#include<stdio.h>
main()
{
unsigned int a=1,b=2;
printf("%d\n",a-b>=0);
getch();
}
Why does this program output "1" ?
Compile with warnings enabled, and it will become quite obvious:
unsigned.c:5:22: warning: comparison of unsigned expression >= 0 is always true
[-Wtautological-compare]
printf("%d\n",a-b>=0);
~~~^ ~
You are subtracting two unsigned integers. Even though the subtraction would equal -1, since they are unsigned it wraps around and you get some very large value. There is no way that an unsigned integer could ever not be greater than or equal to zero.
Quoting ANSI C 89 standard on subtraction generating a negative number using unsigned operands:
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 unsigned integer type
So, unsigned subtraction never generates a negative number, making your condition true, and thus prints 1.
try to print parts of your logic expression
printf("%u >= %u\n",a-b, 0);
and you will see why a-b>=0 is true.
Related
I am not very good at C language and just met a problem I don't understand. The code is:
int main()
{
unsigned int a = 100;
unsigned int b = 200;
float c = 2;
int result_i;
unsigned int result_u;
float result_f;
result_i = (a - b)*2;
result_u = (a - b);
result_f = (a-b)*c;
printf("%d\n", result_i);
printf("%d\n", result_u);
printf("%f\n", result_f);
return 0;
}
And the output is:
-200
-100
8589934592.000000
Program ended with exit code: 0
For (a-b) is negative and a,b are unsigned int type, (a-b) is trivial. And after multiplying a float type number c, the result is 8589934592.000000. I have two questions:
First, why the result is non-trivial after multiplying int type number 2 and assigned to an int type number?
Second, why the result_u is non-trivial even though (a-b) is negative and result_u is unsigned int type?
I am using Xcode to test this code, and the compiler is the default APPLE LLVM 6.0.
Thanks!
Your assumption that a - b is negative is completely incorrect.
Since a and b have unsigned int type all arithmetic operations with these two variables are performed in the domain of unsigned int type. The same applies to mixed "unsigned int with int" arithmetic as well. Such operations implement modulo arithmetic, with the modulo being equal to UINT_MAX + 1.
This means that expression a - b produces a result of type unsigned int. It is a large positive value equal to UINT_MAX + 1 - 100. On a typical platform with 32-bit int it is 4294967296 - 100 = 4294967196.
Expression (a - b) * 2 also produces a result of type unsigned int. It is also a large positive value (UINT_MAX + 1 - 100 multiplied by 2 and taken modulo UINT_MAX + 1). On a typical platform it is 4294967096.
This latter value is too large for type int. Which means that when you force it into a variable result_i, signed integer overflow occurs. The result of signed integer overflow on assignment is implementation defined. In your case result_i ended up being -200. It looks "correct", but this is not guaranteed by the language. (Albeit it might be guaranteed by your implementation.)
Variable result_u receives the correct unsigned result - a positive value UINT_MAX + 1 - 100. But you print that result using %d format specifier in printf, instead of the proper %u. It is illegal to print unsigned int values that do not fit into the range of int using %d specifier. The behavior of your code is undefined for that reason. The -100 value you see in the output is just a manifestation of that undefined behavior. This output is formally meaningless, even though it appears "correct" at the first sight.
Finally, variable result_f receives the "proper" result of (a-b)*c expression, calculated without overflows, since the multiplication is performed in the float domain. What you see is that large positive value I mentioned above, multiplied by 2. It is likely rounded to the precision of float type though, which is implementation-defined. The exact value would be 4294967196 * 2 = 8589934392.
One can argue that the last value you printed is the only one that properly reflects the properties of unsigned arithmetic, i.e. it is "naturally" derived from the actual result of a - b.
You get negative numbers in the printf because you've asked it to print a signed integer with %d. Use %u if you want to see the actual value you ended up with. That will also show you how you ended up with the output for the float multiplication.
Following C code displays the result correctly, -1.
#include <stdio.h>
main()
{
unsigned x = 1;
unsigned y=x-2;
printf("%d", y );
}
But in general, is it always safe to do subtraction involving
unsigned integers?
The reason I ask the question is that I want to do some conditioning
as follows:
unsigned x = 1; // x was defined by someone else as unsigned,
// which I had better not to change.
for (int i=-5; i<5; i++){
if (x+i<0) continue
f(x+i); // f is a function
}
Is it safe to do so?
How are unsigned integers and signed integers different in
representing integers? Thanks!
1: Yes, it is safe to subtract unsigned integers. The definition of arithmetic on unsigned integers includes that if an out-of-range value would be generated, then that value should be adjusted modulo the maximum value for the type, plus one. (This definition is equivalent to truncating high bits).
Your posted code has a bug though: printf("%d", y); causes undefined behaviour because %d expects an int, but you supplied unsigned int. Use %u to correct this.
2: When you write x+i, the i is converted to unsigned. The result of the whole expression is a well-defined unsigned value. Since an unsigned can never be negative, your test will always fail.
You also need to be careful using relational operators because the same implicit conversion will occur. Before I give you a fix for the code in section 2, what do you want to pass to f when x is UINT_MAX or close to it? What is the prototype of f ?
3: Unsigned integers use a "pure binary" representation.
Signed integers have three options. Two can be considered obsolete; the most common one is two's complement. All options require that a positive signed integer value has the same representation as the equivalent unsigned integer value. In two's complement, a negative signed integer is represented the same as the unsigned integer generated by adding UINT_MAX+1, etc.
If you want to inspect the representation, then do unsigned char *p = (unsigned char *)&x; printf("%02X%02X%02X%02X", p[0], p[1], p[2], p[3]);, depending on how many bytes are needed on your system.
Its always safe to subtract unsigned as in
unsigned x = 1;
unsigned y=x-2;
y will take on the value of -1 mod (UINT_MAX + 1) or UINT_MAX.
Is it always safe to do subtraction, addition, multiplication, involving unsigned integers - no UB. The answer will always be the expected mathematical result modded by UINT_MAX+1.
But do not do printf("%d", y ); - that is UB. Instead printf("%u", y);
C11 §6.2.5 9 "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."
When unsigned and int are used in +, the int is converted to an unsigned. So x+i has an unsigned result and never is that sum < 0. Safe, but now if (x+i<0) continue is pointless. f(x+i); is safe, but need to see f() prototype to best explain what may happen.
Unsigned integers are always 0 to power(2,N)-1 and have well defined "overflow" results. Signed integers are 2's complement, 1's complement, or sign-magnitude and have UB on overflow. Some compilers take advantage of that and assume it never occurs when making optimized code.
Rather than really answering your questions directly, which has already been done, I'll make some broader observations that really go to the heart of your questions.
The first is that using unsigned in loop bounds where there's any chance that a signed value might crop up will eventually bite you. I've done it a bunch of times over 20 years and it has ultimately bit me every time. I'm now generally opposed to using unsigned for values that will be used for arithmetic (as opposed to being used as bitmasks and such) without an excellent justification. I have seen it cause too many problems when used, usually with the simple and appealing rationale that “in theory, this value is non-negative and I should use the most restrictive type possible”.
I understand that x, in your example, was decided to be unsigned by someone else, and you can't change it, but you want to do something involving x over an interval potentially involving negative numbers.
The “right” way to do this, in my opinion, is first to assess the range of values that x may take. Suppose that the length of an int is 32 bits. Then the length of an unsigned int is the same. If it is guaranteed to be the case that x can never be larger than 2^31-1 (as it often is), then it is safe in principle to cast x to a signed equivalent and use that, i.e. do this:
int y = (int)x;
// Do your stuff with *y*
x = (unsigned)y;
If you have a long that is longer than unsigned, then even if x uses the full unsigned range, you can do this:
long y = (long)x;
// Do your stuff with *y*
x = (unsigned)y;
Now, the problem with either of these approaches is that before assigning back to x (e.g. x=(unsigned)y; in the immediately preceding example), you really must check that y is non-negative. However, these are exactly the cases where working with the unsigned x would have bitten you anyway, so there's no harm at all in something like:
long y = (long)x;
// Do your stuff with *y*
assert( y >= 0L );
x = (unsigned)y;
At least this way, you'll catch the problems and find a solution, rather than having a strange bug that takes hours to find because a loop bound is four billion unexpectedly.
No, it's not safe.
Integers usually are 4 bytes long, which equals to 32 bits. Their difference in representation is:
As far as signed integers is concerned, the most significant bit is used for sign, so they can represent values between -2^31 and 2^31 - 1
Unsigned integers don't use any bit for sign, so they represent values from 0 to 2^32 - 1.
Part 2 isn't safe either for the same reason as Part 1. As int and unsigned types represent integers in a different way, in this case where negative values are used in the calculations, you can't know what the result of x + i will be.
No, it's not safe. Trying to represent negative numbers with unsigned ints smells like bug. Also, you should use %u to print unsigned ints.
If we slightly modify your code to put %u in printf:
#include <stdio.h>
main()
{
unsigned x = 1;
unsigned y=x-2;
printf("%u", y );
}
The number printed is 4294967295
The reason the result is correct is because C doesn't do any overflow checks and you are printing it as a signed int (%d). This, however, does not mean it is safe practice. If you print it as it really is (%u) you won't get the correct answer.
An Unsigned integer type should be thought of not as representing a number, but as a member of something called an "abstract algebraic ring", specifically the equivalence class of integers congruent modulo (MAX_VALUE+1). For purposes of examples, I'll assume "unsigned int" is 16 bits for numerical brevity; the principles would be the same with 32 bits, but all the numbers would be bigger.
Without getting too deep into the abstract-algebraic nitty-gritty, when assigning a number to an unsigned type [abstract algebraic ring], zero maps to the ring's additive identity (so adding zero to a value yields that value), one means the ring's multiplicative identity (so multiplying a value by one yields that value). Adding a positive integer N to a value is equivalent to adding the multiplicative identity, N times; adding a negative integer -N, or subtracting a positive integer N, will yield the value which, when added to +N, would yield the original value.
Thus, assigning -1 to a 16-bit unsigned integer yields 65535, precisely because adding 1 to 65535 will yield 0. Likewise -2 yields 65534, etc.
Note that in an abstract algebraic sense, every integer can be uniquely assigned into to algebraic rings of the indicated form, and a ring member can be uniquely assigned into a smaller ring whose modulus is a factor of its own [e.g. a 16-bit unsigned integer maps uniquely to one 8-bit unsigned integer], but ring members are not uniquely convertible to larger rings or to integers. Unfortunately, C sometimes pretends that ring members are integers, and implicitly converts them; that can lead to some surprising behavior.
Subtracting a value, signed or unsigned, from an unsigned value which is no smaller than int, and no smaller than the value being subtracted, will yield a result according to the rules of algebraic rings, rather than the rules of integer arithmetic. Testing whether the result of such computation is less than zero will be meaningless, because ring values are never less than zero. If you want to operate on unsigned values as though they are numbers, you must first convert them to a type which can represent numbers (i.e. a signed integer type). If the unsigned type can be outside the range that is representable with the same-sized signed type, it will need to be upcast to a larger type.
#include<stdio.h>
#include<conio.h>
main()
{
int i=-5;
unsigned int j=i;
printf("%d",j);
getch();
}
O/p
-----
-5
#include<stdio.h>
#include<conio.h>
main()
{
int i=-5;
unsigned int j=i;
printf("%u",j);
getch();
}
O/p
===
4255644633
Here I am not getting any compilation error .
It is giving -5 when print with the identifier %d and when printing with %u it is printing some garbage value .
The things I want to know are
1) Why compiler ignores when assigned integer with negative number to unsigned int.
2) How it is converting signed to unsigned ?
Who are "we?"
There's no "garbage value", it's probably just the result of viewing the bits of the signed integer as an unsigned. Typically two's complement will result in very large values for many a negative values. Try printing the value in hex to see the pattern more clearly, in decimal they're often hard to decipher.
I'd simply add that the concept of signed or unsigned is something that humans appreciate more than machines.
Assuming a 32-bit machine, your value of -5 is going to be represented internally by the 32-bit value 0xFFFFFFFB (two's complement).
When you insert printf("%d",j); into your source code, the compiler couldn't care less whether j is signed or unsigned, it just shoves 0xFFFFFFFB onto the stack and then a pointer to the "%d" string. The printf function when called looks at the format string, sees the %d and knows from that that it has to interpret the 0xFFFFFFFB as a signed value, hence the reason for it displaying -5 despite j being an unsigned int.
On the other hand, when you write printf("%u",j);, the "%u" makes printf interpret your 0xFFFFFFFB as an unsigned value. That value is 2^32 - 5, or 4294967291.
It's the format string passed to printf that determines how the value will be interpreted, not the type of the variable j.
There's noting unusual in the possibility to assign a negative value to an unsigned variable. The implicit conversion that happens in such cases is perfectly well defined by C language. The value is brought into the range of the target unsigned type in accordance with the rules of modulo arithmetic. The modulo is equal to 2^N, where N is the number of value bits in the unsigned recipient. This is how it has always been in C.
Printing an unsigned int value with %d specifier makes no sense. This specifier requires a signed int argument. Because of this mismatch, the behavior of your first code is undefined.
In other words, you got it completely backwards with regards to which value is garbage and which is not.
Your first code is essentially "printing garbage value" due to undefined behavior. The fact that it happens to match your original value of -5 is just a specific manifestation of undefined behavior.
Meanwhile, the second code is supposed to print a well-defined proper value. It should be result of conversion of -5 to unsigned int type by modulo UINT_MAX + 1. In your case that modulo probably happens to be 2^32 = 4294967296, which is why you are supposed to see 4294967296 - 5 = 4294967291.
How you managed to get 4255644633 is not clear. Your 4255644633 is apparently a result of different code, not the one you posted.
You can and you should get a warning (or perhaps failure) depending on the compiler and the settings.
The value you get is due to twos-complement.
The output in the second case is not a garbage value...
int i=-5;
when converted to binary form the Most Significant Bit is assigned '1' as -5 is a negative number..
but when u use %u the binary form is treated as a normal number and the 1 in MSB is treated a part of normal number..
#include<stdio.h>
int main()
{
unsigned short a=-1;
printf("%d",a);
return 0;
}
This is giving me output 65535. why?
When I increased the value of a in negative side the output is (2^16-1=)65535-a.
I know the range of unsigned short int is 0 to 65535.
But why is rotating in the range 0 to 65535.What is going inside?
#include<stdio.h>
int main()
{
unsigned int a=-1;
printf("%d",a);
return 0;
}
Output is -1.
%d is used for signed decimal integer than why here it is not following the rule of printing the largest value of its(int) range.
Why the output in this part is -1?
I know %u is used for printing unsigned decimal integer.
Why the behavioral is undefined in second code and not in first.?
This I have compiled in gcc compiler. It's a C code
On my machine sizeof short int is 2 bytes and size of int is 4 bytes.
In your implementation, short is 16 bits and int is 32 bits.
unsigned short a=-1;
printf("%d",a);
First, -1 is converted to unsigned short. This results in the value 65535. For the precise definition see the standard "integer conversions". To summarize: the value is taken modulo USHORT_MAX+1.
This value 65535 is assigned to a.
Then for the printf, which uses varargs, the value is promoted back to int. varargs never pass integer types smaller than int, they're always converted to int. This results in the value 65535, which is printed.
unsigned int a=-1;
printf("%d",a);
First line, same as before but modulo UINT_MAX+1. a is 4294967295.
For the printf, a is passed as an unsigned int. Since %d requires an int the behavior is undefined by the C standard. But your implementation appears to have reinterpreted the unsigned value 4294967295, which has all bits set, as as a signed integer with all-bits-set, i.e. the two's-complement value -1. This behavior is common but not guaranteed.
Variable assignment is done to the amount of memory of the type of the variable (e.g., short is 2 bytes, int is 4 bytes, in 32 bit hardware, typically). Sign of the variable is not important in the assignment. What matters here is how you are going to access it. When you assign to a 'short' (signed/unsigned) you assign the value to a '2 bytes' memory. Now if you are going to use '%d' in printf, printf will consider it 'integer' (4 bytes in your hardware) and the two MSBs will be 0 and hence you got [0|0](two MSBs) [-1] (two LSBs). Due to the new MSBs (introduced by %d in printf, migration) your sign bit is hidden in the LSBs and hence printf considers it unsigned (due to the MSBs being 0) and you see the positive value. To get a negative in this you need to use '%hd' in first case. In the second case you assigned to '4 bytes' memory and the MSB got its SIGN bit '1' (means negative) during assignment and hence you see the negative number in '%d' of printf. Hope it explains. For more clarification please comment on the answer.
NB: I used 'MSB' for a shorthand of higher-order byte(s). Please read it according to the context (e.g., 'SIGN bit' will make you read like 'Most Significant Bit'). Thanks.
See this code snippet
int main()
{
unsigned int a = 1000;
int b = -1;
if (a>b) printf("A is BIG! %d\n", a-b);
else printf("a is SMALL! %d\n", a-b);
return 0;
}
This gives the output: a is SMALL: 1001
I don't understand what's happening here. How does the > operator work here? Why is "a" smaller than "b"? If it is indeed smaller, why do i get a positive number (1001) as the difference?
Binary operations between different integral types are performed within a "common" type defined by so called usual arithmetic conversions (see the language specification, 6.3.1.8). In your case the "common" type is unsigned int. This means that int operand (your b) will get converted to unsigned int before the comparison, as well as for the purpose of performing subtraction.
When -1 is converted to unsigned int the result is the maximal possible unsigned int value (same as UINT_MAX). Needless to say, it is going to be greater than your unsigned 1000 value, meaning that a > b is indeed false and a is indeed small compared to (unsigned) b. The if in your code should resolve to else branch, which is what you observed in your experiment.
The same conversion rules apply to subtraction. Your a-b is really interpreted as a - (unsigned) b and the result has type unsigned int. Such value cannot be printed with %d format specifier, since %d only works with signed values. Your attempt to print it with %d results in undefined behavior, so the value that you see printed (even though it has a logical deterministic explanation in practice) is completely meaningless from the point of view of C language.
Edit: Actually, I could be wrong about the undefined behavior part. According to C language specification, the common part of the range of the corresponding signed and unsigned integer type shall have identical representation (implying, according to the footnote 31, "interchangeability as arguments to functions"). So, the result of a - b expression is unsigned 1001 as described above, and unless I'm missing something, it is legal to print this specific unsigned value with %d specifier, since it falls within the positive range of int. Printing (unsigned) INT_MAX + 1 with %d would be undefined, but 1001u is fine.
On a typical implementation where int is 32-bit, -1 when converted to an unsigned int is 4,294,967,295 which is indeed ≥ 1000.
Even if you treat the subtraction in an unsigned world, 1000 - (4,294,967,295) = -4,294,966,295 = 1,001 which is what you get.
That's why gcc will spit a warning when you compare unsigned with signed. (If you don't see a warning, pass the -Wsign-compare flag.)
You are doing unsigned comparison, i.e. comparing 1000 to 2^32 - 1.
The output is signed because of %d in printf.
N.B. sometimes the behavior when you mix signed and unsigned operands is compiler-specific. I think it's best to avoid them and do casts when in doubt.
#include<stdio.h>
int main()
{
int a = 1000;
signed int b = -1, c = -2;
printf("%d",(unsigned int)b);
printf("%d\n",(unsigned int)c);
printf("%d\n",(unsigned int)a);
if(1000>-1){
printf("\ntrue");
}
else
printf("\nfalse");
return 0;
}
For this you need to understand the precedence of operators
Relational Operators works left to right ...
so when it comes
if(1000>-1)
then first of all it will change -1 to unsigned integer because int is by default treated as unsigned number and it range it greater than the signed number
-1 will change into the unsigned number ,it changes into a very big number
Find a easy way to compare, maybe useful when you can not get rid of unsigned declaration, (for example, [NSArray count]), just force the "unsigned int" to an "int".
Please correct me if I am wrong.
if (((int)a)>b) {
....
}
The hardware is designed to compare signed to signed and unsigned to unsigned.
If you want the arithmetic result, convert the unsigned value to a larger signed type first. Otherwise the compiler wil assume that the comparison is really between unsigned values.
And -1 is represented as 1111..1111, so it a very big quantity ... The biggest ... When interpreted as unsigned.
while comparing a>b where a is unsigned int type and b is int type, b is type casted to unsigned int so, signed int value -1 is converted into MAX value of unsigned**(range: 0 to (2^32)-1 )**
Thus, a>b i.e., (1000>4294967296) becomes false. Hence else loop printf("a is SMALL! %d\n", a-b); executed.