Unsigned int not working C++ - c++

Following are different programs/scenarios using unsigned int with respective outputs. I don't know why some of them are not working as intended.
Expected output: 2
Program 1:
int main()
{
int value = -2;
std::cout << (unsigned int)value;
return 0;
}
// OUTPUT: 4294967294
Program 2:
int main()
{
int value;
value = -2;
std::cout << (unsigned int)value;
return 0;
}
// OUTPUT: 4294967294
Program 3:
int main()
{
int value;
std::cin >> value; // 2
std::cout << (unsigned int)value;
return 0;
}
// OUTPUT: 2
Can someone explain why Program 1 and Program 2 don't work? Sorry, I'm new at coding.

You are expecting the cast from int to unsigned int to simply change the sign of a negative value while maintaining its magnitude. But that isn't how it works in C or C++. when it comes to overflow, unsigned integers follow modular arithmetic, meaning that assigning or initializing from negatives values such as -1 or -2 wraps around to the largest and second largest unsigned values, and so on. So, for example, these two are equivalent:
unsigned int n = -1;
unsigned int m = -2;
and
unsigned int n = std::numeric_limits<unsigned int>::max();
unsigned int m = std::numeric_limits<unsigned int>::max() - 1;
See this working example.
Also note that there is no substantial difference between programs 1 and 2. It is all down to the sign of the value used to initialize or assign to the unsigned integer.

Casting a value from signed to unsigned changes how the single bits of the value are interpreted. Lets have a look at a simple example with an 8 bit value like char and unsigned char.
The values of a character value range from -128 to 127. Including the 0 these are 256 (2^8) values. Usually the first bit indicates wether the value is negativ or positive. Therefore only the last 7 bits can be used to describe the actual value.
An unsigned character can't take any negative values because there is no bit to determine wether the value should be negative or positiv. Therfore its value ranges from 0 to 256.
When all bits are set (1111 1111) the unsigned character will have the value 256. However the simple character value will treat the first bit as an indicator for a negative value. Sticking to the two's complement this value will be -1.
This is the reason the cast from int to unsigned int does not what you expected it to do, but it does exactly what its supposed to do.
EDIT
If you just want to switch from negative to positive values write yourself a simple function like that
uint32_t makeUnsigned(int32_t toCast)
{
if (toCast < 0)
toCast *= -1;
return static_cast<uint32_t>(toCast);
}
This way you will convert your incoming int to an unsigned int with an maximal value of 2^32 - 1

Related

Sum signed 32-bit int with unsigned 64bit int

On my application, I receive two signed 32-bit int and I have to store them. I have to create a sort of counter and I don't know when it will be reset, but I'll receive big values and frequently. Beacause of that, in order to store these values, I decided to use two unsigned 64-bit int.
The following could be a simple version of the counter.
struct Counter
{
unsigned int elementNr;
unsigned __int64 totalLen1;
unsigned __int64 totalLen2;
void UpdateCounter(int len1, int len2)
{
if(len1 > 0 && len2 > 0)
{
++elementNr;
totalLen1 += len1;
totalLen2 += len2;
}
}
}
I know that if a smaller type is casted to a bigger one (e.g. int to long) there should be no issues. However, passing from 32 bit rappresentation to 64 bit rappresentation and from signed to unsigned at the same time, is something new for me.
Reading around, I undertood that len1 should be expanded from 32 bit to 64 bit and then applied sign extension. Because the unsigned int and signen int have the same rank (Section 4.13), the latter should be converted.
If len1 stores a negative value, passing from signed to unsigned will return a wrong value, this is why I check the positivy at the beginning of the function. However, for positive values, there
should be no issues I think.
For clarity I could revrite UpdateCounter(int len1, int len2) like this
void UpdateCounter(int len1, int len2)
{
if(len1 > 0 && len2 > 0)
{
++elementNr;
__int64 tmp = len1;
totalLen1 += static_cast<unsigned __int64>(tmp);
tmp = len2;
totalLen2 += static_cast<unsigned __int64>(tmp);
}
}
Might there be some side effects that I have not considered.
Is there another better and safer way to do that?
A little background, just for reference: binary operators such arithmetic addition work on operands of the same type (the specific CPU instruction to which is translated depends on the number representation that must be the same for both instruction operands).
When you write something like this (using fixed width integer types to be explicit):
int32_t a = <some value>;
uint64_t sum = 0;
sum += a;
As you already know this involves an implicit conversion, more specifically an
integral promotion according to integer conversion rank.
So the expression sum += a; is equivalent to sum += static_cast<uint64_t>(a);, so a is promoted having the lesser rank.
Let's see what happens in this example:
int32_t a = 60;
uint64_t sum = 100;
sum += static_cast<uint64_t>(a);
std::cout << "a=" << static_cast<uint64_t>(a) << " sum=" << sum << '\n';
The output is:
a=60 sum=160
So all is all ok as expected. Let's se what happens adding a negative number:
int32_t a = -60;
uint64_t sum = 100;
sum += static_cast<uint64_t>(a);
std::cout << "a=" << static_cast<uint64_t>(a) << " sum=" << sum << '\n';
The output is:
a=18446744073709551556 sum=40
The result is 40 as expected: this relies on the two's complement integer representation (note: unsigned integer overflow is not undefined behaviour) and all is ok, of course as long as you ensure that the sum does not become negative.
Coming back to your question you won't have any surprises if you always add positive numbers or at least ensuring that sum will never be negative... until you reach the maximum representable value std::numeric_limits<uint64_t>::max() (2^64-1 = 18446744073709551615 ~ 1.8E19).
If you continue to add numbers indefinitely sooner or later you'll reach that limit (this is valid also for your counter elementNr).
You'll overflow the 64 bit unsigned integer by adding 2^31-1 (2147483647) every millisecond for approximately three months, so in this case it may be advisable to check:
#include <limits>
//...
void UpdateCounter(const int32_t len1, const int32_t len2)
{
if( len1>0 )
{
if( static_cast<decltype(totalLen1)>(len1) <= std::numeric_limits<decltype(totalLen1)>::max()-totalLen1 )
{
totalLen1 += len1;
}
else
{// Would overflow!!
// Do something
}
}
}
When I have to accumulate numbers and I don't have particular requirements about accuracy I often use double because the maximum representable value is incredibly high (std::numeric_limits<double>::max() 1.79769E+308) and to reach overflow I would need to add 2^32-1=4294967295 every picoseconds for 1E+279 years.

Returning a big number in for loop

I was working on a problem and got stuck into this silly error and I can't solve it.
Basically I am using a for loop and reading a character from stream. When the character is '-' I am decreasing my integer by one and when it is '+' I am increasing it by one.
I used an unsigned int because I don't want negative numbers. Here is an example of code:
char x;
unsigned int number = 0;
for (int i = 0; i < n; i++){
cin >> x;
if (x == '-'){
number--;
}else if (x == '+'){
number++;
}
}
cout << number;
And it shows a number something like this 4294967293.
Where is the problem?
If your problem statement says that final resultant number will not be negative, it does not mean that the intermediate numbers will also be positive only. Because, there can be a sequence of streams with characters like - + + which leads to values like: 0, -1, 0, 1. Here, final answer is positive but intermediate numbers are still negative.
Now, you are trying to hold both positive and negative numbers in unsigned int datatype which is leads to wrong output here in above example. Because, if number is 0 and you try to apply negation, it will be 4294967295 (maximum value for a variable of type unsigned int) instead of being -1.
So, instead of using unsigned int datatype , you can use int datatype as suggested by Kaidul.
This is due to wrapping around of unsigned data type. Since it's unsigned, it can't be negative. So negation operation wraps around and yields boundary values of 32 bit integer.
Replace
unsigned int number = 0;
with
int number = 0;

what happens with the negative of an unsigned integer?

Since the literal 0xffffffff needs 32 digits, it can be represented as an unsigned int but not as a signed int, and is of type unsigned int. But what happens with the negative of an unsigned integer?
#include <iostream>
#include <limits>
int main()
{
int N[] = {0,0,0};
if ( std::numeric_limits<long int>::digits==63 and
std::numeric_limits<int>::digits==31 and
std::numeric_limits<unsigned int>::digits==32 )
{
for (long int i = -0xffffffff; i ; --i)
{
N[i] = 1;
}
}
else
{
N[1]=1;
}
std::cout << N[0] <<N [1] << N[2];
}
output : 010
There is no such thing as a negative unsigned integer, by definition.
When you go beyond the lower bounds of an unsigned integer, the value "wraps around" starting from the highest possible value. (The same occurs vice versa).
This mechanism is also triggered when converting a negative "signed" value to an unsigned one.
So, the signed value -1 is converted to the unsigned value $maximumUnsignedValue. Similarly, the signed value -$maximumSignedValue is converted to the unsigned value $maximumUnsignedValue - $maximumSignedValue + 1.

Binary-Decimal Negative bit set

How can I tell if a binary number is negative?
Currently I have the code below. It works fine converting to Binary. When converting to decimal, I need to know if the left most bit is 1 to tell if it is negative or not but I cannot seem to figure out how to do that.
Also, instead of making my Bin2 function print 1's an 0's, how can I make it return an integer? I didn't want to store it in a string and then convert to int.
EDIT: I'm using 8 bit numbers.
int Bin2(int value, int Padding = 8)
{
for (int I = Padding; I > 0; --I)
{
if (value & (1 << (I - 1)))
std::cout<< '1';
else
std::cout<<'0';
}
return 0;
}
int Dec2(int Value)
{
//bool Negative = (Value & 10000000);
int Dec = 0;
for (int I = 0; Value > 0; ++I)
{
if(Value % 10 == 1)
{
Dec += (1 << I);
}
Value /= 10;
}
//if (Negative) (Dec -= (1 << 8));
return Dec;
}
int main()
{
Bin2(25);
std::cout<<"\n\n";
std::cout<<Dec2(11001);
}
You are checking for negative value incorrectly. Do the following instead:
bool Negative = (value & 0x80000000); //It will work for 32-bit platforms only
Or may be just compare it with 0.
bool Negative = (value < 0);
Why don't you just compare it to 0. Should work fine and almost certainly you can't do this in a manner more efficient than the compiler.
I am entirely unclear if this is what the OP is looking for, but its worth a toss:
If you know you have a value in a signed int that is supposed to be representing a signed 8-bit value, you can pull it apart, store it in a signed 8-bit value, then promote it back to a native int signed value like this:
#include <stdio.h>
int main(void)
{
// signed integer, value is 245. 8bit signed value is (-11)
int num = 0xF5;
// pull out the low 8 bits, storing them in a signed char.
signed char ch = (signed char)(num & 0xFF);
// now let the signed char promote to a signed int.
int res = ch;
// finally print both.
printf("%d ==> %d\n",num, res);
// do it again for an 8 bit positive value
// this time with just direct casts.
num = 0x70;
printf("%d ==> %d\n", num, (int)((signed char)(num & 0xFF)));
return 0;
}
Output
245 ==> -11
112 ==> 112
Is that what you're trying to do? In short, the code above will take the 8bits sitting at the bottom of num, treat them as a signed 8-bit value, then promote them to a signed native int. The result is you can now "know" not only whether the 8-bits were a negative number (since res will be negative if they were), you also get the 8-bit signed number as a native int in the process.
On the other hand, if all you care about is whether the 8th bit is set in the input int, and is supposed to denote a negative value state, then why not just :
int IsEightBitNegative(int val)
{
return (val & 0x80) != 0;
}

>> and << and data types in C++

I have looked over the guide given in this answer, but I still don't understand bit-shifting. In particular I am confused about the data types come into play.
The following:
unsigned int a = pow(2,31);
cout << (a << 1);
indeed produces 0 as I expect because the int is 32 bits, so moving the 1 to the left, pushes it into nothing.
But the following
unsigned int a = 1;
unsigned char b = (unsigned char)a;
cout << (unsigned int)(b<<8);
produces 256. Why is that? My guess would have been that a char is 8 bit and so moving 1 left 8 places should give zero.
Is there a function/shift that does this? (i.e. evaluates 1<<8 to 0).
Narrow integral values are promoted to int or unsigned int before being used. It's called integral promotion.