This code gives the meaningful output
#include <iostream>
int main() {
unsigned int ui = 100;
unsigned int negative_ui = -22u;
std::cout << ui + negative_ui << std::endl;
}
Output:
78
The variable negative_ui stores -22, but is an unsigned int.
My question is why does unsigned int negative_ui = -22u; work.
How can an unsigned int store a negative number? Is it save to be used or does this yield undefined behaviour?
I use the intel compiler 18.0.3. With the option -Wall no warnings occurred.
Ps. I have read What happens if I assign a negative value to an unsigned variable? and Why unsigned int contained negative number
How can an unsigned int store a negative number?
It doesn't. Instead, it stores a representable number that is congruent with that negative number modulo the number of all representable values. The same is also true with results that are larger than the largest representable value.
Is it save to be used or does this yield undefined behaviour?
There is no UB. Unsigned arithmetic overflow is well defined.
It is safe to rely on the result. However, it can be brittle. For example, if you add -22u and 100ull, then you get UINT_MAX + 79 (i.e. a large value assuming unsigned long long is a larger type than unsigned) which is congruent with 78 modulo UINT_MAX + 1 that is representable in unsigned long long but not representable in unsigned.
Note that signed arithmetic overflow is undefined.
Signed/Unsigned is a convention. It uses the last bit of the variable (in case of x86 int, the last 31th bit). What you store in the variable takes the full bit length.
It's the calculations that follow that take the upper bit as a sign indicator or ignore it. Therefore, any "unsigned" variable can contain a signed value which will be converted to the unsigned form when the unsigned variable participates in a calculation.
unsigned int x = -1; // x is now 0xFFFFFFFF.
x -= 1; // x is now 0xFFFFFFFE.
if (x < 0) // false. x is compared as 0xFFFFFFFE.
int x = -1; // x stored as 0xFFFFFFFF
x -= 1; // x stored as 0xFFFFFFFE
if (x < 0) // true, x is compared as -2.
Technically valid, bad programming.
Related
I'm new to C++ and is trying to learn the concept of array. I saw this code snippet online. For the sample code below, does it make any difference to declare:
unsigned scores[11] = {};
unsigned grade;
as:
int scores[11] = {};
int grade;
I guess there must be a reason why score[11] = {}; and grade is declared as unsigned, but what is the reason behind it?
int main() {
unsigned scores[11] = {};
unsigned grade;
while (cin >> grade) {
if (0 <= grade <= 100) {
++scores[grade / 10];
}
}
for (int i = 0; i < 11; i++) {
cout << scores[i] << endl;
}
}
unsigned means that the variable will not hold a negative values (or even more accurate - It will not care about the sign-). It seems obvious that scores and grades are signless values (no one scores -25). So, it is natural to use unsigned.
But note that: if (0 <= grade <= 100) is redundant. if (grade <= 100) is enough since no negative values are allowed.
As Blastfurnace commented, if (0 <= grade <= 100) is not right even. if you want it like this you should write it as:
if (0 <= grade && grade <= 100)
Unsigned variables
Declaring a variable as unsigned int instead of int has 2 consequences:
It can't be negative. It provides you a guarantee that it never will be and therefore you don't need to check for it and handle special cases when writing code that only works with positive integers
As you have a limited size, it allows you to represent bigger numbers. On 32 bits, the biggest unsigned int is 4294967295 (2^32-1) whereas the biggest int is 2147483647 (2^31-1)
One consequence of using unsigned int is that arithmetic will be done in the set of unsigned int. So 9 - 10 = 4294967295 instead of -1 as no negative number can be encoded on unsigned int type. You will also have issues if you compare them to negative int.
More info on how negative integer are encoded.
Array initialization
For the array definition, if you just write:
unsigned int scores[11];
Then you have 11 uninitialized unsigned int that have potentially values different than 0.
If you write:
unsigned int scores[11] = {};
Then all int are initialized with their default value that is 0.
Note that if you write:
unsigned int scores[11] = { 1, 2 };
You will have the first int intialized to 1, the second to 2 and all the others to 0.
You can easily play a little bit with all these syntax to gain a better understanding of it.
Comparison
About the code:
if(0 <= grade <= 100)
as stated in the comments, this does not do what you expect. In fact, this will always evaluate to true and therefore execute the code in the if. Which means if you enter a grade of, say, 20000, you should have a core dump. The reason is that this:
0 <= grade <= 100
is equivalent to:
(0 <= grade) <= 100
And the first part is either true (implicitly converted to 1) or false (implicitly converted to 0). As both values are lower than 100, the second comparison is always true.
unsigned integers have some strange properties and you should avoid them unless you have a good reason. Gaining 1 extra bit of positive size, or expressing a constraint that a value may not be negative, are not good reasons.
unsigned integers implement arithmetic modulo UINT_MAX+1. By contrast, operations on signed integers represent the natural arithmetic that we are familiar with from school.
Overflow semantics
unsigned has well defined overflow; signed does not:
unsigned u = UINT_MAX;
u++; // u becomes 0
int i = INT_MAX;
i++; // undefined behaviour
This has the consequence that signed integer overflow can be caught during testing, while an unsigned overflow may silently do the wrong thing. So use unsigned only if you are sure you want to legalize overflow.
If you have a constraint that a value may not be negative, then you need a way to detect and reject negative values; int is perfect for this. An unsigned will accept a negative value and silently overflow it into a positive value.
Bit shift semantics
Bit shift of unsigned by an amount not greater than the number of bits in the data type is always well defined. Until C++20, bit shift of signed was undefined if it would cause a 1 in the sign bit to be shifted left, or implementation-defined if it would cause a 1 in the sign bit to be shifted right. Since C++20, signed right shift always preserves the sign, but signed left shift does not. So use unsigned for some kinds of bit twiddling operations.
Mixed sign operations
The built-in arithmetic operations always operate on operands of the same type. If they are supplied operands of different types, the "usual arithmetic conversions" coerce them into the same type, sometimes with surprising results:
unsigned u = 42;
std::cout << (u * -1); // 4294967254
std::cout << std::boolalpha << (u >= -1); // false
What's the difference?
Subtracting an unsigned from another unsigned yields an unsigned result, which means that the difference between 2 and 1 is 4294967295.
Double the max value
int uses one bit to represent the sign of the value. unsigned uses this bit as just another numerical bit. So typically, int has 31 numerical bits and unsigned has 32. This extra bit is often cited as a reason to use unsigned. But if 31 bits are insufficient for a particular purpose, then most likely 32 bits will also be insufficient, and you should be considering 64 bits or more.
Function overloading
The implicit conversion from int to unsigned has the same rank as the conversion from int to double, so the following example is ill formed:
void f(unsigned);
void f(double);
f(42); // error: ambiguous call to overloaded function
Interoperability
Many APIs (including the standard library) use unsigned types, often for misguided reasons. It is sensible to use unsigned to avoid mixed-sign operations when interacting with these APIs.
Appendix
The quoted snippet includes the expression 0 <= grade <= 100. This will first evaluate 0 <= grade, which is always true, because grade can't be negative. Then it will evaluate true <= 100, which is always true, because true is converted to the integer 1, and 1 <= 100 is true.
Yes it does make a difference. In the first case you declare an array of 11 elements a variable of type "unsigned int". In the second case you declare them as ints.
When the int is on 32 bits you can have values from the following ranges
–2,147,483,648 to 2,147,483,647 for plain int
0 to 4,294,967,295 for unsigned int
You normally declare something unsigned when you don't need negative numbers and you need that extra range given by unsigned. In your case I assume that that by declaring the variables unsigned, the developer doesn't accept negative scores and grades. You basically do a statistic of how many grades between 0 and 10 were introduced at the command line. So it looks like something to simulate a school grading system, therefore you don't have negative grades. But this is my opinion after reading the code.
Take a look at this post which explains what unsigned is:
what is the unsigned datatype?
As the name suggests, signed integers can be negative and unsigned cannot be. If we represent an integer with N bits then for unsigned the minimum value is 0 and the maximum value is 2^(N-1). If it is a signed integer of N bits then it can take the values from -2^(N-2) to 2^(N-2)-1. This is because we need 1-bit to represent the sign +/-
Ex: signed 3-bit integer (yes there are such things)
000 = 0
001 = 1
010 = 2
011 = 3
100 = -4
101 = -3
110 = -2
111 = -1
But, for unsigned it just represents the values [0,7]. The most significant bit (MSB) in the example signifies a negative value. That is, all values where the MSB is set are negative. Hence the apparent loss of a bit in its absolute values.
It also behaves as one might expect. If you increment -1 (111) we get (1 000) but since we don't have a fourth bit it simply "falls off the end" and we are left with 000.
The same applies to subtracting 1 from 0. First take the two's complement
111 = twos_complement(001)
and add it to 000 which yields 111 = -1 (from the table) which is what one might expect. What happens when you increment 011(=3) yielding 100(=-4) is perhaps not what one might expect and is at odds with our normal expectations. These overflows are troublesome with fixed point arithmetic and have to be dealt with.
One other thing worth pointing out is the a signed integer can take one negative value more than it can positive which has a consequence for rounding (when using integer to represent fixed point numbers for example) but am sure that's better covered in the DSP or signal processing forums.
I have two functions add and main as follows.
int add(unsigned int a, unsigned int b)
{
return a+b;
}
int main()
{
unsigned int a,b;
cout << "Enter a value for a: ";
cin >> a;
cout << "Enter a value for b: ";
cin >> b;
cout << "a: " << a << " b: "<<b <<endl;
cout << "Result is: " << add(a,b) <<endl;
return 0;
}
When I run this program I get the following results:
Enter a value for a: -1
Enter a value for b: -2
a: 4294967295 b: 4294967294
Result is: -3
Why is the result -3?
Because add returns an int (no unsigned int) which cannot represent 4294967295 + 4294967294 = 4294967293 (unsigned integer arithmetic is defined mod 2^n with n = 32 in this case) because the result is too big.
Thus, you have signed integer overflow (or, more precisely, an implicit conversion from a source integer that cannot be represented as int) which has an implementation defined result, i.e. any output (that is representable as int) would be "correct".
The reason for getting exactly -3 is that the result is 2^32 - 3 and that gets converted to -3 on your system. But still, note that any result would be equally legal.
int add(unsigned int a, unsigned int b)
{
return a+b;
}
The expression a+b adds two operands of type unsigned int, yielding an unsigned int result. Unsigned addition, strictly speaking, does not "overflow"; rather than result is reduced modulo MAX + 1, where MAX is the maximum value of the unsigned type. In this case, assuming 32-bit unsigned int, the result of adding 4294967295 + 4294967294 is well defined: it's 4294967293, or 232-3.
Since add is defined to return an int result, the unsigned value is implicitly converted from unsigned int to int. Unlike an arithmetic overflow, an unsigned-to-signed conversion that can't be represented in the target type yields an implementation-defined result. On a typical implementation, such a conversion (where the source and target have the same size) will reinterpret the representation, yielding -3. Other results are possible, depending on the implementation, but not particularly likely.
As for why a and b were set to those values in the first place, apparently that's how cin >> a behaves when a is an unsigned value and the input is negative. I'm not sure whether that behavior is defined by the language, implementation-defined, or undefined. In any case, once you have those values, the result returned by add follows as described above.
If you are intending to return an unsigned int then you need to add unsigned to your function declaration. If you changed your return type to an unsigned int
and you use the values -1 & -2 then this will be your output:
a: 4294967295 b: 4294967294
Result: 4294967293
unsigned int ranges from [0, 4294967295] provided an unsigned int is 4bytes in size on your local machine. When you input -1 for an unsigned int you have buffer overflow and what happens here is the compiler will set -1 to be the largest possible valid number in an unsigned int. When you pass -2 into your function the same thing happens but you are being index back to the second largest value. With unsigned int there is no "sign" for negatives stored. If you take the largest possible value of an unsigned as stated above by (a) and add 1 it will give you a value of 0. These values are passed into the function, and the function creates 2 stack variables of local scope within this function.
So now you have a copy of a & b on the stack and a has a value of unsigned int max size and b has a value of unsigned int max size - 1. Now you are performing an addition on these two values which exceeds the max value size of an unsigned int and wrapping occurs in the opposite direction. So if the index value starts at .....95 and you add 1 this gives you 0 for the last digit, so if you take the second value which is max value - 1 : .....94 subtract one from it because we already reached 0 now we are in the positive direction. This will give you a result of ......93 which the function is returning if your return type is unsigned int.
This same concept applies if your return type is int, however what happens here
is this: The addition of the two unsigned values are the same result giving you
.....93 then the compiler will implicitly cast this to an int. Well here the range value for a signed int is -2147483648 to 2147483647 - one bit is used to store the (signed value) but it also depends on if two's compliment is being used etc.
The compiler here is smart enough to know that the signed int has a range of these values but the wrapping still occurs. What happens when we store 4294967293 into a singed int? Well the max value of int in the positive is 2147483647 so if we subtract the two (4294967293 - 2147483647) we would get 2147483646 left over. At this point does this value fit in the range of a signed int max value? It does however because of the implicit casting being done from an unsigned to a signed value we have 2 bits to account for, the signed itself and the wrapping value meaning max_value + 1 = 0 to account for, except this doesn't happen with signed values when you add 1 to max_value you get the largest possible -max_value.
For the unsigned values:
- ...95 = -1
- ...94 = -2
- ...93 = -3
With signed values the negative is preserved in its own bit, or if twos compliment is used etc., pending on the definitions to a signed int within the compiler. Here the compiler recognizes this unsigned value as being negative even though negative values are not stored in an unsigned for the sign is not preserved. So when it explicitly casts from an unsigned to signed one bit is used to store the signed value then the calculations are done and the wrapping occurs from a buffer overflow. So as we subtracted the actual unsigned value that would represent -3 : 4294967293 with the max+ value for a signed int 2147483647 we got the value 2147483646 now with signed ints if you add 1 to the max value it does not give you 0 like an unsigned does, it will give you the largest -signed value which for a 4byte int is -2147483648, since this does fit in the max+ value and the compiler knows that this is supposed to be a negative value if we add our remainder by the -max_value for a signed int : 2147483646 + (-2147483648) this will give us -2, but because of the fact that the wrapping with int is different we have to subtract 1 from this -2 - 1 = -3.
When it converts from an unsigned int to an int, the number is over what a signed integer can hold. This causes it to overflow as a negative value.
If you change the return value to an unsigned integer, your problem should be solved.
I am going through C++ Primer 5th Edition and am currently doing the signed/unsigned section. A quick question I have is when there is a wrap-around, say, in this block of code:
unsigned u = 10;
int i = -42;
std::cout << i + i << std::endl; // prints -84
std::cout << u + i << std::endl; // if 32-bit ints, prints 4294967264
I thought that the max range was 4294967295 with the 0 being counted, so I was wondering why the wrap-around seems to be done from 4294967296 in this problem.
Unsigned arithmetic is modulo (maximum value of the type plus 1).
If maximum value of an unsigned is 4294967295 (2^32 - 1), the result will be mathematically equal to (10-42) modulo 4294967296 which equals 10-42+4294967296 i.e. 4294967264
When an out-of-range value is converted to an unsigned type, the result is the remainder of it modulo the number of values the target unsigned type can hold. For instance, the result of n converted to unsigned char is n % 256, because unsigned char can hold values 0 to 255.
It's similar in your example, the wrap-around is done using 4294967296, the number of values that a 32-bit unsigned integer can hold.
Given unsigned int that is 32 bits you're correct that the range is [0, 4294967295].
Therefore -1 is 4294967295. Which is logically equivalent to 4294967296 - 1 which should explain the behavior you're seeing.
This question is regarding the modulo operator %. We know in general a % b returns the remainder when a is divided by b and the remainder is greater than or equal to zero and strictly less than b. But does the above hold when a and b are of magnitude 10^9 ?
I seem to be getting a negative output for the following code for input:
74 41 28
However changing the final output statement does the work and the result becomes correct!
#include<iostream>
using namespace std;
#define m 1000000007
int main(){
int n,k,d;
cin>>n>>k>>d;
if(d>n)
cout<<0<<endl;
else
{
long long *dp1 = new long long[n+1], *dp2 = new long long[n+1];
//build dp1:
dp1[0] = 1;
dp1[1] = 1;
for(int r=2;r<=n;r++)
{
dp1[r] = (2 * dp1[r-1]) % m;
if(r>=k+1) dp1[r] -= dp1[r-k-1];
dp1[r] %= m;
}
//build dp2:
for(int r=0;r<d;r++) dp2[r] = 0;
dp2[d] = 1;
for(int r = d+1;r<=n;r++)
{
dp2[r] = ((2*dp2[r-1]) - dp2[r-d] + dp1[r-d]) % m;
if(r>=k+1) dp2[r] -= dp1[r-k-1];
dp2[r] %= m;
}
cout<<dp2[n]<<endl;
}
}
changing the final output statement to:
if(dp2[n]<0) cout<<dp2[n]+m<<endl;
else cout<<dp2[n]<<endl;
does the work, but why was it required?
By the way, the code is actually my solution to this question
This is a limit imposed by the range of int.
int can only hold values between –2,147,483,648 to 2,147,483,647.
Consider using long long for your m, n, k, d & r variables. If possible use unsigned long long if your calculations should never have a negative value.
long long can hold values from –9,223,372,036,854,775,808 to 9,223,372,036,854,775,807
while unsigned long long can hold values from 0 to 18,446,744,073,709,551,615. (2^64)
The range of positive values is approximately halved in signed types compared to unsigned types, due to the fact that the most significant bit is used for the sign; When you try to assign a positive value greater than the range imposed by the specified Data Type the most significant bit is raised and it gets interpreted as a negative value.
Well, no, modulo with positive operands does not produce negative results.
However .....
The int type is only guaranteed by the C standards to support values in the range -32767 to 32767, which means your macro m is not necessarily expanding to a literal of type int. It will fit in a long though (which is guaranteed to have a large enough range).
If that's happening (e.g. a compiler that has a 16-bit int type and a 32-bit long type) the results of your modulo operations will be computed as long, and may have values that exceed what an int can represent. Converting that value to an int (as will be required with statements like dp1[r] %= m since dp1 is a pointer to int) gives undefined behaviour.
Mathematically, there is nothing special about big numbers, but computers only have a limited width to write down numbers in, so when things get too big you get "overflow" errors. A common analogy is the counter of miles traveled on a car dashboard - eventually it will show as all 9s and roll round to 0. Because of the way negative numbers are handled, standard signed integers don't roll round to zero, but to a very large negative number.
You need to switch to larger variable types so that they overflow less quickly - "long int" or "long long int" instead of just "int", the range doubling with each extra bit of width. You can also use unsigned types for a further doubling, since no range is used for negatives.
I have this code.
#include <iostream>
int main()
{
unsigned long int i = 1U << 31;
std::cout << i << std::endl;
unsigned long int uwantsum = 1 << 31;
std::cout << uwantsum << std::endl;
return 0;
}
It prints out.
2147483648
18446744071562067968
on Arch Linux 64 bit, gcc, ivy bridge architecture.
The first result makes sense, but I don't understand where the second number came from. 1 represented as a 4byte int signed or unsigned is
00000000000000000000000000000001
When you shift it 31 times to the left, you end up with
10000000000000000000000000000000
no? I know shifting left for positive numbers is essentially 2^k where k is how many times you shift it, assuming it still fits within bounds. Why is it I get such a bizarre number?
Presumably you're interested in why this: unsigned long int uwantsum = 1 << 31; produces a "strange" value.
The problem is pretty simple: 1 is a plain int, so the shift is done on a plain int, and only after it's complete is the result converted to unsigned long.
In this case, however, 1<<31 overflows the range of a 32-bit signed int, so the result is undefined1. After conversion to unsigned, the result remains undefined.
That said, in most typical cases, what's likely to happen is that 1<<31 will give a bit pattern of 10000000000000000000000000000000. When viewed as a signed 2's complement2 number, this is -2147483648. Since that's negative, when it's converted to a 64-bit type, it'll be sign extended, so the top 32 bits will be filled with copies of what's in bit 31. That gives: 1111111111111111111111111111111110000000000000000000000000000000 (33 1-bits followed by 31 0-bits).
If we then treat that as an unsigned 64-bit number, we get 18446744071562067968.
§5.8/2:
The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned type, the value of the result is E1 × 2E2, reduced modulo one more than the maximum value representable in the result type. Otherwise, if E1 has a signed type and non-negative value, and E1×2E2 is representable in the corresponding unsigned type of the result type, then that value, converted to the result type, is the resulting value; otherwise, the behavior is undefined.
In theory, the computer could use 1's complement or signed magnitude for signed numbers--but 2's complement is currently much more common than either of those. If it did use one of those, we'd expect a different final result.
The literal 1 with no U is a signed int, so when you shift << 31, you get integer overflow, generating a negative number (under the umbrella of undefined behavior).
Assigning this negative number to an unsigned long causes sign extension, because long has more bits than int, and it translates the negative number into a large positive number by taking its modulus with 264, which is the rule for signed-to-unsigned conversion.
It's not "bizarre".
Try printing the number in hex and see if it's any more recognizable:
std::cout << std::hex << i << std::endl;
And always remember to qualify your literals with "U", "L" and/or "LL" as appropriate:
http://en.cppreference.com/w/cpp/language/integer_literal
unsigned long long l1 = 18446744073709550592ull;
unsigned long long l2 = 18'446'744'073'709'550'592llu;
unsigned long long l3 = 1844'6744'0737'0955'0592uLL;
unsigned long long l4 = 184467'440737'0'95505'92LLU;
I think it is compiler dependent .
It gives same value
2147483648
2147483648
on my machiene (g++) .
Proof : http://ideone.com/cvYzxN
And if overflow is there , then because uwantsum is unsigned long int and unsigned values are ALWAYS positive , conversion is done from signed to unsigned by using (uwantsum)%2^64 .
Hope this helps !
Its in the way you printed it out.
using formar specifier %lu should represent a proper long int