Is it possible to do a bitwise assignment in C? (Assigning the bits of a variable to another, assuming for simplicity that the source and the target of assignment have the same number of bits.)
For example, assign the number int 1 (which has bits 0...01) to a float variable, obtaining not the float number 1.0f but the number (assuming IEEE-754 representation and assuming a float is 4 bytes as the int) with bits:
0 (sign) 0000'0000 (exponent) 0...01 (mantissa)
which would be a subnormal number (cause the exponent bits are all 0's and the mantissa is not zero), hence representing the number
+ 2^-126 2^-23 (assuming mantissa has 23 bits, then 0..(23 zeroes in total)..1 is 2^-23), that is 2^-149, that is approx. 1.4 10^-45.
NOTE: I'm in the process of learning. I am not trying to do this in a real-life scenario.
Given two objects a and b that are known to be the same size, you can copy the bits of b into a with memcpy(&a, &b, sizeof a);.
You could use a union for that:
int source;
float target;
union Data {
int i;
float f;
} data;
source = 42;
data.i = source;
target = data.f; // target should now have the bitwise equivalent of 42.
Be mindful about the sizes of the union members. If they are not equal I think they are padded to the right, but to be sure check with the documentation.
You can access the bits of other types via pointers (including memcpy, which takes them as arguments) or via unions. There is already another answer about the former, so I'll focus on the union approach.
Union members share the same memory, so you could use bit fields or integer types to access the individual bits, and then view the same bits by using a member of another type. However, note that both accessing the value of another type via a union and bit fields themselves are implementation defined, so this is inherently non-portable. In particular it is not specified how the bits end up being aligned in relation to other union members…
An example for the case of floats:
#include <stdio.h>
union float_bits {
float value;
struct {
unsigned mantissa:23;
unsigned exponent:8;
unsigned sign:1;
};
struct {
unsigned bits:32;
};
};
static void print_float_bits(union float_bits f) {
printf("(%c %02x %06x) (%08x) %f\n", f.sign ? '-' : '+', (unsigned) f.exponent, (unsigned) f.mantissa, (unsigned) f.bits, f.value);
}
int main(void) {
union float_bits f;
f.value = 1;
print_float_bits(f);
f.sign = 1;
print_float_bits(f);
// Largest normal number
f.sign = 0; f.exponent = 0xFE; f.mantissa = 0x7FFFFF;
print_float_bits(f);
// Infinity
f.exponent = 0xFF; f.mantissa = 0;
print_float_bits(f);
return 0;
}
On my x86-64 machine with 32-bit IEEE-754 floats, compiled with clang, outputs:
(+ 7f 000000) (3f800000) 1.000000
(- 7f 000000) (bf800000) -1.000000
(+ fe 7fffff) (7f7fffff) 340282346638528859811704183484516925440.000000
(+ ff 000000) (7f800000) inf
Disclaimer: Very much implementation defined behaviour, non-portable and dangerous. Bitfields used for readability of the example. Other alternatives would be to put an array of char or some integer type like uint32_t in the union instead of bitfields, but it's still very much implementation defined behaviour.
Related
I am asked to convert a float number into a 32 bit unsigned integer. I then have to check if all the bits are zero but I am having trouble with this. Sorry I am new to C
This is what I am doing
float number = 12.5;
// copying number into a 32-bit unsigned int
unsigned int val_number = *((unsigned int*) &number);
At this point I'm very confused on how to check if all bits are zero.
I think I need to loop through all the bits but I don't know how to do that.
To copy the bytes of a 32-bit float to an integer, best to copy to an integer type that is certainly 32-bit. unsigned may be less, same or more than 32-bits.
#include <inttypes.h>
float number = 12.5;
uint32_t val_number32; // 32-bit type
memcpy(&val_number32, &number, sizeof val_number32);
Avoid the cast and assign. It leads to aliasing problems with modern compilers #Andrew.
"... need cast the addresses of a and b to type (unsigned int *) and then dereference the addresses" reflects a risky programing technique.
To test if the bits of the unsigned integer are all zero, simply test with the constant 0.
int bit_all_zero = (val_number32 == 0);
An alternative is to use a union to access the bytes from 2 different encodings.
union {
float val_f;
uint32_t val_u;
} x = { .val_f = 12.5f };
int bit_all_zero = (x.val_u == 0);
Checking if all the bits are zero is equivalent to checking if the number is zero.
So it would be int is_zero = (val_number == 0);
Background:
I am playing around with bit-level coding (this is not homework - just curious). I found a lot of good material online and in a book called Hacker's Delight, but I am having trouble with one of the online problems.
It asks to convert an integer to a float. I used the following links as reference to work through the problem:
How to manually (bitwise) perform (float)x?
How to convert an unsigned int to a float?
http://locklessinc.com/articles/i2f/
Problem and Question:
I thought I understood the process well enough (I tried to document the process in the comments), but when I test it, I don't understand the output.
Test Cases:
float_i2f(2) returns 1073741824
float_i2f(3) returns 1077936128
I expected to see something like 2.0000 and 3.0000.
Did I mess up the conversion somewhere? I thought maybe this was a memory address, so I was thinking maybe I missed something in the conversion step needed to access the actual number? Or maybe I am printing it incorrectly? I am printing my output like this:
printf("Float_i2f ( %d ): ", 3);
printf("%u", float_i2f(3));
printf("\n");
But I thought that printing method was fine for unsigned values in C (I'm used to programming in Java).
Thanks for any advice.
Code:
/*
* float_i2f - Return bit-level equivalent of expression (float) x
* Result is returned as unsigned int, but
* it is to be interpreted as the bit-level representation of a
* single-precision floating point values.
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
unsigned float_i2f(int x) {
if (x == 0){
return 0;
}
//save the sign bit for later and get the asolute value of x
//the absolute value is needed to shift bits to put them
//into the appropriate position for the float
unsigned int signBit = 0;
unsigned int absVal = (unsigned int)x;
if (x < 0){
signBit = 0x80000000;
absVal = (unsigned int)-x;
}
//Calculate the exponent
// Shift the input left until the high order bit is set to form the mantissa.
// Form the floating exponent by subtracting the number of shifts from 158.
unsigned int exponent = 158; //158 possibly because of place in byte range
while ((absVal & 0x80000000) == 0){//this checks for 0 or 1. when it reaches 1, the loop breaks
exponent--;
absVal <<= 1;
}
//find the mantissa (bit shift to the right)
unsigned int mantissa = absVal >> 8;
//place the exponent bits in the right place
exponent = exponent << 23;
//get the mantissa
mantissa = mantissa & 0x7fffff;
//return the reconstructed float
return signBit | exponent | mantissa;
}
Continuing from the comment. Your code is correct, and you are simply looking at the equivalent unsigned integer made up by the bits in your IEEE-754 single-precision floating point number. The IEEE-754 single-precision number format (made up of the sign, extended exponent, and mantissa), can be interpreted as a float, or those same bits can be interpreted as an unsigned integer (just the number that is made up by the 32-bits). You are outputting the unsigned equivalent for the floating point number.
You can confirm with a simple union. For example:
#include <stdio.h>
#include <stdint.h>
typedef union {
uint32_t u;
float f;
} u2f;
int main (void) {
u2f tmp = { .f = 2.0 };
printf ("\n u : %u\n f : %f\n", tmp.u, tmp.f);
return 0;
}
Example Usage/Output
$ ./bin/unionuf
u : 1073741824
f : 2.000000
Let me know if you have any further questions. It's good to see that your study resulted in the correct floating point conversion. (also note the second comment regarding truncation/rounding)
I'll just chime in here, because nothing specifically about endianness has been addressed. So let's talk about it.
The construction of the value in the original question was endianness-agnostic, using shifts and other bitwise operations. This means that regardless of whether your system is big- or little-endian, the actual value will be the same. The difference will be its byte order in memory.
The generally accepted convention for IEEE-754 is that the byte order is big-endian (although I believe there is no formal specification of this, and therefore no requirement on implementations to follow it). This means if you want to directly interpret your integer value as a float, it needs to be laid out in big-endian byte order.
So, you can use this approach combined with a union if and only if you know that the endianness of floats and integers on your system is the same.
On the common Intel-based architectures this is not okay. On those architectures, integers are little-endian and floats are big-endian. You need to convert your value to big-endian. A simple approach to this is to repack its bytes even if they are already big-endian:
uint32_t n = float_i2f( input_val );
uint8_t char bytes[4] = {
(uint8_t)((n >> 24) & 0xff),
(uint8_t)((n >> 16) & 0xff),
(uint8_t)((n >> 8) & 0xff),
(uint8_t)(n & 0xff)
};
float fval;
memcpy( &fval, bytes, sizeof(float) );
I'll stress that you only need to worry about this if you are trying to reinterpret your integer representation as a float or the other way round.
If you're only trying to output what the representation is in bits, then you don't need to worry. You can just display your integer in a useful form such as hex:
printf( "0x%08x\n", n );
I am implementing four basic arithmetic functions(add, sub, division, multiplication) in C.
the basic structure of these functions I imagined is
the program gets two operands by user using scanf,
and the program split these values into bytes and compute!
I've completed addition and subtraction,
but I forgot that I shouldn't use arithmetic functions,
so when splitting integer into single bytes,
I wrote codes like
while(quotient!=0){
bin[i]=quotient%2;
quotient=quotient/2;
i++;
}
but since there is arithmetic functions that i shouldn't use..
so i have to rewrite that splitting parts,
but i really have no idea how can i split integer into single byte without using
% or /.
To access the bytes of a variable type punning can be used.
According to the Standard C (C99 and C11), only unsigned char brings certainty to perform this operation in a safe way.
This could be done in the following way:
typedef unsigned int myint_t;
myint_t x = 1234;
union {
myint_t val;
unsigned char byte[sizeof(myint_t)];
} u;
Now, you can of course access to the bytes of x in this way:
u.val = x;
for (int j = 0; j < sizeof(myint_t); j++)
printf("%d ",u.byte[j]);
However, as WhozCrag has pointed out, there are issues with endianness.
It cannot be assumed that the bytes are in determined order.
So, before doing any computation with bytes, your program needs to check how the endianness works.
#include <limits.h> /* To use UCHAR_MAX */
unsigned long int ByteFactor = 1u + UCHAR_MAX; /* 256 almost everywhere */
u.val = 0;
for (int j = sizeof(myint_t) - 1; j >= 0 ; j--)
u.val = u.val * ByteFactor + j;
Now, when you print the values of u.byte[], you will see the order in that bytes are arranged for the type myint_t.
The less significant byte will have value 0.
I assume 32 bit integers (if not the case then just change the sizes) there are more approaches:
BYTE pointer
#include<stdio.h>
int x; // your integer or whatever else data type
BYTE *p=(BYTE*)&x;
x=0x11223344;
printf("%x\n",p[0]);
printf("%x\n",p[1]);
printf("%x\n",p[2]);
printf("%x\n",p[3]);
just get the address of your data as BYTE pointer
and access the bytes directly via 1D array
union
#include<stdio.h>
union
{
int x; // your integer or whatever else data type
BYTE p[4];
} a;
a.x=0x11223344;
printf("%x\n",a.p[0]);
printf("%x\n",a.p[1]);
printf("%x\n",a.p[2]);
printf("%x\n",a.p[3]);
and access the bytes directly via 1D array
[notes]
if you do not have BYTE defined then change it for unsigned char
with ALU you can use not only %,/ but also >>,& which is way faster but still use arithmetics
now depending on the platform endianness the output can be 11,22,33,44 of 44,33,22,11 so you need to take that in mind (especially for code used in multiple platforms)
you need to handle sign of number, for unsigned integers there is no problem
but for signed the C uses 2'os complement so it is better to separate the sign before spliting like:
int s;
if (x<0) { s=-1; x=-x; } else s=+1;
// now split ...
[edit2] logical/bit operations
x<<n,x>>n - is bit shift left and right of x by n bits
x&y - is bitwise logical and (perform logical AND on each bit separately)
so when you have for example 32 bit unsigned int (called DWORD) yu can split it to BYTES like this:
DWORD x; // input 32 bit unsigned int
BYTE a0,a1,a2,a3; // output BYTES a0 is the least significant a3 is the most significant
x=0x11223344;
a0=DWORD((x )&255); // should be 0x44
a1=DWORD((x>> 8)&255); // should be 0x33
a2=DWORD((x>>16)&255); // should be 0x22
a3=DWORD((x>>24)&255); // should be 0x11
this approach is not affected by endianness
but it uses ALU
the point is shift the bits you want to position of 0..7 bit and mask out the rest
the &255 and DWORD() overtyping is not needed on all compilers but some do weird stuff without them especially on signed variables like char or int
x>>n is the same as x/(pow(2,n))=x/(1<<n)
x&((1<<n)-1) is the same as x%(pow(2,n))=x%(1<<n)
so (x>>8)=x/256 and (x&255)=x%256
I am using the HIDAPI to send some data to a USB device. This data can be sent only as byte array and I need to send some float numbers inside this data array. I know floats have 4 bytes. So I thought this might work:
float f = 0.6;
char data[4];
data[0] = (int) f >> 24;
data[1] = (int) f >> 16;
data[2] = (int) f >> 8;
data[3] = (int) f;
And later all I had to do is:
g = (float)((data[0] << 24) | (data[1] << 16) | (data[2] << 8) | (data[3]) );
But testing this shows me that the lines like data[0] = (int) f >> 24; returns always 0. What is wrong with my code and how may I do this correctly (i.e. break a float inner data in 4 char bytes and rebuild the same float later)?
EDIT:
I was able to accomplish this with the following codes:
float f = 0.1;
unsigned char *pc;
pc = (unsigned char*)&f;
// 0.6 in float
pc[0] = 0x9A;
pc[1] = 0x99;
pc[2] = 0x19;
pc[3] = 0x3F;
std::cout << f << std::endl; // will print 0.6
and
*(unsigned int*)&f = (0x3F << 24) | (0x19 << 16) | (0x99 << 8) | (0x9A << 0);
I know memcpy() is a "cleaner" way of doing it, but this way I think the performance is somewhat better.
You can do it like this:
char data[sizeof(float)];
float f = 0.6f;
memcpy(data, &f, sizeof f); // send data
float g;
memcpy(&g, data, sizeof g); // receive data
In order for this to work, both machines need to use the same floating point representations.
As was rightly pointed out in the comments, you don't necessarily need to do the extra memcpy; instead, you can treat f directly as an array of characters (of any signedness). You still have to do memcpy on the receiving side, though, since you may not treat an arbitrary array of characters as a float! Example:
unsigned char const * const p = (unsigned char const *)&f;
for (size_t i = 0; i != sizeof f; ++i)
{
printf("Byte %zu is %02X\n", i, p[i]);
send_over_network(p[i]);
}
In standard C is guaranted that any type can be accessed as an array of bytes.
A straight way to do this is, of course, by using unions:
#include <stdio.h>
int main(void)
{
float x = 0x1.0p-3; /* 2^(-3) in hexa */
union float_bytes {
float val;
unsigned char bytes[sizeof(float)];
} data;
data.val = x;
for (int i = 0; i < sizeof(float); i++)
printf("Byte %d: %.2x\n", i, data.bytes[i]);
data.val *= 2; /* Doing something with the float value */
x = data.val; /* Retrieving the float value */
printf("%.4f\n", data.val);
getchar();
}
As you can see, it is not necessary at all to use memcpy or pointers...
The union approach is easy to understand, standard and fast.
EDIT.
I will explain why this approach is valid in C (C99).
[5.2.4.2.1(1)] A byte has CHAR_BIT bits (an integer constant >= 8, in almost cases is 8).
[6.2.6.1(3)] The unsigned char type uses all its bits to represent the value of the object, which is an nonnegative integer, in a pure binary representation. This means that there are not padding bits or bits used for any other extrange purpouse. (The same thing is not guaranted for signed char or char types).
[6.2.6.1(2)] Every non-bitfield type is represented in memory as a contiguous sequence of bytes.
[6.2.6.1(4)] (Cited) "Values stored in non-bit-field objects of any other object type consist of n × CHAR_BIT bits, where n is the size of an object of that type, in bytes. The value may be copied into an object of type unsigned char [n] (e.g., by memcpy); [...]"
[6.7.2.1(14)] A pointer to a structure object (in particular, unions), suitably converted, points to its initial member. (Thus, there is no padding bytes at the beginning of a union).
[6.5(7)] The content of an object can be accessed by a character type:
An object shall have its stored value accessed only by an lvalue expression that has one of
the following types:
— a type compatible with the effective type of the object,
— a qualified version of a type compatible with the effective type of the object,
— a type that is the signed or unsigned type corresponding to the effective type of the
object,
— a type that is the signed or unsigned type corresponding to a qualified version of the
effective type of the object,
— an aggregate or union type that includes one of the aforementioned types among its
members (including, recursively,amember of a subaggregate or contained union), or
— a character type
More information:
A discussion in google groups
Type-punning
EDIT 2
Another detail of the standard C99:
[6.5.2.3(3) footnote 82] Type-punning is allowed:
If the member used to access the contents of a union object is not the same as the member last used to
store a value in the object, the appropriate part of the object representation of the value is reinterpreted
as an object representation in the new type as described in 6.2.6 (a process sometimes called "type
punning"). This might be a trap representation.
The C language guarantees that any value of any type¹ can be accessed as an array of bytes. The type of bytes is unsigned char. Here's a low-level way of copying a float to an array of bytes. sizeof(f) is the number of bytes used to store the value of the variable f; you can also use sizeof(float) (you can either pass sizeof a variable or more complex expression, or its type).
float f = 0.6;
unsigned char data[sizeof(float)];
size_t i;
for (i = 0; i < sizeof(float); i++) {
data[i] = (unsigned char*)f + i;
}
The functions memcpy or memmove do exactly that (or an optimized version thereof).
float f = 0.6;
unsigned char data[sizeof(float)];
memcpy(data, f, sizeof(f));
You don't even need to make this copy, though. You can directly pass a pointer to the float to your write-to-USB function, and tell it how many bytes to copy (sizeof(f)). You'll need an explicit cast if the function takes a pointer argument other than void*.
int write_to_usb(unsigned char *ptr, size_t size);
result = write_to_usb((unsigned char*)f, sizeof(f))
Note that this will work only if the device uses the same representation of floating point numbers, which is common but not universal. Most machines use the IEEE floating point formats, but you may need to switch endianness.
As for what is wrong with your attempt: the >> operator operates on integers. In the expression (int) f >> 24, f is cast to an int; if you'd written f >> 24 without the cast, f would still be automatically converted to an int. Converting a floating point value to an integer approximates it by truncating or rounding it (usually towards 0, but the rule depends on the platform). 0.6 rounded to an integer is 0 or 1, so data[0] is 0 or 1 and the others are all 0.
You need to act on the bytes of the float object, not on its value.
¹ Excluding functions which can't really be manipulated in C, but including function pointers which functions decay to automatically.
Assuming that both devices have the same notion of how floats are represented then why not just do a memcpy. i.e
unsigned char payload[4];
memcpy(payload, &f, 4);
the safest way to do this, if you control both sides is to send some sort of standardized representation... this isn't the most efficient, but it isn't too bad for small numbers.
hostPort writes char * "34.56\0" byte by byte
client reads char * "34.56\0"
then converts to float with library function atof or atof_l.
of course that isn't the most optimized, but it sure will be easy to debug.
if you wanted to get more optimized and creative, first byte is length then the exponent, then each byte represents 2 decimal places... so
34.56 becomes char array[] = {4,-2,34,56}; something like that would be portable... I would just try not to pass binary float representations around... because it can get messy fast.
It might be safer to union the float and char array. Put in the float member, pull out the 4 (or whatever the length is) bytes.
I created a structure to represent a fixed-point positive number. I want the numbers in both sides of the decimal point to consist 2 bytes.
typedef struct Fixed_t {
unsigned short floor; //left side of the decimal point
unsigned short fraction; //right side of the decimal point
} Fixed;
Now I want to add two fixed point numbers, Fixed x and Fixed y. To do so I treat them like integers and add.
(Fixed) ( (int)x + (int)y );
But as my visual studio 2010 compiler says, I cannot convert between Fixed and int.
What's the right way to do this?
EDIT: I'm not committed to the {short floor, short fraction} implementation of Fixed.
You could attempt a nasty hack, but there's a problem here with endian-ness. Whatever you do to convert, how is the compiler supposed to know that you want floor to be the most significant part of the result, and fraction the less significant part? Any solution that relies on re-interpreting memory is going to work for one endian-ness but not another.
You should either:
(1) define the conversion explicitly. Assuming short is 16 bits:
unsigned int val = (x.floor << 16) + x.fraction;
(2) change Fixed so that it has an int member instead of two shorts, and then decompose when required, rather than composing when required.
If you want addition to be fast, then (2) is the thing to do. If you have a 64 bit type, then you can also do multiplication without decomposing: unsigned int result = (((uint64_t)x) * y) >> 16.
The nasty hack, by the way, would be this:
unsigned int val;
assert(sizeof(Fixed) == sizeof(unsigned int)) // could be a static test
assert(2 * sizeof(unsigned short) == sizeof(unsigned int)) // could be a static test
memcpy(&val, &x, sizeof(unsigned int));
That would work on a big-endian system, where Fixed has no padding (and the integer types have no padding bits). On a little-endian system you'd need the members of Fixed to be in the other order, which is why it's nasty. Sometimes casting through memcpy is the right thing to do (in which case it's a "trick" rather than a "nasty hack"). This just isn't one of those times.
If you have to you can use a union but beware of endian issues. You might find the arithmetic doesn't work and certainly is not portable.
typedef struct Fixed_t {
union {
struct { unsigned short floor; unsigned short fraction };
unsigned int whole;
};
} Fixed;
which is more likely (I think) to work big-endian (which Windows/Intel isn't).
Some magic:
typedef union Fixed {
uint16_t w[2];
uint32_t d;
} Fixed;
#define Floor w[((Fixed){1}).d==1]
#define Fraction w[((Fixed){1}).d!=1]
Key points:
I use fixed-size integer types so you're not depending on short being 16-bit and int being 32-bit.
The macros for Floor and Fraction (capitalized to avoid clashing with floor() function) access the two parts in an endian-independent way, as foo.Floor and foo.Fraction.
Edit: At OP's request, an explanation of the macros:
Unions are a way of declaring an object consisting of several different overlapping types. Here we have uint16_t w[2]; overlapping uint32_t d;, making it possible to access the value as 2 16-bit units or 1 32-bit unit.
(Fixed){1} is a compound literal, and could be written more verbosely as (Fixed){{1,0}}. Its first element (uint16_t w[2];) gets initialized with {1,0}. The expression ((Fixed){1}).d then evaluates to the 32-bit integer whose first 16-bit half is 1 and whose second 16-bit half is 0. On a little-endian system, this value is 1, so ((Fixed){1}).d==1 evaluates to 1 (true) and ((Fixed){1}).d!=1 evaluates to 0 (false). On a big-endian system, it'll be the other way around.
Thus, on a little-endian system, Floor is w[1] and Fraction is w[0]. On a big-endian system, Floor is w[0] and Fraction is w[1]. Either way, you end up storing/accessing the correct half of the 32-bit value for the endian-ness of your platform.
In theory, a hypothetical system could use a completely different representation for 16-bit and 32-bit values (for instance interleaving the bits of the two halves), breaking these macros. In practice, that's not going to happen. :-)
This is not possible portably, as the compiler does not guarantee a Fixed will use the same amount of space as an int. The right way is to define a function Fixed add(Fixed a, Fixed b).
Just add the pieces separately. You need to know the value of the fraction that means "1" - here I'm calling that FRAC_MAX:
// c = a + b
void fixed_add( Fixed* a, Fixed* b, Fixed* c){
unsigned short carry = 0;
if((int)(a->floor) + (int)(b->floor) > FRAC_MAX){
carry = 1;
c->fraction = a->floor + b->floor - FRAC_MAX;
}
c->floor = a->floor + b->floor + carry;
}
Alternatively, if you're just setting the fixed point as being at the 2 byte boundary you can do something like:
void fixed_add( Fixed* a, Fixed *b, Fixed *c){
int ia = a->floor << 16 + a->fraction;
int ib = b->floor << 16 + b->fraction;
int ic = ia + ib;
c->floor = ic >> 16;
c->fraction = ic - c->floor;
}
Try this:
typedef union {
struct Fixed_t {
unsigned short floor; //left side of the decimal point
unsigned short fraction; //right side of the decimal point
} Fixed;
int Fixed_int;
}
If your compiler puts the two short on 4 bytes, then you can use memcpy to copy your int in your struct, but as said in another answer, this is not portable... and quite ugly.
Do you really care adding separately each field in a separate method?
Do you want to keep the integer for performance reason?
// add two Fixed
Fixed operator+( Fixed a, Fixed b )
{
...
}
//add Fixed and int
Fixed operator+( Fixed a, int b )
{
...
}
You may cast any addressable type to another one by using:
*(newtype *)&var