I am writing a code that has to compile in single and double precision. The original version was only in double precision, but I am trying to enable single precision now by using templates.
My question is: is it necessary to cast the 1. and the 7.8 to the specified type with static_cast<TF>(1.) for instance, or will the compiler take care of it? I find the casting not remarkably pretty and prefer to stay away from it. (I have other functions that are much longer and that contain many more literal constants).
template<typename TF>
inline TF phih_stable(const TF zeta)
{
// Hogstrom, 1988
return 1. + 7.8*zeta;
}
Casting and implicit conversions are two things. For this example, you could treat the template function as if it were two overloaded functions, but with the same code within. At the interface level (parameters, return value) the compiler will generate implicit conversions.
Now, the question you will have to ask yourself is this: Do those implicit conversions do what I want? If they do, just leave it as it is. If they don't, you could try to add explicit conversions (maybe use the function-style casting like TF(1.)) or, you could specialize this function for double and float.
Another option, less general but maybe it works here, is that you switch the code around, i.e. that you write the code for single-precision float and then let the compiler apply its implicit conversions. Since the conversions usually only go to the bigger type, it should fit for both double and float without incurring any overhead for float.
When you do:
return 1. + 7.8*zeta;
The literals 1. and 7.8 are double, so when if zeta is a float, it will first be converted to double, then the whole computation will be done in double-precision, and the result will be cast back to float, this is equivalent to:
return (float)(1. + 7.8 * (double)zeta);
Otherwise said, this is equivalent to calling phih_stable(double) and storing the result in a float, so your template would be useless for float.
If you want the computation to be made in single precision, you need the casts1:
return TF(1.) + TF(7.8) * zeta;
What about using 1.f and 7.8f? The problem is that (double)7.8f != 7.8 due to floating point precision. The difference is around 1e-7, the actual stored value for 7.8f (assuming 32-bits float) is:
7.80000019073486328125
While the actual stored value for 7.8 (assuming 64-bits double) is:
7.79999999999999982236431605997
So you have to ask yourself if you accept this loss of precision.
You can compare the two following implementations:
template <class T>
constexpr T phih_stable_cast(T t) {
return T(1l) + T(7.8l) * t;
}
template <class T>
constexpr T phih_stable_float(T t) {
return 1.f + 7.8f * t;
}
And the following assertions:
static_assert(phih_stable_cast(3.4f) == 1. + 7.8f * 3.4f, "");
static_assert(phih_stable_cast(3.4) == 1. + 7.8 * 3.4, "");
static_assert(phih_stable_cast(3.4l) == 1. + 7.8l * 3.4l, "");
static_assert(phih_stable_float(3.4f) == 1.f + 7.8f * 3.4f, "");
static_assert(phih_stable_float(3.4) == 1. + 7.8 * 3.4, "");
static_assert(phih_stable_float(3.4l) == 1.l + 7.8l * 3.4l, "");
The two last assertions fail due to the loss of precision when doing the computation.
1 You should even downcast from long double to not lose precision when using your function with long double: return TF(1.l) + TF(7.8l) * zeta;.
This is not a duplicate of the famous Is floating point math broken, even if it looks like one at first sight.
I'm reading a double from a text file using fscanf(file, "%lf", &value); and comparing it with the == operator against a double literal. If the string is the same as the literal, will the comparision using == be true in all cases?
Example
Text file content:
7.7
Code snippet:
double value;
fscanf(file, "%lf", &value); // reading "7.7" from file into value
if (value == 7.7)
printf("strictly equal\n");
The expected and actual output is
strictly equal
But this supposes that the compiler converts the double literal 7.7 into a double exactly the same way as does the fscanf function, but the compiler may or may not use the same library for converting strings to double.
Or asked otherwise: does the conversion from string to double result in a unique binary representation or may there be slight implementation dependent differences?
Live demonstration
From the c++ standard:
[lex.fcon]
... If the scaled value is in the range
of representable values for its type, the result is the scaled value if representable, else the larger or smaller
representable value nearest the scaled value, chosen in an implementation-defined manner...
emphasis mine.
So you can only rely on equality if the value is strictly representable by a double.
About C++, from cppreference one can read:
[lex.fcon] (§6.4.4.2)
The result of evaluating a floating constant is either the nearest representable value or the larger or smaller representable value immediately adjacent to the nearest representable value, chosen in an implementation-defined manner (in other words, default rounding direction during translation is implementation-defined).
Since the representation of a floating literal is unspecified, I guess you cannot conclude about its comparison with a scanf result.
About C11 (standard ISO/IEC 9899:2011):
[lex.fcon] (§6.4.4.2)
Recommended practice
7 The translation-time conversion of floating constants should match the execution-time conversion of character strings by library functions, such as strtod, given matching inputs suitable for both conversions, the same result format, and default execution-time
rounding.
So clearly for C11, this is not guaranteed to match.
If the string is the same as the literal, will the comparison using == be true in all cases?
A common consideration not yet explored: FLT_EVAL_METHOD
#include <float.h>
...
printf("%d\n", FLT_EVAL_METHOD);
2 evaluate all operations and constants to the range and precision of the
long double type.
If this returns 2, then the math used in value == 7.7 is long double and 7.7 treated as 7.7L. In OP's case, this may evaluate to false.
To account for this wider precision, assign values which will removes all extra range and precision.
scanf(file, "%lf", &value);
double seven_seven = 7.7;
if (value == seven_seven)
printf("strictly equal\n");
IMO, this is a more likely occurring problem than variant rounding modes or variations in library/compiler conversions.
Note that this case is akin to the below, a well known issue.
float value;
fscanf(file, "%f", &value);
if (value == 7.7)
printf("strictly equal\n");
Demonstration
#include <stdio.h>
#include <float.h>
int main() {
printf("%d\n", FLT_EVAL_METHOD);
double value;
sscanf("7.7", "%lf", &value);
double seven_seven = 7.7;
if (value == seven_seven) {
printf("value == seven_seven\n");
} else {
printf("value != seven_seven\n");
}
if (value == 7.7) {
printf("value == 7.7\n");
} else {
printf("value != 7.7\n");
}
return 0;
}
Output
2
value == seven_seven
value != 7.7
Alternative Compare
To compare 2 double that are "near" each other, we need a definition of "near". A useful approach is to consider all the finite double values sorted into a ascending sequence and then compare their sequence numbers from each other. double_distance(x, nextafter(x, 2*x) --> 1
Following code makes various assumptions about double layout and size.
#include <assert.h>
unsigned long long double_order(double x) {
union {
double d;
unsigned long long ull;
} u;
assert(sizeof(double) == sizeof(unsigned long long));
u.d = x;
if (u.ull & 0x8000000000000000) {
u.ull ^= 0x8000000000000000;
return 0x8000000000000000 - u.ull;
}
return u.ull + 0x8000000000000000;
}
unsigned long long double_distance(double x, double y) {
unsigned long long ullx = double_order(x);
unsigned long long ully = double_order(y);
if (x > y) return ullx - ully;
return ully - ullx;
}
....
printf("%llu\n", double_distance(value, 7.7)); // 0
printf("%llu\n", double_distance(value, nextafter(value,value*2))); // 1
printf("%llu\n", double_distance(value, nextafter(value,value/2))); // 1
Or just use
if (nextafter(7.7, -INF) <= value && value <= nextafter(7.7, +INF)) {
puts("Close enough");
}
There's no guarantee.
You can hope that the compiler uses a high quality algorithm for the conversion of literals, and that the standard library implementation uses a high quality conversion as well, and two high quality algorithms should agree quite often.
It's also possible that both use the exact same algorithm (for example, the compiler converts the literal by putting the characters into a char array and calling sscanf.
BTW. I had one bug caused by the fact that a compiler didn't convert the literal 999999999.5 exactly. Replaced it with 9999999995 / 10.0 and everything was fine.
I am writing a protocol, that uses RFC 7049 as its binary representation. The standard states, that the protocol may use 32-bit floating point representation of numbers, if their numeric value is equivalent to respective 64-bit numbers. The conversion must not lead to lose of precision.
What 32-bit float numbers can be bigger than 64-bit integer and numerically equivalent with them?
Is comparing float x; uint64_t y; (float)x == (float)y enough for ensuring, that the values are equivalent? Will this comparison ever be true?
RFC 7049 §3.6. Numbers
For the purposes of this specification, all number representations
for the same numeric value are equivalent. This means that an
encoder can encode a floating-point value of 0.0 as the integer 0.
It, however, also means that an application that expects to find
integer values only might find floating-point values if the encoder
decides these are desirable, such as when the floating-point value is
more compact than a 64-bit integer.
There certainly are numbers for which this is true:
2^33 can be perfectly represented as a floating point number, but clearly cannot be represented as a 32-bit integer. The following code should work as expected:
bool representable_as_float(int64_t value) {
float repr = value;
return repr >= -0x1.0p63 && repr < 0x1.0p63 && (int64_t)repr == value;
}
It is important to notice though that we are basically doing (int64_t)(float)value and not the other way around - we are interested if the cast to float loses any precision.
The check to see whether repr is smaller than the maximum value of int64_t is important since we could invoke undefined behavior otherwise, since the cast to float may round up to the next higher number (which could then be larger than the maximum value possible in int64_t). (Thanks to #tmyklebu for pointing this out).
Two samples:
// powers of 2 can easily be represented
assert(representable_as_float(((int64_t)1) << 33));
// Other numbers not so much:
assert(!representable_as_float(std::numeric_limits<int64_t>::max()));
The following is based on Julia's method for comparing floats and integers. This does not require access to 80-bit long doubles or floating point exceptions, and should work under any rounding mode. I believe this should work for any C float type (IEEE754 or not), and not cause any undefined behaviour.
UPDATE: technically this assumes a binary float format, and that the float exponent size is large enough to represent 264: this is certainly true for the standard IEEE754 binary32 (which you refer to in your question), but not, say, binary16.
#include <stdio.h>
#include <stdint.h>
int cmp_flt_uint64(float x,uint64_t y) {
return (x == (float)y) && (x != 0x1p64f) && ((uint64_t)x == y);
}
int main() {
float x = 0x1p64f;
uint64_t y = 0xffffffffffffffff;
if (cmp_flt_uint64(x,y))
printf("true\n");
else
printf("false\n");
;
}
The logic here is as follows:
The first equality can be true only if x is a non-negative integer in the interval [0,264].
The second checks that x (and hence (float)y) is not 264: if this is the case, then y cannot be represented exactly by a float, and so the comparison is false.
Any remaining values of x can be exactly converted to a uint64_t, and so we cast and compare.
No, you need to compare (long double)x == (long double)y on an architecture where the mantissa of a long double can hold 63 bits. This is because some big long long ints will lose precision when you convert them to float, and compare as equal to a non-equivalent float, but if you convert to long double, it will not lose precision on that architecture.
The following program demonstrates this behavior when compiled with gcc -std=c99 -mssse3 -mfpmath=sse on x86, because these settings use wide-enough long doubles but prevent the implicit use of higher-precision types in calculations:
#include <assert.h>
#include <stdint.h>
const int64_t x = (1ULL<<62) - 1ULL;
const float y = (float)(1ULL<<62);
// The mantissa is not wide enough to store
// 63 bits of precision.
int main(void)
{
assert ((float)x == (float)y);
assert ((long double)x != (long double)y);
return 0;
}
Edit: If you don’t have wide enough long doubles, the following might work:
feclearexcept(FE_ALL_EXCEPT);
x == y;
ftestexcept(FE_INEXACT);
I think, although I could be mistaken, that an implementation could round off x during the conversion in a way that loses precision.
Another strategy that could work is to compare
extern uint64_t x;
extern float y;
const float z = (float)x;
y == z && (uint64_t)z == x;
This should catch losses of precision due to round-off error, but it could conceivably cause undefined behavior if the conversion to z rounds up. It will work if the conversion is set to round toward zero when converting x to z.
I want to define a function that takes an unsigned int as argument and returns an int congruent modulo UINT_MAX+1 to the argument.
A first attempt might look like this:
int unsigned_to_signed(unsigned n)
{
return static_cast<int>(n);
}
But as any language lawyer knows, casting from unsigned to signed for values larger than INT_MAX is implementation-defined.
I want to implement this such that (a) it only relies on behavior mandated by the spec; and (b) it compiles into a no-op on any modern machine and optimizing compiler.
As for bizarre machines... If there is no signed int congruent modulo UINT_MAX+1 to the unsigned int, let's say I want to throw an exception. If there is more than one (I am not sure this is possible), let's say I want the largest one.
OK, second attempt:
int unsigned_to_signed(unsigned n)
{
int int_n = static_cast<int>(n);
if (n == static_cast<unsigned>(int_n))
return int_n;
// else do something long and complicated
}
I do not much care about the efficiency when I am not on a typical twos-complement system, since in my humble opinion that is unlikely. And if my code becomes a bottleneck on the omnipresent sign-magnitude systems of 2050, well, I bet someone can figure that out and optimize it then.
Now, this second attempt is pretty close to what I want. Although the cast to int is implementation-defined for some inputs, the cast back to unsigned is guaranteed by the standard to preserve the value modulo UINT_MAX+1. So the conditional does check exactly what I want, and it will compile into nothing on any system I am likely to encounter.
However... I am still casting to int without first checking whether it will invoke implementation-defined behavior. On some hypothetical system in 2050 it could do who-knows-what. So let's say I want to avoid that.
Question: What should my "third attempt" look like?
To recap, I want to:
Cast from unsigned int to signed int
Preserve the value mod UINT_MAX+1
Invoke only standard-mandated behavior
Compile into a no-op on a typical twos-complement machine with optimizing compiler
[Update]
Let me give an example to show why this is not a trivial question.
Consider a hypothetical C++ implementation with the following properties:
sizeof(int) equals 4
sizeof(unsigned) equals 4
INT_MAX equals 32767
INT_MIN equals -232 + 32768
UINT_MAX equals 232 - 1
Arithmetic on int is modulo 232 (into the range INT_MIN through INT_MAX)
std::numeric_limits<int>::is_modulo is true
Casting unsigned n to int preserves the value for 0 <= n <= 32767 and yields zero otherwise
On this hypothetical implementation, there is exactly one int value congruent (mod UINT_MAX+1) to each unsigned value. So my question would be well-defined.
I claim that this hypothetical C++ implementation fully conforms to the C++98, C++03, and C++11 specifications. I admit I have not memorized every word of all of them... But I believe I have read the relevant sections carefully. So if you want me to accept your answer, you either must (a) cite a spec that rules out this hypothetical implementation or (b) handle it correctly.
Indeed, a correct answer must handle every hypothetical implementation permitted by the standard. That is what "invoke only standard-mandated behavior" means, by definition.
Incidentally, note that std::numeric_limits<int>::is_modulo is utterly useless here for multiple reasons. For one thing, it can be true even if unsigned-to-signed casts do not work for large unsigned values. For another, it can be true even on one's-complement or sign-magnitude systems, if arithmetic is simply modulo the entire integer range. And so on. If your answer depends on is_modulo, it's wrong.
[Update 2]
hvd's answer taught me something: My hypothetical C++ implementation for integers is not permitted by modern C. The C99 and C11 standards are very specific about the representation of signed integers; indeed, they only permit twos-complement, ones-complement, and sign-magnitude (section 6.2.6.2 paragraph (2); ).
But C++ is not C. As it turns out, this fact lies at the very heart of my question.
The original C++98 standard was based on the much older C89, which says (section 3.1.2.5):
For each of the signed integer types, there is a corresponding (but
different) unsigned integer type (designated with the keyword
unsigned) that uses the same amount of storage (including sign
information) and has the same alignment requirements. The range of
nonnegative values of a signed integer type is a subrange of the
corresponding unsigned integer type, and the representation of the
same value in each type is the same.
C89 says nothing about only having one sign bit or only allowing twos-complement/ones-complement/sign-magnitude.
The C++98 standard adopted this language nearly verbatim (section 3.9.1 paragraph (3)):
For each of the signed integer types, there exists a corresponding
(but different) unsigned integer type: "unsigned char", "unsigned
short int", "unsigned int", and "unsigned long int", each of
which occupies the same amount of storage and has the same alignment
requirements (3.9) as the corresponding signed integer type ; that
is, each signed integer type has the same object representation as
its corresponding unsigned integer type. The range of nonnegative
values of a signed integer type is a subrange of the corresponding
unsigned integer type, and the value representation of each
corresponding signed/unsigned type shall be the same.
The C++03 standard uses essentially identical language, as does C++11.
No standard C++ spec constrains its signed integer representations to any C spec, as far as I can tell. And there is nothing mandating a single sign bit or anything of the kind. All it says is that non-negative signed integers must be a subrange of the corresponding unsigned.
So, again I claim that INT_MAX=32767 with INT_MIN=-232+32768 is permitted. If your answer assumes otherwise, it is incorrect unless you cite a C++ standard proving me wrong.
Expanding on user71404's answer:
int f(unsigned x)
{
if (x <= INT_MAX)
return static_cast<int>(x);
if (x >= INT_MIN)
return static_cast<int>(x - INT_MIN) + INT_MIN;
throw x; // Or whatever else you like
}
If x >= INT_MIN (keep the promotion rules in mind, INT_MIN gets converted to unsigned), then x - INT_MIN <= INT_MAX, so this won't have any overflow.
If that is not obvious, take a look at the claim "If x >= -4u, then x + 4 <= 3.", and keep in mind that INT_MAX will be equal to at least the mathematical value of -INT_MIN - 1.
On the most common systems, where !(x <= INT_MAX) implies x >= INT_MIN, the optimizer should be able (and on my system, is able) to remove the second check, determine that the two return statements can be compiled to the same code, and remove the first check too. Generated assembly listing:
__Z1fj:
LFB6:
.cfi_startproc
movl 4(%esp), %eax
ret
.cfi_endproc
The hypothetical implementation in your question:
INT_MAX equals 32767
INT_MIN equals -232 + 32768
is not possible, so does not need special consideration. INT_MIN will be equal to either -INT_MAX, or to -INT_MAX - 1. This follows from C's representation of integer types (6.2.6.2), which requires n bits to be value bits, one bit to be a sign bit, and only allows one single trap representation (not including representations that are invalid because of padding bits), namely the one that would otherwise represent negative zero / -INT_MAX - 1. C++ doesn't allow any integer representations beyond what C allows.
Update: Microsoft's compiler apparently does not notice that x > 10 and x >= 11 test the same thing. It only generates the desired code if x >= INT_MIN is replaced with x > INT_MIN - 1u, which it can detect as the negation of x <= INT_MAX (on this platform).
[Update from questioner (Nemo), elaborating on our discussion below]
I now believe this answer works in all cases, but for complicated reasons. I am likely to award the bounty to this solution, but I want to capture all the gory details in case anybody cares.
Let's start with C++11, section 18.3.3:
Table 31 describes the header <climits>.
...
The contents are the same as the Standard C library header <limits.h>.
Here, "Standard C" means C99, whose specification severely constrains the representation of signed integers. They are just like unsigned integers, but with one bit dedicated to "sign" and zero or more bits dedicated to "padding". The padding bits do not contribute to the value of the integer, and the sign bit contributes only as twos-complement, ones-complement, or sign-magnitude.
Since C++11 inherits the <climits> macros from C99, INT_MIN is either -INT_MAX or -INT_MAX-1, and hvd's code is guaranteed to work. (Note that, due to the padding, INT_MAX could be much less than UINT_MAX/2... But thanks to the way signed->unsigned casts work, this answer handles that fine.)
C++03/C++98 is trickier. It uses the same wording to inherit <climits> from "Standard C", but now "Standard C" means C89/C90.
All of these -- C++98, C++03, C89/C90 -- have the wording I give in my question, but also include this (C++03 section 3.9.1 paragraph 7):
The representations of integral types shall define values by use of a
pure binary numeration system.(44) [Example: this International
Standard permits 2’s complement, 1’s complement and signed magnitude
representations for integral types.]
Footnote (44) defines "pure binary numeration system":
A positional representation for integers that uses the binary digits 0
and 1, in which the values represented by successive bits are
additive, begin with 1, and are multiplied by successive integral
power of 2, except perhaps for the bit with the highest position.
What is interesting about this wording is that it contradicts itself, because the definition of "pure binary numeration system" does not permit a sign/magnitude representation! It does allow the high bit to have, say, the value -2n-1 (twos complement) or -(2n-1-1) (ones complement). But there is no value for the high bit that results in sign/magnitude.
Anyway, my "hypothetical implementation" does not qualify as "pure binary" under this definition, so it is ruled out.
However, the fact that the high bit is special means we can imagine it contributing any value at all: A small positive value, huge positive value, small negative value, or huge negative value. (If the sign bit can contribute -(2n-1-1), why not -(2n-1-2)? etc.)
So, let's imagine a signed integer representation that assigns a wacky value to the "sign" bit.
A small positive value for the sign bit would result in a positive range for int (possibly as large as unsigned), and hvd's code handles that just fine.
A huge positive value for the sign bit would result in int having a maximum larger than unsigned, which is is forbidden.
A huge negative value for the sign bit would result in int representing a non-contiguous range of values, and other wording in the spec rules that out.
Finally, how about a sign bit that contributes a small negative quantity? Could we have a 1 in the "sign bit" contribute, say, -37 to the value of the int? So then INT_MAX would be (say) 231-1 and INT_MIN would be -37?
This would result in some numbers having two representations... But ones-complement gives two representations to zero, and that is allowed according to the "Example". Nowhere does the spec say that zero is the only integer that might have two representations. So I think this new hypothetical is allowed by the spec.
Indeed, any negative value from -1 down to -INT_MAX-1 appears to be permissible as a value for the "sign bit", but nothing smaller (lest the range be non-contiguous). In other words, INT_MIN might be anything from -INT_MAX-1 to -1.
Now, guess what? For the second cast in hvd's code to avoid implementation-defined behavior, we just need x - (unsigned)INT_MIN less than or equal to INT_MAX. We just showed INT_MIN is at least -INT_MAX-1. Obviously, x is at most UINT_MAX. Casting a negative number to unsigned is the same as adding UINT_MAX+1. Put it all together:
x - (unsigned)INT_MIN <= INT_MAX
if and only if
UINT_MAX - (INT_MIN + UINT_MAX + 1) <= INT_MAX
-INT_MIN-1 <= INT_MAX
-INT_MIN <= INT_MAX+1
INT_MIN >= -INT_MAX-1
That last is what we just showed, so even in this perverse case, the code actually works.
That exhausts all of the possibilities, thus ending this extremely academic exercise.
Bottom line: There is some seriously under-specified behavior for signed integers in C89/C90 that got inherited by C++98/C++03. It is fixed in C99, and C++11 indirectly inherits the fix by incorporating <limits.h> from C99. But even C++11 retains the self-contradictory "pure binary representation" wording...
This code relies only on behavior, mandated by the spec, so requirement (a) is easily satisfied:
int unsigned_to_signed(unsigned n)
{
int result = INT_MAX;
if (n > INT_MAX && n < INT_MIN)
throw runtime_error("no signed int for this number");
for (unsigned i = INT_MAX; i != n; --i)
--result;
return result;
}
It's not so easy with requirement (b). This compiles into a no-op with gcc 4.6.3 (-Os, -O2, -O3) and with clang 3.0 (-Os, -O, -O2, -O3). Intel 12.1.0 refuses to optimize this. And I have no info about Visual C.
The original answer solved the problem only for unsigned => int. What if we want to solve the general problem of "some unsigned type" to its corresponding signed type? Furthermore, the original answer was excellent at citing sections of the standard and analyzing some corner cases, but it did not really help me get a feel for why it worked, so this answer will try to give a strong conceptual basis. This answer will try to help explain "why", and use modern C++ features to try to simplify the code.
C++20 answer
The problem has simplified dramatically with P0907: Signed Integers are Two’s Complement and the final wording P1236 that was voted into the C++20 standard. Now, the answer is as simple as possible:
template<std::unsigned_integral T>
constexpr auto cast_to_signed_integer(T const value) {
return static_cast<std::make_signed_t<T>>(value);
}
That's it. A static_cast (or C-style cast) is finally guaranteed to do the thing you need for this question, and the thing many programmers thought it always did.
C++17 answer
In C++17, things are much more complicated. We have to deal with three possible integer representations (two's complement, ones' complement, and sign-magnitude). Even in the case where we know it must be two's complement because we checked the range of possible values, the conversion of a value outside the range of the signed integer to that signed integer still gives us an implementation-defined result. We have to use tricks like we have seen in other answers.
First, here is the code for how to solve the problem generically:
template<typename T, typename = std::enable_if_t<std::is_unsigned_v<T>>>
constexpr auto cast_to_signed_integer(T const value) {
using result = std::make_signed_t<T>;
using result_limits = std::numeric_limits<result>;
if constexpr (result_limits::min() + 1 != -result_limits::max()) {
if (value == static_cast<T>(result_limits::max()) + 1) {
throw std::runtime_error("Cannot convert the maximum possible unsigned to a signed value on this system");
}
}
if (value <= result_limits::max()) {
return static_cast<result>(value);
} else {
using promoted_unsigned = std::conditional_t<sizeof(T) <= sizeof(unsigned), unsigned, T>;
using promoted_signed = std::make_signed_t<promoted_unsigned>;
constexpr auto shift_by_window = [](auto x) {
// static_cast to avoid conversion warning
return x - static_cast<decltype(x)>(result_limits::max()) - 1;
};
return static_cast<result>(
shift_by_window( // shift values from common range to negative range
static_cast<promoted_signed>(
shift_by_window( // shift large values into common range
static_cast<promoted_unsigned>(value) // cast to avoid promotion to int
)
)
)
);
}
}
This has a few more casts than the accepted answer, and that is to ensure there are no signed / unsigned mismatch warnings from your compiler and to properly handle integer promotion rules.
We first have a special case for systems that are not two's complement (and thus we must handle the maximum possible value specially because it doesn't have anything to map to). After that, we get to the real algorithm.
The second top-level condition is straightforward: we know the value is less than or equal to the maximum value, so it fits in the result type. The third condition is a little more complicated even with the comments, so some examples would probably help understand why each statement is necessary.
Conceptual basis: the number line
First, what is this window concept? Consider the following number line:
| signed |
<.........................>
| unsigned |
It turns out that for two's complement integers, you can divide the subset of the number line that can be reached by either type into three equally sized categories:
- => signed only
= => both
+ => unsigned only
<..-------=======+++++++..>
This can be easily proven by considering the representation. An unsigned integer starts at 0 and uses all of the bits to increase the value in powers of 2. A signed integer is exactly the same for all of the bits except the sign bit, which is worth -(2^position) instead of 2^position. This means that for all n - 1 bits, they represent the same values. Then, unsigned integers have one more normal bit, which doubles the total number of values (in other words, there are just as many values with that bit set as without it set). The same logic holds for signed integers, except that all the values with that bit set are negative.
The other two legal integer representations, ones' complement and sign-magnitude, have all of the same values as two's complement integers except for one: the most negative value. C++ defines everything about integer types, except for reinterpret_cast (and the C++20 std::bit_cast), in terms of the range of representable values, not in terms of the bit representation. This means that our analysis will hold for each of these three representations as long as we do not ever try to create the trap representation. The unsigned value that would map to this missing value is a rather unfortunate one: the one right in the middle of the unsigned values. Fortunately, our first condition checks (at compile time) whether such a representation exists, and then handles it specially with a runtime check.
The first condition handles the case where we are in the = section, which means that we are in the overlapping region where the values in one can be represented in the other without change. The shift_by_window function in the code moves all values down by the size of each of these segments (we have to subtract the max value then subtract 1 to avoid arithmetic overflow issues). If we are outside of that region (we are in the + region), we need to jump down by one window size. This puts us in the overlapping range, which means we can safely convert from unsigned to signed because there is no change in value. However, we are not done yet because we have mapped two unsigned values to each signed value. Therefore, we need to shift down to the next window (the - region) so that we have a unique mapping again.
Now, does this give us a result congruent mod UINT_MAX + 1, as requested in the question? UINT_MAX + 1 is equivalent to 2^n, where n is the number of bits in the value representation. The value we use for our window size is equal to 2^(n - 1) (the final index in a sequence of values is one less than the size). We subtract that value twice, which means we subtract 2 * 2^(n - 1) which is equal to 2^n. Adding and subtracting x is a no-op in arithmetic mod x, so we have not affected the original value mod 2^n.
Properly handling integer promotions
Because this is a generic function and not just int and unsigned, we also have to concern ourselves with integral promotion rules. There are two possibly interesting cases: one in which short is smaller than int and one in which short is the same size as int.
Example: short smaller than int
If short is smaller than int (common on modern platforms) then we also know that unsigned short can fit in an int, which means that any operations on it will actually happen in int, so we explicitly cast to the promoted type to avoid this. Our final statement is pretty abstract and becomes easier to understand if we substitute in real values. For our first interesting case, with no loss of generality let us consider a 16-bit short and a 17-bit int (which is still allowed under the new rules, and would just mean that at least one of those two integer types have some padding bits):
constexpr auto shift_by_window = [](auto x) {
return x - static_cast<decltype(x)>(32767) - 1;
};
return static_cast<int16_t>(
shift_by_window(
static_cast<int17_t>(
shift_by_window(
static_cast<uint17_t>(value)
)
)
)
);
Solving for the greatest possible 16-bit unsigned value
constexpr auto shift_by_window = [](auto x) {
return x - static_cast<decltype(x)>(32767) - 1;
};
return int16_t(
shift_by_window(
int17_t(
shift_by_window(
uint17_t(65535)
)
)
)
);
Simplifies to
return int16_t(
int17_t(
uint17_t(65535) - uint17_t(32767) - 1
) -
int17_t(32767) -
1
);
Simplifies to
return int16_t(
int17_t(uint17_t(32767)) -
int17_t(32767) -
1
);
Simplifies to
return int16_t(
int17_t(32767) -
int17_t(32767) -
1
);
Simplifies to
return int16_t(-1);
We put in the largest possible unsigned and get back -1, success!
Example: short same size as int
If short is the same size as int (uncommon on modern platforms), the integral promotion rule are slightly different. In this case, short promotes to int and unsigned short promotes to unsigned. Fortunately, we explicitly cast each result to the type we want to do the calculation in, so we end up with no problematic promotions. With no loss of generality let us consider a 16-bit short and a 16-bit int:
constexpr auto shift_by_window = [](auto x) {
return x - static_cast<decltype(x)>(32767) - 1;
};
return static_cast<int16_t>(
shift_by_window(
static_cast<int16_t>(
shift_by_window(
static_cast<uint16_t>(value)
)
)
)
);
Solving for the greatest possible 16-bit unsigned value
auto x = int16_t(
uint16_t(65535) - uint16_t(32767) - 1
);
return int16_t(
x - int16_t(32767) - 1
);
Simplifies to
return int16_t(
int16_t(32767) - int16_t(32767) - 1
);
Simplifies to
return int16_t(-1);
We put in the largest possible unsigned and get back -1, success!
What if I just care about int and unsigned and don't care about warnings, like the original question?
constexpr int cast_to_signed_integer(unsigned const value) {
using result_limits = std::numeric_limits<int>;
if constexpr (result_limits::min() + 1 != -result_limits::max()) {
if (value == static_cast<unsigned>(result_limits::max()) + 1) {
throw std::runtime_error("Cannot convert the maximum possible unsigned to a signed value on this system");
}
}
if (value <= result_limits::max()) {
return static_cast<int>(value);
} else {
constexpr int window = result_limits::min();
return static_cast<int>(value + window) + window;
}
}
See it live
https://godbolt.org/z/74hY81
Here we see that clang, gcc, and icc generate no code for cast and cast_to_signed_integer_basic at -O2 and -O3, and MSVC generates no code at /O2, so the solution is optimal.
You can explicitly tell the compiler what you want to do:
int unsigned_to_signed(unsigned n) {
if (n > INT_MAX) {
if (n <= UINT_MAX + INT_MIN) {
throw "no result";
}
return static_cast<int>(n + INT_MIN) - (UINT_MAX + INT_MIN + 1);
} else {
return static_cast<int>(n);
}
}
Compiles with gcc 4.7.2 for x86_64-linux (g++ -O -S test.cpp) to
_Z18unsigned_to_signedj:
movl %edi, %eax
ret
If x is our input...
If x > INT_MAX, we want to find a constant k such that 0 < x - k*INT_MAX < INT_MAX.
This is easy -- unsigned int k = x / INT_MAX;. Then, let unsigned int x2 = x - k*INT_MAX;
We can now cast x2 to int safely. Let int x3 = static_cast<int>(x2);
We now want to subtract something like UINT_MAX - k * INT_MAX + 1 from x3, if k > 0.
Now, on a 2s complement system, so long as x > INT_MAX, this works out to:
unsigned int k = x / INT_MAX;
x -= k*INT_MAX;
int r = int(x);
r += k*INT_MAX;
r -= UINT_MAX+1;
Note that UINT_MAX+1 is zero in C++ guaranteed, the conversion to int was a noop, and we subtracted k*INT_MAX then added it back on "the same value". So an acceptable optimizer should be able to erase all that tomfoolery!
That leaves the problem of x > INT_MAX or not. Well, we create 2 branches, one with x > INT_MAX, and one without. The one without does a strait cast, which the compiler optimizes to a noop. The one with ... does a noop after the optimizer is done. The smart optimizer realizes both branches to the same thing, and drops the branch.
Issues: if UINT_MAX is really large relative to INT_MAX, the above might not work. I am assuming that k*INT_MAX <= UINT_MAX+1 implicitly.
We could probably attack this with some enums like:
enum { divisor = UINT_MAX/INT_MAX, remainder = UINT_MAX-divisor*INT_MAX };
which work out to 2 and 1 on a 2s complement system I believe (are we guaranteed for that math to work? That's tricky...), and do logic based on these that easily optimize away on non-2s complement systems...
This also opens up the exception case. It is only possible if UINT_MAX is much larger than (INT_MIN-INT_MAX), so you can put your exception code in an if block asking exactly that question somehow, and it won't slow you down on a traditional system.
I'm not exactly sure how to construct those compile-time constants to deal correctly with that.
std::numeric_limits<int>::is_modulo is a compile time constant. so you can use it for template specialization. problem solved, at least if compiler plays along with inlining.
#include <limits>
#include <stdexcept>
#include <string>
#ifdef TESTING_SF
bool const testing_sf = true;
#else
bool const testing_sf = false;
#endif
// C++ "extensions"
namespace cppx {
using std::runtime_error;
using std::string;
inline bool hopefully( bool const c ) { return c; }
inline bool throw_x( string const& s ) { throw runtime_error( s ); }
} // namespace cppx
// C++ "portability perversions"
namespace cppp {
using cppx::hopefully;
using cppx::throw_x;
using std::numeric_limits;
namespace detail {
template< bool isTwosComplement >
int signed_from( unsigned const n )
{
if( n <= unsigned( numeric_limits<int>::max() ) )
{
return static_cast<int>( n );
}
unsigned const u_max = unsigned( -1 );
unsigned const u_half = u_max/2 + 1;
if( n == u_half )
{
throw_x( "signed_from: unsupported value (negative max)" );
}
int const i_quarter = static_cast<int>( u_half/2 );
int const int_n1 = static_cast<int>( n - u_half );
int const int_n2 = int_n1 - i_quarter;
int const int_n3 = int_n2 - i_quarter;
hopefully( n == static_cast<unsigned>( int_n3 ) )
|| throw_x( "signed_from: range error" );
return int_n3;
}
template<>
inline int signed_from<true>( unsigned const n )
{
return static_cast<int>( n );
}
} // namespace detail
inline int signed_from( unsigned const n )
{
bool const is_modulo = numeric_limits< int >::is_modulo;
return detail::signed_from< is_modulo && !testing_sf >( n );
}
} // namespace cppp
#include <iostream>
using namespace std;
int main()
{
int const x = cppp::signed_from( -42u );
wcout << x << endl;
}
EDIT: Fixed up code to avoid possible trap on non-modular-int machines (only one is known to exist, namely the archaically configured versions of the Unisys Clearpath). For simplicity this is done by not supporting the value -2n-1 where n is the number of int value bits, on such machine (i.e., on the Clearpath). in practice this value will not be supported by the machine either (i.e., with sign-and-magnitude or 1’s complement representation).
I think the int type is at least two bytes, so the INT_MIN and INT_MAX may change in different platforms.
Fundamental types
≤climits≥ header
My money is on using memcpy. Any decent compiler knows to optimise it away:
#include <stdio.h>
#include <memory.h>
#include <limits.h>
static inline int unsigned_to_signed(unsigned n)
{
int result;
memcpy( &result, &n, sizeof(result));
return result;
}
int main(int argc, const char * argv[])
{
unsigned int x = UINT_MAX - 1;
int xx = unsigned_to_signed(x);
return xx;
}
For me (Xcode 8.3.2, Apple LLVM 8.1, -O3), that produces:
_main: ## #main
Lfunc_begin0:
.loc 1 21 0 ## /Users/Someone/main.c:21:0
.cfi_startproc
## BB#0:
pushq %rbp
Ltmp0:
.cfi_def_cfa_offset 16
Ltmp1:
.cfi_offset %rbp, -16
movq %rsp, %rbp
Ltmp2:
.cfi_def_cfa_register %rbp
##DEBUG_VALUE: main:argc <- %EDI
##DEBUG_VALUE: main:argv <- %RSI
Ltmp3:
##DEBUG_VALUE: main:x <- 2147483646
##DEBUG_VALUE: main:xx <- 2147483646
.loc 1 24 5 prologue_end ## /Users/Someone/main.c:24:5
movl $-2, %eax
popq %rbp
retq
Ltmp4:
Lfunc_end0:
.cfi_endproc