Aside from %hn and %hhn (where the h or hh specifies the size of the pointed-to object), what is the point of the h and hh modifiers for printf format specifiers?
Due to default promotions which are required by the standard to be applied for variadic functions, it is impossible to pass arguments of type char or short (or any signed/unsigned variants thereof) to printf.
According to 7.19.6.1(7), the h modifier:
Specifies that a following d, i, o, u, x, or X conversion specifier applies to a
short int or unsigned short int argument (the argument will
have been promoted according to the integer promotions, but its value shall
be converted to short int or unsigned short int before printing);
or that a following n conversion specifier applies to a pointer to a short
int argument.
If the argument was actually of type short or unsigned short, then promotion to int followed by a conversion back to short or unsigned short will yield the same value as promotion to int without any conversion back. Thus, for arguments of type short or unsigned short, %d, %u, etc. should give identical results to %hd, %hu, etc. (and likewise for char types and hh).
As far as I can tell, the only situation where the h or hh modifier could possibly be useful is when the argument passed it an int outside the range of short or unsigned short, e.g.
printf("%hu", 0x10000);
but my understanding is that passing the wrong type like this results in undefined behavior anyway, so that you could not expect it to print 0.
One real world case I've seen is code like this:
char c = 0xf0;
printf("%hhx", c);
where the author expects it to print f0 despite the implementation having a plain char type that's signed (in which case, printf("%x", c) would print fffffff0 or similar). But is this expectation warranted?
(Note: What's going on is that the original type was char, which gets promoted to int and converted back to unsigned char instead of char, thus changing the value that gets printed. But does the standard specify this behavior, or is it an implementation detail that broken software might be relying on?)
One possible reason: for symmetry with the use of those modifiers in the formatted input functions? I know it wouldn't be strictly necessary, but maybe there was value seen for that?
Although they don't mention the importance of symmetry for the "h" and "hh" modifiers in the C99 Rationale document, the committee does mention it as a consideration for why the "%p" conversion specifier is supported for fscanf() (even though that wasn't new for C99 - "%p" support is in C90):
Input pointer conversion with %p was added to C89, although it is obviously risky, for symmetry with fprintf.
In the section on fprintf(), the C99 rationale document does discuss that "hh" was added, but merely refers the reader to the fscanf() section:
The %hh and %ll length modifiers were added in C99 (see §7.19.6.2).
I know it's a tenuous thread, but I'm speculating anyway, so I figured I'd give whatever argument there might be.
Also, for completeness, the "h" modifier was in the original C89 standard - presumably it would be there even if it wasn't strictly necessary because of widespread existing use, even if there might not have been a technical requirement to use the modifier.
In %...x mode, all values are interpreted as unsigned. Negative numbers are therefore printed as their unsigned conversions. In 2's complement arithmetic, which most processors use, there is no difference in bit patterns between a signed negative number and its positive unsigned equivalent, which is defined by modulus arithmetic (adding the maximum value for the field plus one to the negative number, according to the C99 standard). Lots of software- especially the debugging code most likely to use %x- makes the silent assumption that the bit representation of a signed negative value and its unsigned cast is the same, which is only true on a 2's complement machine.
The mechanics of this cast are such that hexidecimal representations of value always imply, possibly inaccurately, that a number has been rendered in 2's complement, as long as it didn't hit an edge condition of where the different integer representations have different ranges. This even holds true for arithmetic representations where the value 0 is not represented with the binary pattern of all 0s.
A negative short displayed as an unsigned long in hexidecimal will therefore, on any machine, be padded with f, due to implicit sign extension in the promotion, which printf will print. The value is the same, but it is truly visually misleading as to the size of the field, implying a significant amount of range that simply isn't present.
%hx truncates the displayed representation to avoid this padding, exactly as you concluded from your real-world use case.
The behavior of printf is undefined when passed an int outside the range of short that should be printed as a short, but the easiest implementation by far simply discards the high bit by a raw downcast, so while the spec doesn't require any specific behavior, pretty much any sane implementation is going to just perform the truncation. There're generally better ways to do that, though.
If printf isn't padding values or displaying unsigned representations of signed values, %h isn't very useful.
The only use I can think of is for passing an unsigned short or unsigned char and using the %x conversion specifier. You cannot simply use a bare %x - the value may be promoted to int rather than unsigned int, and then you have undefined behaviour.
Your alternatives are either to explicitly cast the argument to unsigned; or to use %hx / %hhx with a bare argument.
The variadic arguments to printf() et al are automatically promoted using the default conversions, so any short or char values are promoted to int when passed to the function.
In the absence of the h or hh modifiers, you would have to mask the values passed to get the correct behaviour reliably. With the modifiers, you no longer have to mask the values; the printf() implementation does the job properly.
Specifically, for the format %hx, the code inside printf() can do something like:
va_list args;
va_start(args, format);
...
int i = va_arg(args, int);
unsigned short s = (unsigned short)i;
...print s correctly, as 4 hex digits maximum
...even on a machine with 64-bit `int`!
I'm blithely assuming that short is a 16-bit quantity; the standard does not actually guarantee that, of course.
I found it useful to avoid casting when formatting unsigned chars to hex:
sprintf_s(tmpBuf, 3, "%2.2hhx", *(CEKey + i));
It's a minor coding convenience, and looks cleaner than multiple casts (IMO).
another place it's handy is snprintf size check.
gcc7 added size check when using snprintf
so this will fail
char arr[4];
char x='r';
snprintf(arr,sizeof(arr),"%d",r);
so it forces you to use bigger char when using %d when formatting a char
here is a commit that shows those fixes instead of increasing the char array size they changed %d to %h. this also give more accurate description
https://github.com/Mellanox/libvma/commit/b5cb1e34a04b40427d195b14763e462a0a705d23#diff-6258d0a11a435aa372068037fe161d24
I agree with you that it is not strictly necessary, and so by that reason alone is no good in a C library function :)
It might be "nice" for the symmetry of the different flags, but it is mostly counter-productive because it hides the "conversion to int" rule.
Related
I know that in order to get the 4 least significant bytes of a number of type long I can cast it to int/unsigned int or use a bitwise AND (& 0xFFFFFFFF).
This code produces the following output:
#include <stdio.h>
int main()
{
long n = 0x8899AABBCCDDEEFF;
printf("0x%016lX\n", n);
printf("0x%016X\n", (int)n);
printf("0x%016X\n", (unsigned int)n);
printf("0x%016lX\n", n & 0xFFFFFFFF);
}
Output:
0x8899AABBCCDDEEFF
0x00000000CCDDEEFF
0x00000000CCDDEEFF
0x00000000CCDDEEFF
Does that mean that the two methods used are equivalent? If so, do they always produce the same output regardless of the platform/compiler?
Also, is there any catch or pitfall while casting to unsigned int rather than int for the purpose of this question?
Finally, why is the output the same if you change the number n to be an unsigned long instead?
The methods are definitely different.
According to integral conversion rules (cf, for example, this online c++11 standard), a conversion (e.g. through an explicit cast) from one integral type to another depends on whether the destination type is signed or unsigned. If the destination type is unsigned, one can rely on a "modulo 2n" truncation, whereas with signed destination types one could tap into implementation defined behaviour:
4.7 Integral conversions [conv.integral]
2 If the destination type is unsigned, the resulting value is the
least unsigned integer congruent to the source integer (modulo 2n
where n is the number of bits used to represent the unsigned type). [
Note: In a two's complement representation, this conversion is
conceptual and there is no change in the bit pattern (if there is no
truncation). — end note ]
3 If the destination type is signed, the value is unchanged if it can
be represented in the destination type (and bit-field width);
otherwise, the value is implementation-defined.
For your first question, as others have pointed out, the size of int and long is dependent on the platform, so the methods are not equivalent. In C data types, check that the types say "at least XX bits in size"
For the second question, it comes down to this: long and int are signed, meaning that one bit is reserved for sign (take a look also to two's complement). If you were the compiler, what can you do with negative values (especially the long ones)? As Stepahn Lechner mentioned, this is implementation defined (that is, is up to the compiler).
Finally, in the spirit of "your code must do what it says it does", the best thing to do if you need to do masks is to use masks (and, if you use masks, use unsigned types). Don't try to use cleaver answers. Believe me, they always bite you in the rear. I've dealt with a lot of legacy code to know that by heart.
What's the difference between casting a long to int versus using a bitwise AND in order to get the 4 least significant bytes?
Type. Casting makes the value an int. And'ing does not change the type.
Range. Depending on int,long range, a cast may not change the value at all.
IDB and UB. implementation defined behavior and undefined behavior are present with mixing signed-ness.
To "get" the 4 LSBytes, use & 0xFFFFFFFFu or cast to uint32_t.
OP's question is unnecessarily convoluted.
long n = 0x8899AABBCCDDEEFF; --> Converting a value outside the range of a signed integer type is implementation-defined.
Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
C11 §6.3.1.3 3
printf("0x%016lX\n", n); --> Printing a long with a "%lX" outside the the common range of long/unsigned long is undefined behavior.
Let's go forward with unsigned long:
unsigned long n = 0x8899AABBCCDDEEFF; // no problem,
printf("0x%016lX\n", n); // no problem,
printf("0x%016X\n", (int)n); // problem, C11 6.3.1.3 3
printf("0x%016X\n", (unsigned int)n); // no problem,
printf("0x%016lX\n", n & 0xFFFFFFFF); // no problem,
The "no problem" are OK even is unsigned long is 32-bit or 64-bit. The output will differ, yet is OK.
Recall that int,long are not always 32,64 bit. (16,32), (32,32), (32,64) are common.
int is at least 16 bit.
long is at least that of int and at least 32 bit.
I was going through this example which has a function outputting a hex bit pattern to represent an arbitrary float.
void ExamineFloat(float fValue)
{
printf("%08lx\n", *(unsigned long *)&fValue);
}
Why take the address of fValue, cast to unsigned long pointer, then dereference? Isn't all that work just equivalent to a direct cast to unsigned long?
printf("%08lx\n", (unsigned long)fValue);
I tried it and the answer isn't the same, so confused.
(unsigned long)fValue
This converts the float value to an unsigned long value, according to the "usual arithmetic conversions".
*(unsigned long *)&fValue
The intention here is to take the address at which fValue is stored, pretend that there is not a float but an unsigned long at this address, and to then read that unsigned long. The purpose is to examine the bit pattern which is used to store the float in memory.
As shown, this causes undefined behavior though.
Reason: You may not access an object through a pointer to a type that is not "compatible" to the object's type. "Compatible" types are for example (unsigned) char and every other type, or structures that share the same initial members (speaking of C here). See §6.5/7 N1570 for the detailed (C11) list (Note that my use of "compatible" is different - more broad - than in the referenced text.)
Solution: Cast to unsigned char *, access the individual bytes of the object and assemble an unsigned long out of them:
unsigned long pattern = 0;
unsigned char * access = (unsigned char *)&fValue;
for (size_t i = 0; i < sizeof(float); ++i) {
pattern |= *access;
pattern <<= CHAR_BIT;
++access;
}
Note that (as #CodesInChaos pointed out) the above treats the floating point value as being stored with its most significant byte first ("big endian"). If your system uses a different byte order for floating point values you'd need to adjust to that (or rearrange the bytes of above unsigned long, whatever's more practical to you).
Floating-point values have memory representations: for example the bytes can represent a floating-point value using IEEE 754.
The first expression *(unsigned long *)&fValue will interpret these bytes as if it was the representation of an unsigned long value. In fact in C standard it results in an undefined behavior (according to the so-called "strict aliasing rule"). In practice, there are issues such as endianness that have to be taken into account.
The second expression (unsigned long)fValue is C standard compliant. It has a precise meaning:
C11 (n1570), § 6.3.1.4 Real floating and integer
When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero). If the value of the integral part cannot be represented by the integer type, the behavior is undefined.
*(unsigned long *)&fValue is not equivalent to a direct cast to an unsigned long.
The conversion to (unsigned long)fValue converts the value of fValue into an unsigned long, using the normal rules for conversion of a float value to an unsigned long value. The representation of that value in an unsigned long (for example, in terms of the bits) can be quite different from how that same value is represented in a float.
The conversion *(unsigned long *)&fValue formally has undefined behaviour. It interprets the memory occupied by fValue as if it is an unsigned long. Practically (i.e. this is what often happens, even though the behaviour is undefined) this will often yield a value quite different from fValue.
Typecasting in C does both a type conversion and a value conversion. The floating point → unsigned long conversion truncates the fractional portion of the floating point number and restricts the value to the possible range of an unsigned long. Converting from one type of pointer to another has no required change in value, so using the pointer typecast is a way to keep the same in-memory representation while changing the type associated with that representation.
In this case, it's a way to be able to output the binary representation of the floating point value.
As others have already noted, casting a pointer to a non-char type to a pointer to a different non-char type and then dereferencing is undefined behavior.
That printf("%08lx\n", *(unsigned long *)&fValue) invokes undefined behavior does not necessarily mean that running a program that attempts to perform such a travesty will result in hard drive erasure or make nasal demons erupt from ones nose (the two hallmarks of undefined behavior). On a computer in which sizeof(unsigned long)==sizeof(float) and on which both types have the same alignment requirements, that printf will almost certainly do what one expects it to do, which is to print the hex representation of the floating point value in question.
This shouldn't be surprising. The C standard openly invites implementations to extend the language. Many of these extensions are in areas that are, strictly speaking, undefined behavior. For example, the POSIX function dlsym returns a void*, but this function is typically used to find the address of a function rather than a global variable. This means the void pointer returned by dlsym needs to be cast to a function pointer and then dereferenced to call the function. This is obviously undefined behavior, but it nonetheless works on any POSIX compliant platform. This will not work on a Harvard architecture machine on which pointers to functions have different sizes than do pointers to data.
Similarly, casting a pointer to a float to a pointer to an unsigned integer and then dereferencing happens to work on almost any computer with almost any compiler in which the size and alignment requirements of that unsigned integer are the same as that of a float.
That said, using unsigned long might well get you into trouble. On my computer, an unsigned long is 64 bits long and has 64 bit alignment requirements. This is not compatible with a float. It would be better to use uint32_t -- on my computer, that is.
The union hack is one way around this mess:
typedef struct {
float fval;
uint32_t ival;
} float_uint32_t;
Assigning to a float_uint32_t.fval and accessing from a ``float_uint32_t.ival` used to be undefined behavior. That is no longer the case in C. No compiler that I know of blows nasal demons for the union hack. This was not UB in C++. It was illegal. Until C++11, a compliant C++ compiler had to complain to be compliant.
Any even better way around this mess is to use the %a format, which has been part of the C standard since 1999:
printf ("%a\n", fValue);
This is simple, easy, portable, and there is no chance of undefined behavior. This prints the hexadecimal/binary representation of the double precision floating point value in question. Since printf is an archaic function, all float arguments are converted to double prior to the call to printf. This conversion must be exact per the 1999 version of the C standard. One can pick up that exact value via a call to scanf or its sisters.
I have the below simple program:
#include <iostream>
#include <stdio.h>
void SomeFunction(int a)
{
std::cout<<"Value in function: a = "<<a<<std::endl;
}
int main(){
size_t a(0);
std::cout<<"Value in main: "<<a-1<<std::endl;
SomeFunction(a-1);
return 0;
}
Upon executing this I get:
Value in main: 18446744073709551615
Value in function: a = -1
I think I roughly understand why the function gets the 'correct' value of -1: there is an implicit conversion from the unsigned type to the signed one i.e. 18446744073709551615(unsigned) = -1(signed).
Is there any situation where the function will not get the 'correct' value?
Since size_t type is unsigned, subtracting 1 is well defined:
A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.
However, the resultant value of 264-1 is out of ints range, so you get implementation-defined behavior:
[when] the new type is signed and the value cannot be represented in it, either the result is implementation-defined or an implementation-defined signal is raised.
Therefore, the answer to your question is "yes": there are platforms where the value of a would be different; there are also platforms where instead of calling SomeFunction the program will raise a signal.
Not on your computer... but technically yes, there is a situation where things can go wrong.
All modern PCs use the "two's complement" system for signed integer arithmetic (read Wikipedia for details). Two's complement has many advantages, but one of the biggest is this: unsaturated addition and subtraction of signed integers is identical to that of unsigned integers. As long as overflow/underflow causes the result to "wrap around" (i.e., 0-1 = UINT_MAX), the computer can add and subtract without even knowing whether you're interpreting the numbers as signed or unsigned.
BUT! C/C++ do not technically require two's complement for signed integers. There are two other permissible systems, known as "sign-magnitude" and "one's complement". These are unusual systems, never found outside antique architectures and embedded processors (and rarely even there). But in those systems, signed and unsigned arithmetic do not match up, and (signed)(a+b) will not necessarily equal (signed)a + (signed) b.
There's another, more mundane caveat when you're also narrowing types, as is the case between size_t and int on x64, because C/C++ don't require compilers to follow a particular rule when narrowing out-of-range values to signed types. This is likewise more a matter of language lawyering than actual unsafeness, though: VC++, GCC, Clang, and all other compilers I'm aware of narrow through truncation, leading to the expected behavior.
It's easier to compare and contrast signed and insigned of type same basic type, say signed int and unsigned int.
On a system that uses 32 bits for int, the range of unsigned int is [0 - 4294967295] and the range of signed int is [-2147483647 - 2147483647].
Say you have a variable of type unsigned int and its value is greater than 2147483647. If you pass such a variable to SomeFunction, you will see an incorrect value in the function.
Conversely, say you have a variable of type signed int and its value is less than zero. If you pass such a variable to a function that expects an unsigned int, you will see an incorrect value in the function.
This question already has answers here:
sizeof() operator in if-statement
(5 answers)
Closed 4 years ago.
#include <stdio.h>
int arr[] = {1,2,3,4,5,6,7,8};
#define SIZE (sizeof(arr)/sizeof(int))
int main()
{
printf("SIZE = %d\n", SIZE);
if ((-1) < SIZE)
printf("less");
else
printf("more");
}
The output after compiling with gcc is "more". Why the if condition fails even when -1 < 8?
The problem is in your comparison:
if ((-1) < SIZE)
sizeof typically returns an unsigned long, so SIZE will be unsigned long, whereas -1 is just an int. The rules for promotion in C and related languages mean that -1 will be converted to size_t before the comparison, so -1 will become a very large positive value (the maximum value of an unsigned long).
One way to fix this is to change the comparison to:
if (-1 < (long long)SIZE)
although it's actually a pointless comparison, since an unsigned value will always be >= 0 by definition, and the compiler may well warn you about this.
As subsequently noted by #Nobilis, you should always enable compiler warnings and take notice of them: if you had compiled with e.g. gcc -Wall ... the compiler would have warned you of your bug.
TL;DR
Be careful with mixed signed/unsigned operations (use -Wall compiler warnings). The Standard has a long section about it. In particular, it is often but not always true that signed is value-converted to unsigned (although it does in your particular example). See this explanation below (taken from this Q&A)
Relevant quote from the C++ Standard:
5 Expressions [expr]
10 Many binary operators that expect operands of arithmetic or
enumeration type cause conversions and yield result types in a similar
way. The purpose is to yield a common type, which is also the type of
the result. This pattern is called the usual arithmetic conversions,
which are defined as follows:
[2 clauses about equal types or types of equal sign omitted]
— Otherwise, if the operand that has unsigned integer type has rank
greater than or equal to the rank of the type of the other operand,
the operand with signed integer type shall be converted to the type of
the operand with unsigned integer type.
— Otherwise, if the type of
the operand with signed integer type can represent all of the values
of the type of the operand with unsigned integer type, the operand
with unsigned integer type shall be converted to the type of the
operand with signed integer type.
— Otherwise, both operands shall be
converted to the unsigned integer type corresponding to the type of
the operand with signed integer type.
Your actual example
To see into which of the 3 cases your program falls, modify it slightly to this
#include <stdio.h>
int arr[] = {1,2,3,4,5,6,7,8};
#define SIZE (sizeof(arr)/sizeof(int))
int main()
{
printf("SIZE = %zu, sizeof(-1) = %zu, sizeof(SIZE) = %zu \n", SIZE, sizeof(-1), sizeof(SIZE));
if ((-1) < SIZE)
printf("less");
else
printf("more");
}
On the Coliru online compiler, this prints 4 and 8 for the sizeof() of -1 and SIZE, respectively, and selects the "more" branch (live example).
The reason is that the unsigned type is of greater rank than the signed type. Hence, clause 1 applies and the signed type is value-converted to the unsigned type (on most implementation, typically by preserving the bit-representation, so wrapping around to a very large unsigned number), and the comparison then proceeds to select the "more" branch.
Variations on a theme
Rewriting the condition to if ((long long)(-1) < (unsigned)SIZE) would take the "less" branch (live example).
The reason is that the signed type is of greater rank than the unsigned type and can also accomodate all the unsigned values. Hence, clause 2 applies and the unsigned type is converted to the signed type, and the comparison then proceeds to select the "less" branch.
Of course, you would never write such a contrived if() statement with explicit casts, but the same effect could happen if you compare variables with types long long and unsigned. So it illustrates the point that mixed signed/unsigned arithmetic is very subtle and depends on the relative sizes ("ranking" in the words of the Standard). In particular, there is no fixed rules saying that signed will always be converted to unsigned.
When you do comparison between signed and unsigned where unsigned has at least an equal rank to that of the signed type (see TemplateRex's answer for the exact rules), the signed is converted to the type of the unsigned.
With regards to your case, on a 32bit machine the binary representation of -1 as unsigned is 4294967295. So in effect you are comparing if 4294967295 is smaller than 8 (it isn't).
If you had enabled warnings, you would have been warned by the compiler that something fishy is going on:
warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
Since the discussion has shifted a bit on how appropriate the use of unsigned is, let me put a quote by James Gosling with regards to the lack of unsigned types in Java (and I will shamelessly link to another post of mine on the subject):
Gosling: For me as a language designer, which I don't really count
myself as these days, what "simple" really ended up meaning was could
I expect J. Random Developer to hold the spec in his head. That
definition says that, for instance, Java isn't -- and in fact a lot of
these languages end up with a lot of corner cases, things that nobody
really understands. Quiz any C developer about unsigned, and pretty
soon you discover that almost no C developers actually understand what
goes on with unsigned, what unsigned arithmetic is. Things like that
made C complex. The language part of Java is, I think, pretty simple.
The libraries you have to look up.
This is an historical design bug of C that was also repeated in C++.
It dates back to 16-bit computers and the error was deciding to use all 16 bits to represent sizes up to 65536 giving up the possibility to represent negative sizes.
This in se wouldn't have been an error if unsigned meaning was "non-negative integer" (a size cannot logically be negative) but it's a problem with the conversion rules of the language.
Given the conversion rules of the language the unsigned type in C doesn't represent a non-negative number, but it's instead more like a bitmask (the mathematical term is actually "a member of the ℤ/n ring"). To see why consider that for the C and C++ language
unsigned - unsigned gives an unsigned result
signed + unsigned gives and unsigned result
both of them clearly make no sense at all if you read unsigned as "non-negative number".
Of course saying that the size of an object is a member of ℤ/n ring doesn't make any sense at all and here it's where the error resides.
Practical implications:
Every time you deal with the size of an object be careful because the value is unsigned and that type in C/C++ has a lot of properties that are illogical for a number. Please always remember that unsigned doesn't mean "non-negative integer" but "member of ℤ/n algebraic ring" and that, most dangerous, in case of a mixed operation an int is converted to unsigned int and not the opposite.
For example:
void drawPolyline(const std::vector<P2d>& pts) {
for (int i=0; i<pts.size()-1; i++) {
drawLine(pts[i], pts[i+1]);
}
}
is buggy, because if passed an empty vector of points it will do illegal (UB) operations. The reason is that pts.size() is an unsigned.
The rules of the language will convert 1 (an integer) to 1{mod n}, will perform the subtraction in ℤ/n resulting in (size-1){mod n}, will convert i also to a {mod n} representation and will do the comparison in ℤ/n.
C/C++ actually defines a < operator in ℤ/n (rarely done in math) and you will end up accessing pts[0], pts[1] ... and so on until huge numbers even if the input vector was empty.
A correct loop could be
void drawPolyline(const std::vector<P2d>& pts) {
for (int i=1; i<pts.size(); i++) {
drawLine(pts[i-1], pts[i]);
}
}
but I normally prefer
void drawPolyline(const std::vector<P2d>& pts) {
for (int i=0,n=pts.size(); i<n-1; i++) {
drawLine(pts[i], pts[i+1]);
}
}
in other words getting rid of unsigned as soon as possible, and just working with regular ints.
Never use unsigned to represent size of containers or counters because unsigned means "member of ℤ/n" and the size of a container is not one of those things. Unsigned types are useful, but NOT to represent size of objects.
The standard C/C++ library unfortunately made this wrong choice, and it's too late to fix it. You are not forced to do the same mistake however.
In the words of Bjarne Stroustrup:
Using an unsigned instead of an int to gain one more bit to represent
positive integers is almost never a good idea. Attempts to ensure that
some values are positive by declaring variables unsigned will
typically be defeated by the implicit conversion rules
well, i'm not going to repeat the strong words Paul R said, but when you are comparing unsigned and integers you are going to experience dome bad things.
do if ((-1) < (int)SIZE)
instead of your if condition
Convert the unsigned type returned from sizeof operator to signed
when you compare two unsigned and signed number compiler implicitly converts signed to unsigned.
-1 signed representation in 4 byte int is 11111111 11111111 11111111 11111111 when converted to unsigned this representation would refer to 2^16-1
So basically your are comparing that 2^16-1>SIZE, which would be true.
You have to override that by explicitly casting the unsigned value to signed.
Since sizeof operator returns unsigned long long you should cast it to signed long long
if((-1)<(signed long long)SIZE)
use this if condition in your code
Code:
typedef signed short SIGNED_SHORT; //16 bit
typedef signed int SIGNED_INT; //32 bit
SIGNED_SHORT x;
x = (SIGNED_SHORT)(SIGNED_INT) 45512; //or any value over 32,767
Here is what I know:
Signed 16 bits:
Signed: From −32,768 to 32,767
Unsigned: From 0 to 65,535
Don't expect 45512 to fit into x as x is declared a 16 bit signed integer.
How and what does the double casting above do?
Thank You!
typedef signed short SIGNED_SHORT; //16 bit
typedef signed int SIGNED_INT; //32 bit
These typedefs are not particularly useful. A typedef does nothing more than provide a new name for an existing type. Type signed short already has a perfectly good name: "signed short"; calling it SIGNED_SHORT as well doesn't buy you anything. (It would make sense if it abstracted away some information about the type, or if the type were likely to change -- but using the name SIGNED_SHORT for a type other than signed short would be extremely confusing.)
Note also that short and int are both guaranteed to be at least 16 bits wide, and int is at least as wide as short, but different sizes are possible. For example, a compiler could make both short and int 16 bits -- or 64 bits for that matter. But I'll assume the sizes for your compiler are as you state.
In addition, signed short and short are names for the same type, as are signed int and int.
SIGNED_SHORT x;
x = (SIGNED_SHORT)(SIGNED_INT) 45512; //or any value over 32,767
A cast specifies a conversion to a specified type. Two casts specify two such conversions. The value 45512 is converted to signed int, and then to signed short.
The constant 45512 is already of type int (another name for signed int), so the innermost cast is fairly pointless. (Note that if int is only 16 bits, then 45512 will be of type long.)
When you assign a value of one numeric type to an object of another numeric type, the value is implicitly converted to the object's type, so the outermost cast is also redundant.
So the above code snippet is exactly equivalent to:
short x = 45512;
Given the ranges of int and short on your system, the mathematical value 45512 cannot be represented in type short. The language rules state that the result of such a conversion is implementation-defined, which means that it's up to each implementation to determine what the result is, and it must document that choice, but different implementations can do it differently. (Actually that's not quite the whole story; the 1999 ISO C standard added permission for such a conversion to raise an implementation-defined signal. I don't know of any compiler that does this.)
The most common semantics for this kind of conversion is that the result gets the low-order bits of the source value. This will probably result in the value -20024 being assigned to x. But you shouldn't depend on that if you want your program to be maximally portable.
When you cast twice, the casts are applied in sequence.
int a = 45512;
int b = (int) a;
short x = (short) b;
Since 45512 does not fit in a short on most (but not all!) platforms, the cast overflows on those platforms. This will either raise an implementation-defined signal or result in an implementation-defined value.
In practice, many platforms define the result as the truncated value, which is -20024 in this case. However, there are platforms which raise a signal, which will probably terminate your program if uncaught.
Citation: n1525 §6.3.1.3
Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
The double casting is equivalent to:
short x = static_cast<short>(static_cast<int>(45512));
which is equivalent to:
short x = 45512;
which will likely wrap around so x equals -20024, but technically it's implementation defined behavior if a short has a maximum value less than 45512 on your platform. The literal 45512 is of type int.
You can assume it does two type conversions (although signed int and int are only separated once in the C standard, IIRC).
If SIGNED_SHORT is too small to handle 45512, the result is either implementation-defined or an implementation-defined signal is raised. (In C++ only the former applies.)