(x)
obj.x = x;
end
function out = plus(a,b)
[a,b] = promote(a, b); % for convenience, and to mimic Matlab implicit promotion
if ~isequal(class(a.x), class(b.x))
error('inputs must have same wrapped type');
end
out_x = c_plus(a.x, b.x);
out = cnumeric(out_x);
end
% You'd have to define the math operations that you want normal
% Matlab behavior on, too
function out = minus(a,b)
[a,b] = promote(a, b);
out = cnumeric(a.x - b.x);
end
function display(obj)
fprintf('%s = \ncnumeric: %s\n', inputname(1), num2str(obj.x));
end
function [a,b] = promote(a,b)
%PROMOTE Implicit promotion of numeric to cnumeric and doubles to int
if isnumeric(a); a = cnumeric(a); end
if isnumeric(b); b = cnumeric(b); end
if isinteger(a.x) && isa(b.x, 'double')
b.x = cast(b.x, class(a.x));
end
if isinteger(b.x) && isa(a.x, 'double')
a.x = cast(a.x, class(b.x));
end
end
end
end
Then wrap your numbers in the #cnumeric where you want C-style int behavior and do math with them.
>> cnumeric(int32(intmax))
ans =
cnumeric: 2147483647
>> cnumeric(int32(intmax)) - 1
ans =
cnumeric: 2147483646
>> cnumeric(int32(intmax)) + 1
ans =
cnumeric: -2147483648
>> cnumeric(int16(intmax('int16')))
ans =
cnumeric: 32767
>> cnumeric(int16(intmax('int16'))) + 1
ans =
cnumeric: -32768
There's your C-style overflow behavior, isolated from breaking the primitive #int32 type. Plus, you can pass a #cnumeric object in to other functions that are expecting regular numerics and it'll "work" as long as they treat their inputs polymorphically.
Performance caveat: because this is an object, + will have the slower speed of a method dispatch instead of a builtin. If you have few calls on large arrays, this'll be fast, because the actual numeric operations are in C. Lots of calls on small arrays, could slow things down, because you're paying the per method call overhead a lot.
I ran the following snippet of code
test = int32(2^31-12);
for i = 1:24
test = test + int32(1)
end
with unexpected results. It seems that, for Matlab, intmax('int32')+1==intmax('int32'). I'm running 2010a on a 64-bit Mac OS X.
Not sure that this as an answer, more confirmation that Matlab behaves counterintuitively. However, the documentation for the intmax() function states:
Any value larger than the value returned by intmax saturates to the intmax value when cast to a 32-bit integer.
So I guess Matlab is behaving as documented.
Hm, yes...
Actually, I was able to solve the problem with my custom "overflow"-Subroutine... Now it runs painfully slow, but without unexpected behaviour! My mistake was a missing round(), since Matlab/Octave will introduce small errors.
But if someone knows a faster solution, I would be glad to try it!
function ret = overflow_sg(arg,bw)
% remove possible rounding errors, and prepare returnvalue (if number is inside boundaries, nothing will happen)
ret = round(arg);
argsize = size(ret);
for i = 1:argsize(1)
for j = 1:argsize(2)
ret(i,j) = flow_sg(ret(i,j),bw);
end
end
end%function
%---
function ret = flow_sg(arg,bw)
ret = arg;
while (ret < (-2^(bw-1)))
ret = ret + 2^bw;
end
% Check for overflows:
while (ret > (2^(bw-1)-1))
ret = ret - 2^bw;
end
end%function
If 64 bits is enough to not overflow, and you need a lot of these, perhaps do this:
function ret = overflow_sg(arg,bw)
mask = int64(0);
for i=1:round(bw)
mask = bitset(mask,i);
end
topbit = bitshift(int64(1),round(bw-1));
subfrom = double(bitshift(topbit,1))
ret = bitand( int64(arg) , mask );
i = (ret >= topbit);
ret(i) = int64(double(ret(i))-subfrom);
if (bw<=32)
ret = int32(ret);
end
end
Almost everything is done as a matrix calculation, and a lot is done with bits, and everything is done in one step (no while loops), so it should be pretty fast. If you're going to populate it with rand, subtract 0.5 since it assumes it should round to integer values (rather than truncate).
I'm not a Java expert, but underlying Java classes available in Matlab should allow handling of overflows like C would. One solution I found, works only for single value, but it converts a number to the int16 (Short) or int32 (Integer) representation. You must do your math using Matlab double, then convert to Java int16 or int32, then convert back to Matlab double. Unfortunately Java doesn't appear to support unsigned types in this way, only signed.
double(java.lang.Short(hex2dec('7FFF')))
<br>ans = 32767
double(java.lang.Short(hex2dec('7FFF')+1))
<br>ans = -32768
double(java.lang.Short(double(intmax('int16'))+1))
<br>ans = -32768
double(java.lang.Integer(hex2dec('7FFF')+1))
<br>ans = 32768
https://www.tutorialspoint.com/java/lang/java_lang_integer.htm
I'm working on a verification-tool for some VHDL-Code in MATLAB/Octave. Therefore I need data types which generate "real" overflows:
intmax('int32') + 1
ans = -2147483648
Later on, it would be helpful if I can define the bit width of a variable, but that is not so important right now.
When I build a C-like example, where a variable gets increased until it's smaller than zero, it spins forever and ever:
test = int32(2^30);
while (test > 0)
test = test + int32(1);
end
Another approach I tried was a custom "overflow"-routine which was called every time after a number is changed. This approach was painfully slow, not practicable and not working in all cases at all. Any suggestions?
In MATLAB, one option you have is to overload the methods that handle arithmetic operations for integer data types, creating your own custom overflow behavior that will result in a "wrap-around" of the integer value. As stated in the documentation:
You can define or overload your own
methods for int* (as you can for any
object) by placing the appropriately
named method in an #int* folder within
a folder on your path. Type help
datatypes for the names of the methods
you can overload.
This page of the documentation lists the equivalent methods for the arithmetic operators. The binary addition operation A+B is actually handled by the function plus(A,B). Therefore, you can create a folder called #int32 (placed in another folder on your MATLAB path) and put a function plus.m in there that will be used instead of the built-in method for int32 data types.
Here's an example of how you could design your overloaded plus function in order to create the overflow/underflow behavior you want:
function C = plus(A,B)
%# NOTE: This code sample is designed to work for scalar values of
%# the inputs. If one or more of the inputs is non-scalar,
%# the code below will need to be vectorized to accommodate,
%# and error checking of the input sizes will be needed.
if (A > 0) && (B > (intmax-A)) %# An overflow condition
C = builtin('plus',intmin,...
B-(intmax-A)-1); %# Wraps around to negative
elseif (A < 0) && (B < (intmin-A)) %# An underflow condition
C = builtin('plus',intmax,...
B-(intmin-A-1)); %# Wraps around to positive
else
C = builtin('plus',A,B); %# No problems; call the built-in plus.m
end
end
Notice that I call the built-in plus method (using the BUILTIN function) to perform addition of int32 values that I know will not suffer overflow/underflow problems. If I were to instead perform the integer addition using the operation A+B it would result in a recursive call to my overloaded plus method, which could lead to additional computational overhead or (in the worst-case scenario where the last line was C = A+B;) infinite recursion.
Here's a test, showing the wrap-around overflow behavior in action:
>> A = int32(2147483642); %# A value close to INTMAX
>> for i = 1:10, A = A+1; disp(A); end
2147483643
2147483644
2147483645
2147483646
2147483647 %# INTMAX
-2147483648 %# INTMIN
-2147483647
-2147483646
-2147483645
-2147483644
If you want to get C style numeric operations, you can use a MEX function to call the C operators directly, and by definition they'll work like C data types.
This method is a lot more work than gnovice's overrides, but it should integrate better into a large codebase and is safer than altering the definition for built-in types, so I think it should be mentioned for completeness.
Here's a MEX file which performs the C "+" operation on a Matlab array. Make one of these for each operator you want C-style behavior on.
/* c_plus.c - MEX function: C-style (not Matlab-style) "+" operation */
#include "mex.h"
#include "matrix.h"
#include <stdio.h>
void mexFunction(
int nlhs, mxArray *plhs[],
int nrhs, const mxArray *prhs[]
)
{
mxArray *out;
/* In production code, input/output type and bounds checks would go here. */
const mxArray *a = prhs[0];
const mxArray *b = prhs[1];
int i, n;
int *a_int32, *b_int32, *out_int32;
short *a_int16, *b_int16, *out_int16;
mxClassID datatype = mxGetClassID(a);
int n_a = mxGetNumberOfElements(a);
int n_b = mxGetNumberOfElements(b);
int a_is_scalar = n_a == 1;
int b_is_scalar = n_b == 1;
n = n_a >= n_b ? n_a : n_b;
out = mxCreateNumericArray(mxGetNumberOfDimensions(a), mxGetDimensions(a),
datatype, mxIsComplex(a));
switch (datatype) {
case mxINT32_CLASS:
a_int32 = (int*) mxGetData(a);
b_int32 = (int*) mxGetData(b);
out_int32 = (int*) mxGetData(out);
for (i=0; i<n; i++) {
if (a_is_scalar) {
out_int32[i] = a_int32[i] + b_int32[i];
} else if (b_is_scalar) {
out_int32[i] = a_int32[i] + b_int32[0];
} else {
out_int32[i] = a_int32[i] + b_int32[i];
}
}
break;
case mxINT16_CLASS:
a_int16 = (short*) mxGetData(a);
b_int16 = (short*) mxGetData(b);
out_int16 = (short*) mxGetData(out);
for (i=0; i<n; i++) {
if (a_is_scalar) {
out_int16[i] = a_int16[0] + b_int16[i];
} else if (b_is_scalar) {
out_int16[i] = a_int16[i] + b_int16[0];
} else {
out_int16[i] = a_int16[i] + b_int16[i];
}
}
break;
/* Yes, you'd have to add a separate case for every numeric mxClassID... */
/* In C++ you could do it with a template. */
default:
mexErrMsgTxt("Unsupported array type");
break;
}
plhs[0] = out;
}
Then you have to figure out how to invoke it from your Matlab code. If you're writing all the code, you could just call "c_plus(a, b)" instead of "a + b" everywhere. Alternately, you could create your own numeric wrapper class, e.g. #cnumeric, that holds a Matlab numeric array in its field and defines plus() and other operations that invoke the approprate C style MEX function.
classdef cnumeric
properties
x % the underlying Matlab numeric array
end
methods
function obj = cnumeric(x)
obj.x = x;
end
function out = plus(a,b)
[a,b] = promote(a, b); % for convenience, and to mimic Matlab implicit promotion
if ~isequal(class(a.x), class(b.x))
error('inputs must have same wrapped type');
end
out_x = c_plus(a.x, b.x);
out = cnumeric(out_x);
end
% You'd have to define the math operations that you want normal
% Matlab behavior on, too
function out = minus(a,b)
[a,b] = promote(a, b);
out = cnumeric(a.x - b.x);
end
function display(obj)
fprintf('%s = \ncnumeric: %s\n', inputname(1), num2str(obj.x));
end
function [a,b] = promote(a,b)
%PROMOTE Implicit promotion of numeric to cnumeric and doubles to int
if isnumeric(a); a = cnumeric(a); end
if isnumeric(b); b = cnumeric(b); end
if isinteger(a.x) && isa(b.x, 'double')
b.x = cast(b.x, class(a.x));
end
if isinteger(b.x) && isa(a.x, 'double')
a.x = cast(a.x, class(b.x));
end
end
end
end
Then wrap your numbers in the #cnumeric where you want C-style int behavior and do math with them.
>> cnumeric(int32(intmax))
ans =
cnumeric: 2147483647
>> cnumeric(int32(intmax)) - 1
ans =
cnumeric: 2147483646
>> cnumeric(int32(intmax)) + 1
ans =
cnumeric: -2147483648
>> cnumeric(int16(intmax('int16')))
ans =
cnumeric: 32767
>> cnumeric(int16(intmax('int16'))) + 1
ans =
cnumeric: -32768
There's your C-style overflow behavior, isolated from breaking the primitive #int32 type. Plus, you can pass a #cnumeric object in to other functions that are expecting regular numerics and it'll "work" as long as they treat their inputs polymorphically.
Performance caveat: because this is an object, + will have the slower speed of a method dispatch instead of a builtin. If you have few calls on large arrays, this'll be fast, because the actual numeric operations are in C. Lots of calls on small arrays, could slow things down, because you're paying the per method call overhead a lot.
I ran the following snippet of code
test = int32(2^31-12);
for i = 1:24
test = test + int32(1)
end
with unexpected results. It seems that, for Matlab, intmax('int32')+1==intmax('int32'). I'm running 2010a on a 64-bit Mac OS X.
Not sure that this as an answer, more confirmation that Matlab behaves counterintuitively. However, the documentation for the intmax() function states:
Any value larger than the value returned by intmax saturates to the intmax value when cast to a 32-bit integer.
So I guess Matlab is behaving as documented.
Hm, yes...
Actually, I was able to solve the problem with my custom "overflow"-Subroutine... Now it runs painfully slow, but without unexpected behaviour! My mistake was a missing round(), since Matlab/Octave will introduce small errors.
But if someone knows a faster solution, I would be glad to try it!
function ret = overflow_sg(arg,bw)
% remove possible rounding errors, and prepare returnvalue (if number is inside boundaries, nothing will happen)
ret = round(arg);
argsize = size(ret);
for i = 1:argsize(1)
for j = 1:argsize(2)
ret(i,j) = flow_sg(ret(i,j),bw);
end
end
end%function
%---
function ret = flow_sg(arg,bw)
ret = arg;
while (ret < (-2^(bw-1)))
ret = ret + 2^bw;
end
% Check for overflows:
while (ret > (2^(bw-1)-1))
ret = ret - 2^bw;
end
end%function
If 64 bits is enough to not overflow, and you need a lot of these, perhaps do this:
function ret = overflow_sg(arg,bw)
mask = int64(0);
for i=1:round(bw)
mask = bitset(mask,i);
end
topbit = bitshift(int64(1),round(bw-1));
subfrom = double(bitshift(topbit,1))
ret = bitand( int64(arg) , mask );
i = (ret >= topbit);
ret(i) = int64(double(ret(i))-subfrom);
if (bw<=32)
ret = int32(ret);
end
end
Almost everything is done as a matrix calculation, and a lot is done with bits, and everything is done in one step (no while loops), so it should be pretty fast. If you're going to populate it with rand, subtract 0.5 since it assumes it should round to integer values (rather than truncate).
I'm not a Java expert, but underlying Java classes available in Matlab should allow handling of overflows like C would. One solution I found, works only for single value, but it converts a number to the int16 (Short) or int32 (Integer) representation. You must do your math using Matlab double, then convert to Java int16 or int32, then convert back to Matlab double. Unfortunately Java doesn't appear to support unsigned types in this way, only signed.
double(java.lang.Short(hex2dec('7FFF')))
<br>ans = 32767
double(java.lang.Short(hex2dec('7FFF')+1))
<br>ans = -32768
double(java.lang.Short(double(intmax('int16'))+1))
<br>ans = -32768
double(java.lang.Integer(hex2dec('7FFF')+1))
<br>ans = 32768
https://www.tutorialspoint.com/java/lang/java_lang_integer.htm