How to convert a byte array into double in C? - c++

I've got a byte array containing 8 bytes and would like to convert and use them as a double precision binary floating-point number.
Could someone please tell me how to convert it?


Try this:
double a;
memcpy(&a, ptr, sizeof(double));
where ptr is the pointer to your byte array. If you want to avoid copying use a union, e.g.
union {
double d;
char bytes[sizeof(double)];
} u;
// Store your data in u.bytes
// Use floating point number from u.d


Here is one solution using memcpy function:
double d = 0;
unsigned char buf[sizeof d] = {0};
memcpy(&d, buf, sizeof d);
Note that a solution like:
d = *((double *) buf);
shoud be avoided. This is undefined behavior because it potentially violates alignment and aliasing rules.


In C++:
double x;
char buf[sizeof(double)]; // your data
#include <algorithm>
// ...
std::copy(buf, buf + sizeof(double), reinterpret_cast<char*>(&x));
In C:
#include <string.h>
/* ... */
memcpy(&x, buf, sizeof(double));
In C++11, you can also use std::begin(buf) and std::end(buf) as the boundaries (include the header <iterator>), and in both languages you can use sizeof(buf) / sizeof(buf[0]) (or simply sizeof(buf)) for the size, all provided buf is actually an array and not just a pointer.

Related

Why is my C++ OpenGL program not rendering a simple rectangle on screen with no errors from OpenGL debugging context? [duplicate]

How do I determine the size of my array in C?
That is, the number of elements the array can hold?

Executive summary:
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);
Full answer:
To determine the size of your array in bytes, you can use the sizeof
operator:
int a[17];
size_t n = sizeof(a);
On my computer, ints are 4 bytes long, so n is 68.
To determine the number of elements in the array, we can divide
the total size of the array by the size of the array element.
You could do this with the type, like this:
int a[17];
size_t n = sizeof(a) / sizeof(int);
and get the proper answer (68 / 4 = 17), but if the type of
a changed you would have a nasty bug if you forgot to change
the sizeof(int) as well.
So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.
int a[17];
size_t n = sizeof(a) / sizeof(a[0]);
Another advantage is that you can now easily parameterize
the array name in a macro and get:
#define NELEMS(x) (sizeof(x) / sizeof((x)[0]))
int a[17];
size_t n = NELEMS(a);

The sizeof way is the right way iff you are dealing with arrays not received as parameters. An array sent as a parameter to a function is treated as a pointer, so sizeof will return the pointer's size, instead of the array's.
Thus, inside functions this method does not work. Instead, always pass an additional parameter size_t size indicating the number of elements in the array.
Test:
#include <stdio.h>
#include <stdlib.h>
void printSizeOf(int intArray[]);
void printLength(int intArray[]);
int main(int argc, char* argv[])
{
int array[] = { 0, 1, 2, 3, 4, 5, 6 };
printf("sizeof of array: %d\n", (int) sizeof(array));
printSizeOf(array);
printf("Length of array: %d\n", (int)( sizeof(array) / sizeof(array[0]) ));
printLength(array);
}
void printSizeOf(int intArray[])
{
printf("sizeof of parameter: %d\n", (int) sizeof(intArray));
}
void printLength(int intArray[])
{
printf("Length of parameter: %d\n", (int)( sizeof(intArray) / sizeof(intArray[0]) ));
}
Output (in a 64-bit Linux OS):
sizeof of array: 28
sizeof of parameter: 8
Length of array: 7
Length of parameter: 2
Output (in a 32-bit windows OS):
sizeof of array: 28
sizeof of parameter: 4
Length of array: 7
Length of parameter: 1

It is worth noting that sizeof doesn't help when dealing with an array value that has decayed to a pointer: even though it points to the start of an array, to the compiler it is the same as a pointer to a single element of that array. A pointer does not "remember" anything else about the array that was used to initialize it.
int a[10];
int* p = a;
assert(sizeof(a) / sizeof(a[0]) == 10);
assert(sizeof(p) == sizeof(int*));
assert(sizeof(*p) == sizeof(int));

The sizeof "trick" is the best way I know, with one small but (to me, this being a major pet peeve) important change in the use of parenthesis.
As the Wikipedia entry makes clear, C's sizeof is not a function; it's an operator. Thus, it does not require parenthesis around its argument, unless the argument is a type name. This is easy to remember, since it makes the argument look like a cast expression, which also uses parenthesis.
So: If you have the following:
int myArray[10];
You can find the number of elements with code like this:
size_t n = sizeof myArray / sizeof *myArray;
That, to me, reads a lot easier than the alternative with parenthesis. I also favor use of the asterisk in the right-hand part of the division, since it's more concise than indexing.
Of course, this is all compile-time too, so there's no need to worry about the division affecting the performance of the program. So use this form wherever you can.
It is always best to use sizeof on an actual object when you have one, rather than on a type, since then you don't need to worry about making an error and stating the wrong type.
For instance, say you have a function that outputs some data as a stream of bytes, for instance across a network. Let's call the function send(), and make it take as arguments a pointer to the object to send, and the number of bytes in the object. So, the prototype becomes:
void send(const void *object, size_t size);
And then you need to send an integer, so you code it up like this:
int foo = 4711;
send(&foo, sizeof (int));
Now, you've introduced a subtle way of shooting yourself in the foot, by specifying the type of foo in two places. If one changes but the other doesn't, the code breaks. Thus, always do it like this:
send(&foo, sizeof foo);
Now you're protected. Sure, you duplicate the name of the variable, but that has a high probability of breaking in a way the compiler can detect, if you change it.

int size = (&arr)[1] - arr;
Check out this link for explanation

I would advise to never use sizeof (even if it can be used) to get any of the two different sizes of an array, either in number of elements or in bytes, which are the last two cases I show here. For each of the two sizes, the macros shown below can be used to make it safer. The reason is to make obvious the intention of the code to maintainers, and difference sizeof(ptr) from sizeof(arr) at first glance (which written this way isn't obvious), so that bugs are then obvious for everyone reading the code.
TL;DR:
#define ARRAY_SIZE(arr) (sizeof(arr) / sizeof((arr)[0]) + must_be_array(arr))
#define ARRAY_BYTES(arr) (sizeof(arr) + must_be_array(arr))
must_be_array(arr) (defined below) IS needed as -Wsizeof-pointer-div is buggy (as of april/2020):
#define is_same_type(a, b) __builtin_types_compatible_p(typeof(a), typeof(b))
#define is_array(arr) (!is_same_type((arr), &(arr)[0]))
#define must_be(e) \
( \
0 * (int)sizeof( \
struct { \
static_assert(e); \
char ISO_C_forbids_a_struct_with_no_members__; \
} \
) \
)
#define must_be_array(arr) must_be(is_array(arr))
There have been important bugs regarding this topic: https://lkml.org/lkml/2015/9/3/428
I disagree with the solution that Linus provides, which is to never use array notation for parameters of functions.
I like array notation as documentation that a pointer is being used as an array. But that means that a fool-proof solution needs to be applied so that it is impossible to write buggy code.
From an array we have three sizes which we might want to know:
The size of the elements of the array
The number of elements in the array
The size in bytes that the array uses in memory
The size of the elements of the array
The first one is very simple, and it doesn't matter if we are dealing with an array or a pointer, because it's done the same way.
Example of usage:
void foo(size_t nmemb, int arr[nmemb])
{
qsort(arr, nmemb, sizeof(arr[0]), cmp);
}
qsort() needs this value as its third argument.
For the other two sizes, which are the topic of the question, we want to make sure that we're dealing with an array, and break the compilation if not, because if we're dealing with a pointer, we will get wrong values. When the compilation is broken, we will be able to easily see that we weren't dealing with an array, but with a pointer instead, and we will just have to write the code with a variable or a macro that stores the size of the array behind the pointer.
The number of elements in the array
This one is the most common, and many answers have provided you with the typical macro ARRAY_SIZE:
#define ARRAY_SIZE(arr) (sizeof(arr) / sizeof((arr)[0]))
Recent versions of compilers, such as GCC 8, will warn you when you apply this macro to a pointer, so it is safe (there are other methods to make it safe with older compilers).
It works by dividing the size in bytes of the whole array by the size of each element.
Examples of usage:
void foo(size_t nmemb)
{
char buf[nmemb];
fgets(buf, ARRAY_SIZE(buf), stdin);
}
void bar(size_t nmemb)
{
int arr[nmemb];
for (size_t i = 0; i < ARRAY_SIZE(arr); i++)
arr[i] = i;
}
If these functions didn't use arrays, but got them as parameters instead, the former code would not compile, so it would be impossible to have a bug (given that a recent compiler version is used, or that some other trick is used), and we need to replace the macro call by the value:
void foo(size_t nmemb, char buf[nmemb])
{
fgets(buf, nmemb, stdin);
}
void bar(size_t nmemb, int arr[nmemb])
{
for (size_t i = nmemb - 1; i < nmemb; i--)
arr[i] = i;
}
The size in bytes that the array uses in memory
ARRAY_SIZE is commonly used as a solution to the previous case, but this case is rarely written safely, maybe because it's less common.
The common way to get this value is to use sizeof(arr). The problem: the same as with the previous one; if you have a pointer instead of an array, your program will go nuts.
The solution to the problem involves using the same macro as before, which we know to be safe (it breaks compilation if it is applied to a pointer):
#define ARRAY_BYTES(arr) (sizeof((arr)[0]) * ARRAY_SIZE(arr))
How it works is very simple: it undoes the division that ARRAY_SIZE does, so after mathematical cancellations you end up with just one sizeof(arr), but with the added safety of the ARRAY_SIZE construction.
Example of usage:
void foo(size_t nmemb)
{
int arr[nmemb];
memset(arr, 0, ARRAY_BYTES(arr));
}
memset() needs this value as its third argument.
As before, if the array is received as a parameter (a pointer), it won't compile, and we will have to replace the macro call by the value:
void foo(size_t nmemb, int arr[nmemb])
{
memset(arr, 0, sizeof(arr[0]) * nmemb);
}
Update (23/apr/2020): -Wsizeof-pointer-div is buggy:
Today I found out that the new warning in GCC only works if the macro is defined in a header that is not a system header. If you define the macro in a header that is installed in your system (usually /usr/local/include/ or /usr/include/) (#include <foo.h>), the compiler will NOT emit a warning (I tried GCC 9.3.0).
So we have #define ARRAY_SIZE(arr) (sizeof(arr) / sizeof((arr)[0])) and want to make it safe. We will need C2X static_assert() and some GCC extensions: Statements and Declarations in Expressions, __builtin_types_compatible_p:
#include <assert.h>
#define is_same_type(a, b) __builtin_types_compatible_p(typeof(a), typeof(b))
#define is_array(arr) (!is_same_type((arr), &(arr)[0]))
#define Static_assert_array(arr) static_assert(is_array(arr))
#define ARRAY_SIZE(arr) \
({ \
Static_assert_array(arr); \
sizeof(arr) / sizeof((arr)[0]); \
})
Now ARRAY_SIZE() is completely safe, and therefore all its derivatives will be safe.
Update: libbsd provides __arraycount():
Libbsd provides the macro __arraycount() in <sys/cdefs.h>, which is unsafe because it lacks a pair of parentheses, but we can add those parentheses ourselves, and therefore we don't even need to write the division in our header (why would we duplicate code that already exists?). That macro is defined in a system header, so if we use it we are forced to use the macros above.
#inlcude <assert.h>
#include <stddef.h>
#include <sys/cdefs.h>
#include <sys/types.h>
#define is_same_type(a, b) __builtin_types_compatible_p(typeof(a), typeof(b))
#define is_array(arr) (!is_same_type((arr), &(arr)[0]))
#define Static_assert_array(arr) static_assert(is_array(arr))
#define ARRAY_SIZE(arr) \
({ \
Static_assert_array(arr); \
__arraycount((arr)); \
})
#define ARRAY_BYTES(arr) (sizeof((arr)[0]) * ARRAY_SIZE(arr))
Some systems provide nitems() in <sys/param.h> instead, and some systems provide both. You should check your system, and use the one you have, and maybe use some preprocessor conditionals for portability and support both.
Update: Allow the macro to be used at file scope:
Unfortunately, the ({}) gcc extension cannot be used at file scope.
To be able to use the macro at file scope, the static assertion must be
inside sizeof(struct {}). Then, multiply it by 0 to not affect
the result. A cast to (int) might be good to simulate a function
that returns (int)0 (in this case it is not necessary, but then it
is reusable for other things).
Additionally, the definition of ARRAY_BYTES() can be simplified a bit.
#include <assert.h>
#include <stddef.h>
#include <sys/cdefs.h>
#include <sys/types.h>
#define is_same_type(a, b) __builtin_types_compatible_p(typeof(a), typeof(b))
#define is_array(arr) (!is_same_type((arr), &(arr)[0]))
#define must_be(e) \
( \
0 * (int)sizeof( \
struct { \
static_assert(e); \
char ISO_C_forbids_a_struct_with_no_members__; \
} \
) \
)
#define must_be_array(arr) must_be(is_array(arr))
#define ARRAY_SIZE(arr) (__arraycount((arr)) + must_be_array(arr))
#define ARRAY_BYTES(arr) (sizeof(arr) + must_be_array(arr))
Notes:
This code makes use of the following extensions, which are completely necessary, and their presence is absolutely necessary to achieve safety. If your compiler doesn't have them, or some similar ones, then you can't achieve this level of safety.
__builtin_types_compatible_p()
typeof()
I also make use of the following C2X feature. However, its absence by using an older standard can be overcome using some dirty tricks (see for example: What is “:-!!” in C code?) (in C11 you also have static_assert(), but it requires a message).
static_assert()

You can use the sizeof operator, but it will not work for functions, because it will take the reference of a pointer.
You can do the following to find the length of an array:
len = sizeof(arr)/sizeof(arr[0])
The code was originally found here:
C program to find the number of elements in an array

If you know the data type of the array, you can use something like:
int arr[] = {23, 12, 423, 43, 21, 43, 65, 76, 22};
int noofele = sizeof(arr)/sizeof(int);
Or if you don't know the data type of array, you can use something like:
noofele = sizeof(arr)/sizeof(arr[0]);
Note: This thing only works if the array is not defined at run time (like malloc) and the array is not passed in a function. In both cases, arr (array name) is a pointer.

The macro ARRAYELEMENTCOUNT(x) that everyone is making use of evaluates incorrectly. This, realistically, is just a sensitive matter, because you can't have expressions that result in an 'array' type.
/* Compile as: CL /P "macro.c" */
# define ARRAYELEMENTCOUNT(x) (sizeof (x) / sizeof (x[0]))
ARRAYELEMENTCOUNT(p + 1);
Actually evaluates as:
(sizeof (p + 1) / sizeof (p + 1[0]));
Whereas
/* Compile as: CL /P "macro.c" */
# define ARRAYELEMENTCOUNT(x) (sizeof (x) / sizeof (x)[0])
ARRAYELEMENTCOUNT(p + 1);
It correctly evaluates to:
(sizeof (p + 1) / sizeof (p + 1)[0]);
This really doesn't have a lot to do with the size of arrays explicitly. I've just noticed a lot of errors from not truly observing how the C preprocessor works. You always wrap the macro parameter, not an expression in might be involved in.
This is correct; my example was a bad one. But that's actually exactly what should happen. As I previously mentioned p + 1 will end up as a pointer type and invalidate the entire macro (just like if you attempted to use the macro in a function with a pointer parameter).
At the end of the day, in this particular instance, the fault doesn't really matter (so I'm just wasting everyone's time; huzzah!), because you don't have expressions with a type of 'array'. But really the point about preprocessor evaluation subtles I think is an important one.

For multidimensional arrays it is a tad more complicated. Oftenly people define explicit macro constants, i.e.
#define g_rgDialogRows 2
#define g_rgDialogCols 7
static char const* g_rgDialog[g_rgDialogRows][g_rgDialogCols] =
{
{ " ", " ", " ", " 494", " 210", " Generic Sample Dialog", " " },
{ " 1", " 330", " 174", " 88", " ", " OK", " " },
};
But these constants can be evaluated at compile-time too with sizeof:
#define rows_of_array(name) \
(sizeof(name ) / sizeof(name[0][0]) / columns_of_array(name))
#define columns_of_array(name) \
(sizeof(name[0]) / sizeof(name[0][0]))
static char* g_rgDialog[][7] = { /* ... */ };
assert( rows_of_array(g_rgDialog) == 2);
assert(columns_of_array(g_rgDialog) == 7);
Note that this code works in C and C++. For arrays with more than two dimensions use
sizeof(name[0][0][0])
sizeof(name[0][0][0][0])
etc., ad infinitum.

Size of an array in C:
int a[10];
size_t size_of_array = sizeof(a); // Size of array a
int n = sizeof (a) / sizeof (a[0]); // Number of elements in array a
size_t size_of_element = sizeof(a[0]); // Size of each element in array a
// Size of each element = size of type

sizeof(array) / sizeof(array[0])

#define SIZE_OF_ARRAY(_array) (sizeof(_array) / sizeof(_array[0]))

If you really want to do this to pass around your array I suggest implementing a structure to store a pointer to the type you want an array of and an integer representing the size of the array. Then you can pass that around to your functions. Just assign the array variable value (pointer to first element) to that pointer. Then you can go Array.arr[i] to get the i-th element and use Array.size to get the number of elements in the array.
I included some code for you. It's not very useful but you could extend it with more features. To be honest though, if these are the things you want you should stop using C and use another language with these features built in.
/* Absolutely no one should use this...
By the time you're done implementing it you'll wish you just passed around
an array and size to your functions */
/* This is a static implementation. You can get a dynamic implementation and
cut out the array in main by using the stdlib memory allocation methods,
but it will work much slower since it will store your array on the heap */
#include <stdio.h>
#include <string.h>
/*
#include "MyTypeArray.h"
*/
/* MyTypeArray.h
#ifndef MYTYPE_ARRAY
#define MYTYPE_ARRAY
*/
typedef struct MyType
{
int age;
char name[20];
} MyType;
typedef struct MyTypeArray
{
int size;
MyType *arr;
} MyTypeArray;
MyType new_MyType(int age, char *name);
MyTypeArray newMyTypeArray(int size, MyType *first);
/*
#endif
End MyTypeArray.h */
/* MyTypeArray.c */
MyType new_MyType(int age, char *name)
{
MyType d;
d.age = age;
strcpy(d.name, name);
return d;
}
MyTypeArray new_MyTypeArray(int size, MyType *first)
{
MyTypeArray d;
d.size = size;
d.arr = first;
return d;
}
/* End MyTypeArray.c */
void print_MyType_names(MyTypeArray d)
{
int i;
for (i = 0; i < d.size; i++)
{
printf("Name: %s, Age: %d\n", d.arr[i].name, d.arr[i].age);
}
}
int main()
{
/* First create an array on the stack to store our elements in.
Note we could create an empty array with a size instead and
set the elements later. */
MyType arr[] = {new_MyType(10, "Sam"), new_MyType(3, "Baxter")};
/* Now create a "MyTypeArray" which will use the array we just
created internally. Really it will just store the value of the pointer
"arr". Here we are manually setting the size. You can use the sizeof
trick here instead if you're sure it will work with your compiler. */
MyTypeArray array = new_MyTypeArray(2, arr);
/* MyTypeArray array = new_MyTypeArray(sizeof(arr)/sizeof(arr[0]), arr); */
print_MyType_names(array);
return 0;
}

The best way is you save this information, for example, in a structure:
typedef struct {
int *array;
int elements;
} list_s;
Implement all necessary functions such as create, destroy, check equality, and everything else you need. It is easier to pass as a parameter.

The function sizeof returns the number of bytes which is used by your array in the memory. If you want to calculate the number of elements in your array, you should divide that number with the sizeof variable type of the array. Let's say int array[10];, if variable type integer in your computer is 32 bit (or 4 bytes), in order to get the size of your array, you should do the following:
int array[10];
size_t sizeOfArray = sizeof(array)/sizeof(int);

A more elegant solution will be
size_t size = sizeof(a) / sizeof(*a);

You can use the & operator. Here is the source code:
#include<stdio.h>
#include<stdlib.h>
int main(){
int a[10];
int *p;
printf("%p\n", (void *)a);
printf("%p\n", (void *)(&a+1));
printf("---- diff----\n");
printf("%zu\n", sizeof(a[0]));
printf("The size of array a is %zu\n", ((char *)(&a+1)-(char *)a)/(sizeof(a[0])));
return 0;
};
Here is the sample output
1549216672
1549216712
---- diff----
4
The size of array a is 10

The simplest answer:
#include <stdio.h>
int main(void) {
int a[] = {2,3,4,5,4,5,6,78,9,91,435,4,5,76,7,34}; // For example only
int size;
size = sizeof(a)/sizeof(a[0]); // Method
printf("size = %d", size);
return 0;
}

"you've introduced a subtle way of shooting yourself in the foot"
C 'native' arrays do not store their size. It is therefore recommended to save the length of the array in a separate variable/const, and pass it whenever you pass the array, that is:
#define MY_ARRAY_LENGTH 15
int myArray[MY_ARRAY_LENGTH];
If you are writing C++, you SHOULD always avoid native arrays anyway (unless you can't, in which case, mind your foot). If you are writing C++, use the STL's 'vector' container. "Compared to arrays, they provide almost the same performance", and they are far more useful!
// vector is a template, the <int> means it is a vector of ints
vector<int> numbers;
// push_back() puts a new value at the end (or back) of the vector
for (int i = 0; i < 10; i++)
numbers.push_back(i);
// Determine the size of the array
cout << numbers.size();
See:
http://www.cplusplus.com/reference/stl/vector/

Beside the answers already provided, I want to point out a special case by the use of
sizeof(a) / sizeof (a[0])
If a is either an array of char, unsigned char or signed char you do not need to use sizeof twice since a sizeof expression with one operand of these types do always result to 1.
Quote from C18,6.5.3.4/4:
"When sizeof is applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1."
Thus, sizeof(a) / sizeof (a[0]) would be equivalent to NUMBER OF ARRAY ELEMENTS / 1 if a is an array of type char, unsigned char or signed char. The division through 1 is redundant.
In this case, you can simply abbreviate and do:
sizeof(a)
For example:
char a[10];
size_t length = sizeof(a);
If you want a proof, here is a link to GodBolt.
Nonetheless, the division maintains safety, if the type significantly changes (although these cases are rare).

To know the size of a fixed array declared explicitly in code and referenced by its variable, you can use sizeof, for example:
int a[10];
int len = sizeof(a)/sizeof(int);
But this is usually useless, because you already know the answer.
But if you have a pointer you can’t use sizeof, its a matter of principle.
But...Since arrays are presented as linear memory for the user, you can calculate the size if you know the last element address and if you know the size of the type, then you can count how many elements it have. For example:
#include <stdio.h>
int main(){
int a[10];
printf("%d\n", sizeof(a)/sizeof(int));
int *first = a;
int *last = &(a[9]);
printf("%d\n", (last-first) + 1);
}
Output:
10
10
Also if you can't take advantage of compile time you can:
#include <stdio.h>
int main(){
int a[10];
printf("%d\n", sizeof(a)/sizeof(int));
void *first = a;
void *last = &(a[9]);
printf("%d\n", (last-first)/sizeof(int) + 1);
}

Note: This one can give you undefined behaviour as pointed out by M.M in the comment.
int a[10];
int size = (*(&a+1)-a);
For more details, see here and also here.

For a predefined array:
int a[] = {1, 2, 3, 4, 5, 6};
Calculating number of elements in the array:
element _count = sizeof(a) / sizeof(a[0]);

concatenating uint16_t and uint32_t values for hashing

I am trying concatenating (not adding) 2 uint16_t struct members and 2 uint32_t struct members and assigning the result to const void *p for the purpose of hashing. The struct and concatenation function that I am trying to implement is as follows.
struct xyz {
....
uint32_t a;
uint32_t b;
....
uint16_t c;
uint16_t d;
....
}
const void *p=concatenation(xyz.a,xyz.b,xyz.c,xyz.d)
Edited:
I have to use pre-defined hash functions. The most suitable hash function for my task seems to be this.
uint32_t hash(const uint32_t p[], size_t n)
{
//Returns the hash of the 'n' 32-bit words at 'p'
}
or
uint32_t hash64(const uint64_t p[], size_t n)
{
//Returns the hash of the 'n' 64-bit words at 'p'
}

for the purpose of hashing
In this case, I'd rather prefer providing a custom hash function – or specialise std::hash for. For use with standard templates, this might look like this:
namespace std // any extension of std namespace is UB
// sole exception: specialising templates, which we are going to do
{
template <>
struct hash<xyz>
{
size_t operator()(xyz const& i) const
{
// TODO: need to calculate the value from a, b, c, and d appropriately
return 0;
};
};
// if xyz is polymorphic, you might need to operate on pointers
// no problem either:
template <>
struct hash<xyz*>
{
size_t operator()(xyz const* i) const
{
return hash<xyz>()(*i);
// or if hash value is type dependent:
return i->hash(); // custom virtual hash member function needs to be implented
}
}
// now you can have
std::unordered_set<xyz> someSet;
void demo()
{
someSet.insert(xyz());
}
(Untested code, in case of errors please fix yourself.)
A list of hashing algorithms which might be used can be found at wikipedia.

If you want the value to fit into a pointer, the full value can be 32 bits on x86 or 64 bits on x64. I'm going to assume you are compiling for 64 bit machines.
This means you can only fit 2 uint16 and one uint32, or 2 uint32s.
Either way, you would shift the values into a uint64 (c | (d << 16) | (c << 32)) and then convert that value to a void*.
Edit: for clarification, you cannot fit all the structs members bit shifted one after another into a single pointer. You need a minimum of 96 bits to hold the packed struct which means at least two 64 bit pointers.

There are a few things to consider:
Does that hash value need to be portable across systems? If it does, then you will need to be careful to order the bytes the same way on different systems. If not, then the implementation can be simpler.
Do you want to hash every member of the class, and the class has no padding, and no value of a member should be hashed equally to another different value?
If both of these simplifications apply, then your function is fast and easy to implement but violating that precondition will break the hash. If not, then you must serialise the the data into a buffer, which practically means that you cannot simply return a pointer.
Here is a super simple implementation for the case that you don't need portability, and you hash all members, and there is no padding:
xyz example;
static_assert(std::has_unique_object_representations_v<xyz>);
const void* p = &example;
Note that this doesn't work with (IEEE-754) float members due to peculiarities of NaN.
A more robust solution that can produce hashes that are portable across systems is to use a general purpose serialisation scheme, and hash the serialised result. There is no standard serialisation functionality in C++.

void* has problems like: Who owns the memory? What's the type you are going to reinterpret the pointer as?
A more typed solution would be to use std::array of std::byte then you at least know that you're looking at an array of raw bytes and nothing else:
#include <cstdint>
#include <array>
#include <cstddef>
#include <cstring>
auto concat(std::uint32_t a, std::uint32_t b, std::uint16_t c, std::uint16_t d) {
std::array<std::byte, sizeof a + sizeof b + sizeof c + sizeof d> res;
std::byte* p = res.data();
std::memcpy(p, &a, sizeof a);
std::memcpy(p += sizeof a, &b, sizeof b);
std::memcpy(p += sizeof b, &c, sizeof c);
std::memcpy(p += sizeof c, &d, sizeof d);
return res;
}
int main() {
std::uint32_t a = 1, b = 0;
std::uint16_t c = 1, d = 0;
auto res = concat(a, b, c, d);
return 0;
}

Cast char* to short*

I want to sum up all bytes of my structure. I read that I should cast pointer of my structure from char to short. Why?
Does casting using (short) from char to short is correct?
My code
#include <stdio.h>
#include <string.h>
struct pseudo_header
{
int var1;
int var2;
char name[25];
};
void csum(const unsigned short* ptr, int nbytes)
{
unsigned long sum = 0;
for(int i = 0; i < sizeof(struct pseudo_header); i++)
{
printf("%#8x\n", ptr[i]);
sum+= ptr[i];
}
printf("%#8x", sum);
}
int main() {
struct pseudo_header psh = {0};
char datagram[4096];
psh.var1 = 10;
psh.var2 = 20;
strcpy(psh.name, "Test");
memcpy(datagram, &psh, sizeof(struct pseudo_header));
csum((unsigned short*)datagram, sizeof(struct pseudo_header));
return 0;
}
It looks like it works, but I can't verify this. Any help is appreciated.

No, the behaviour on dereferencing a pointer that's been set to the result of a cast from a char* to a short* is undefined, unless the data to which char* is pointing was originally a short object or array; which yours isn't.
The well-defined way (in both C and C++) to analyse memory is to use an unsigned char*, but be careful not to traverse your memory so as to reach areas that are not owned by your program.

Basically this works because you cleared the structure with zero. = {0}.
You can give the function a pointer to a structure struct *pseudo_header.
I would see an alignment issue.
I would check sizeof(struct ..) for expected value 33 if I have to add a pragma pack() statement before the structure and then cast to unsigned char* inside the function.
Test your function with a 25 chars length name.

short is at least two bytes. So if you want to sum all the bytes then casting to short* is wrong. Instead cast to unsigned char*.

Casting (aliasing) a pointer with any type other than a char * violates the strict aliasing rule. Gcc will optimize based on assumptions of the strict aliasing rule and can lead to interesting bugs. Tge only truly safe way round the strict aliasing rule is to use a memcpy. Gcc supports memcpy as a compiler intrinsic so can optimize into the copy.
Alternatively, you can disable strict aliasing with the -fno-strict-aliasing flag.
PS - I am unclear if a union provides a suitable way round the strict aliasing rule.

How to convert a byte array of size 64 to a list of double values in Arduino C++?

void Manager::byteArrayToDoubleArray(byte ch[]) {
int counter = 0;
// temp array to break the byte array into size of 8 and read it
byte temp[64];
// double result values
double res[8];
int index = 0;
int size = (sizeof(ch) / sizeof(*ch));
for (int i = 0; i < size; i++) {
counter++;
temp[i] = ch[i];
if (counter % 8 == 0) {
res[index] = *reinterpret_cast<double * const>(temp);
index++;
counter = 0;
}
}
}
Here result would be a list of double values with count = 8.

Your problem is two things. You have some typos and misunderstanding. And the C++ standard is somewhat broken in this area.
I'll try to fix both.
First, a helper function called laundry_pods. It takes raw memory and "launders" it into an array of a type of your choice, so long as you pick a pod type:
template<class T, std::size_t N>
T* laundry_pods( void* ptr ) {
static_assert( std::is_pod<std::remove_cv_t<T>>{} );
char optimized_away[sizeof(T)*N];
std::memcpy( optimized_away, ptr , sizeof(T)*N );
T* r = ::new( ptr ) T[N];
assert( r == ptr );
std::memcpy( r, optimized_away, sizeof(T)*N );
return r;
}
now simply do
void Manager::byteArrayToDoubleArray(byte ch[]) {
double* pdouble = laundry_pods<double, 8>(ch);
}
and pdouble is a pointer to memory of ch interpreted as an array of 8 doubles. (It is not a copy of it, it interprets those bytes in-place).
While laundry_pods appears to copy the bytes around, both g++ and clang optimize it down into a binary noop. The seeming copying of bytes around is a way to get around aliasing restrictions and object lifetime rules in the C++ standard.
It relies on arrays of pod not having extra bookkeeping overhead (which C++ implementations are free to do; none do that I know of. That is what the non-static assert double-checks), but it returns a pointer to a real honest to goodness array of double. If you want to avoid that assumption, you could instead create each doulbe as a separate object. However, then they aren't an array, and pointer arithmetic over non-arrays is fraught as far as the standard is concerned.
The use of the term "launder" has to do with getting around aliasing and object lifetime requirements. The function does nothing at runtime, but in the C++ abstract machine it takes the memory and converts it into binary identical memory that is now a bunch of doubles.

The trick of doing this kind of "conversion" is to always cast the double* to a char* (or unsigned char or std::byte). Never the other way round.
You should be able to do something like this:
void byteArrayToDoubleArray(byte* in, std::size_t n, double* out)
{
for(auto out_bytes = (byte*) out; n--;)
*out_bytes++ = *in++;
}
// ...
byte ch[64];
// .. fill ch with double data somehow
double res[8];
byteArrayToDoubleArray(ch, 64, res);
Assuming that type byte is an alias of char or unsigned char or std::byte.

I am not completly sure what you are trying to achieve here because of the code (sizeof(ch) / sizeof(*ch)) which does not make sense for an array of undefined size.
If you have a byte-Array (POD data type; something like a typedef char byte;) then this most simple solution would be a reinterpret_cast:
double *result = reinterpret_cast<double*>(ch);
This allows you to use result[0]..result[7] as long as ch[] is valid and contains at least 64 bytes. Be aware that this construct does not generate code. It tells the compiler that result[0] corresponds to ch[0..7] and so on. An access to result[] will result in an access to ch[].
But you have to know the number of elements in ch[] to calculate the number of valid double elements in result.
If you need a copy (because - for example - the ch[] is a temporary array) you could use
std::vector<double> result(reinterpret_cast<double*>(ch), reinterpret_cast<double*>(ch) + itemsInCh * sizeof(*ch) / sizeof(double));
So if ch[] is an array with 64 items and a byte is really an 8-bit value, then
std::vector<double> result(reinterpret_cast<double*>(ch), reinterpet_cast<double*>(ch) + 8);
will provide a std::vector containing 8 double values.
There is another possible method using a union:
union ByteToDouble
{
byte b[64];
double d[8];
} byteToDouble;
the 8 double values will occupie the same memory as the 64 byte values. So you can write the byte values to byteToDouble.b[] and read the resultingdouble values from byteToDouble.d[].

Is it possible to store 8 characters (1-byte each) in a variable of type double (8-bytes) in c/c++?

I am migrating some legacy Fortran77 code to C/C++. In the Fortran77 code, if 8 characters are read in from a file, they can be stored in a variable of type real*8 without a problem.
Is it possible to do a similar thing in C or C++? If so, how would I do it? I haven't been able to find any solutions on the internet. I need to read in 8 characters using C/C++ and store them in a variable of type double, which is then passed back to Fortran and corresponds to the original real*8 variable.
Many thanks in advance for any help.
EDIT:
In response to #sixlettervariables, I'll just clarify my use-case a bit more. The issue I have with his suggestion is that I only know the format of each line (i.e. which fields are strings, which numbers) at runtime, and hence I can't know what members the struct should have statically. The fields also need to occupy a contiguous block of memory in the order they are read in.
Concretely, in one run of the program the format of each line might be: f1:string, f2:number, f3:number, f4:string, but in another f1:string, f2:string, f3:string, f4:number, f5:number. For the first case I'd need:
struct { char[8] f1; double f2; double f3; char[8] f4}
For the second I'd need:
struct { char[8] f1; char[8] f2; char[8] f3; double f4; double f5}
Perhaps there is some way to do this with templates?

You do not need to store them in a double just because Fortran needed to do that. In fact, you absolutely should not do that in your C/C++ code.
Simply store the character data in a character array.
If you're mixing Fortran and C/C++, the two have no idea about one another outside of their ABI. From the C side you can simply claim that the Fortran interface takes a character array, when in fact it is expecting an array of doubles. And the same is true from the Fortran side.
From the C side:
extern void FCHARS(char* str, int length);
/* ... */
int flength = 0; /* optional: length of the string in Fortran terms */
char str[9]; /* in C/C++ add one for \0 at the end */
/* read in up to a block of 8 */
fgets(str, sizeof(str), fp);
/* At this point if you know the 8 characters are space padded to their end
* you have no more work to do, otherwise you may need to fill from the right
* with spaces.
*/
{
size_t ii = sizeof(str) - 1;
while (ii > 0 && str[ii - 1] == '\0') {
str[ii - 1] = ' ';
flength = ii--; /* optional: keep track of our unpadded length */
}
}
/* Once you've space padded the string you can call your Fortran method.
* If your Fortran method accepts a string of unknown length, supply
* `flength`. If your Fortran method expects a string of a fixed length, pass
* the size of `str` (excluding '\0') instead.
*/
FCHARS(str, flength);
As long as you follow the ABI requirements of your Fortran compiler (e.g. CDECL, hidden string lengths passed interleaved) from the C/C++ code, you'll be fine.

Sure, just use a cast. You might want to add a static_assert for safety:
double d;
char * b = (char*)&d;
static_assert(sizeof(d) == sizeof(char[8]), "Double must be large enough to hold 8 chars");

union Data{
char c[8];
double d;
};
Save the 8 characters into c, and read it by d. Example:
#include <stdio.h>
// #include <stdlib.h>
union Data{
char c[8];
double d;
};
int main(){
int i;
union Data data;
for(i = 0; i < 8; i++)
scanf("%hhd", data.c + i);
printf("%e\n", data.d);
// system("pause");
return 0;
}

method 1
double a;
char ch;
int i;
for(i=0;i<8;i++)
{
scanf("%c", &ch);
a=a+ch;
if(i!=8)
a<<8; //left shifts by 8 bits, to accomodate another 8 bits on right.
}
method 2
double a
char *ch;
int i;
ch=&a;
for(i=0;i<8;i++)
{
scanf("%c", ch);
ch++;
}

Even if it would be possible in C to store several chars and real numbers in a DOUBLE variable because C DOUBLE is equal to REAL*8 because both are double precision float numbers I would strongly advise against it.
This "trick" to store several chars and reals in REAL*8 feels like a hack to save space. I don't think this is needed nowadays and I do not think this "trick" would generate faster code. I would advise to use a UNION as mentioned above.
You could post the FORTAN code which reads the chars and reals into the REAL*8 this would make it easier to help you. And I must say I am intrigued to find out how this is done.