What will the unsigned int contain when I overflow it? To be specific, I want to do a multiplication with two unsigned ints: what will be in the unsigned int after the multiplication is finished?
unsigned int someint = 253473829*13482018273;
unsigned numbers can't overflow, but instead wrap around using the properties of modulo.
For instance, when unsigned int is 32 bits, the result would be: (a * b) mod 2^32.
As CharlesBailey pointed out, 253473829*13482018273 may use signed multiplication before being converted, and so you should be explicit about unsigned before the multiplication:
unsigned int someint = 253473829U * 13482018273U;
Unsigned integer overflow, unlike its signed counterpart, exhibits well-defined behaviour.
Values basically "wrap" around. It's safe and commonly used for counting down, or hashing/mod functions.
It probably depends a bit on your compiler. I had errors like this years ago, and sometimes you would get runtime error, other times it would basically "wrap" back to a really small number that would result from chopping off the highest level bits and leaving the remainder, i.e if it's a 32 bit unsigned int, and the result of your multiplication would be a 34 bit number, it would chop off the high order 2 bits and give you the remainder. You would probably have to try it on your compiler to see exactly what you get, which may not be the same thing you would get with a different compiler, especially if the overflow happens in the middle of an expression where the end result is within the range of an unsigned int.
Related
I have a problem in which I need to declare some variables as natural numbers. Which is the propper fundamental type that I should use for variables that should be natural numbers ? Like for integers is int ...
The following types resemble natural numbers set with 0 included in C++:
unsigned char
unsigned short int
unsigned int
unsigned long int
unsigned long long int, since C++11.
Each one differs with the other in the range of values it can represent.
Notice that a computer (and perhaps even the entire universe) is a finite machine; it has a finite (but very large number) of bits (my laptop has probably less than 1015 bits).
Of course int are not the mathematical integers. On my machine int is a 32 bits signed integer (and long is a 64 bits signed integer), so int-s have only 232 possible values (and that is much less than the infinite cardinal of mathematical integers).
So a computer can only represent a finite set of numbers, but quite a large one. That is smaller than the infinite set of natural numbers (remember, some of them are not representable on the entire Earth; read about Richard's paradox).
You might want to use unsigned (same as unsigned int, on my machine represents natural numbers up to 232-1), unsigned long, unsigned long long or (from <stdint.h>) types like uint32_t, uint64_t ... you would get unsigned binary numbers of 32 or 64 bits. Some compilers and implementations might know about uint128_t or something similar.
If that is not enough, consider using big ints. You could use a library like GMPlib (but even a big computer is not able to represent extremely large natural numbers -with all their bits-..., and your own brain cannot comprehend them neither).
If you need numbers that can't be negative, your best bet would be unsigned int. If you want to learn more about data types, you can check this site
There's not any particular data type representing natural numbers. But you can use data types for whole numbers and then make some appropriate edits. Here are a few ways to declare whole numbers:
- unsigned short int
- unsigned int
- unsigned long int
- unsigned long long int
Suppose I have declared following variables :
some_dataType a,b,c; //a,b,c are positive
int res;
res=a+b-c;
And I have been provided with the information that the variable res cannot cross integer limit. Just to avoid overflow from the expression a+b in a+b-c I have following options-
declare a,b,c as long int
(a-c)+b
(long int)(a+b-c) //or some other type casting
My question is which of the above is good practice. Also, just to escape from all this, I prefer doing option 1. But it may increase the memory size of the program (not for this program though, but for large applications).
Assumption : long int is greater than int Thanks #Retired Ninja
There's no simple, general, portable way to avoid integer overflow.
Using a wider integer type is not a general solution, because there's no guarantee that there is a wider integer type. It's very common for the types int and long to be the same size on some systems. 32-bit systems typically make both int and long 32 bits; even some 64-bit systems do the same thing. And some 64-bit systems make both int and long 64 bits.
You wrote:
And I have been provided with the information that the variable res cannot cross integer limit.
It's best to be careful with terminology. Although the type name int is obviously an abbreviation of the word "integer", the words are not synonymous. In C and C++, int is just one of several integer types; others include unsigned char and long long.
Presumably what you mean is that the value of res must not be outside the range of type int, which is INT_MIN to INT_MAX.
If you use long long for the result, it's likely that you can avoid overflow; you can then compare the result to the values INT_MIN and INT_MAX, and take whatever corrective action is appropriate if it's out of range. But it's still possible for both int and long long to be 64 bits. What you do with that depends on how portable your code needs to be.
You cannot safely check whether a signed integer addition or subtraction overflowed after the fact. An overflow in signed arithmetic causes undefined behavior. Typically the result wraps around, but in principle your program could crash before you have a chance to examine the result.
Some compilers might provide functions that perform safe signed arithmetic and tell you whether there was an overflow. Consult your compiler's documentation.
There are ways to test the operands before performing the operation. They're complicated, and I'm too lazy to work out the details, but briefly:
If the operands of + are of opposite signs, or if one operand is 0, the addition is safe.
If both operands are positive, the addition is safe if x does not exceed INT_MAX - y.
If both operands are negative, the addition is safe if y is no less than INT_MIN - y.
I do not guarantee that the above is completely correct *I just wrote it off the top of my head), but in most cases it's probably more effort than it's worth.
Trust, but verify:
assert(a > 0);
assert(b > 0);
assert(c > 0);
assert((long)a - (long)c + (long)b <= INT_MAX);
int res = a - c + b;
I'm working on a relatively simple problem based around adding all the primes under a certain value together. I've written a program that should accomplish this task. I am using long type variables. As I get up into higher numbers (~200/300k), the variable I am using to track the sum becomes negative despite the fact that no negative values are being added to it (based on my knowledge and some testing I've done). Is there some issue with the data type or I am missing something.
My code is below (in C++) [Vector is basically a dynamic array in case people are wondering]:
bool checkPrime(int number, vector<long> & primes, int numberOfPrimes) {
for (int i=0; i<numberOfPrimes-1; i++) {
if(number%primes[i]==0) return false;
}
return true;
}
long solveProblem10(int maxNumber) {
long sumOfPrimes=0;
vector<long> primes;
primes.resize(1);
int numberOfPrimes=0;
for (int i=2; i<maxNumber; i++) {
if(checkPrime(i, primes, numberOfPrimes)) {
sumOfPrimes=sumOfPrimes+i;
primes[numberOfPrimes]=long(i);
numberOfPrimes++;
primes.resize(numberOfPrimes+1);
}
}
return sumOfPrimes;
}
Integers represent values use two's complement which means that the highest order bit represents the sign. When you add the number up high enough, the highest bit is set (an integer overflow) and the number becomes negative.
You can resolve this by using an unsigned long (32-bit, and may still overflow with the values you're summing) or by using an unsigned long long (which is 64 bit).
the variable I am using to track the sum becomes negative despite the fact that no negative values are being added to it (based on my knowledge and some testing I've done)
longs are signed integers. In C++ and other lower-level languages, integer types have a fixed size. When you add past their maximum they will overflow and wrap-around to negative numbers. This is due to the behavior of how twos complement works.
check valid integer values: Variables. Data Types.
you're using signed long, which is usually 32 bit, which means -2kkk - 2kkk, you can either use unsigned long, which is 0-4kkk, or use 64 bit (un)signed long long
if you need values bigger 2^64 (unsigned long long), you will need to use bignum math
long is probably only 32 bits on your system - use uint64_t for the sum - this gives you a guaranteed 64 bit unsigned integer.
#include <cstdint>
uint64_t sumOfPrimes=0;
You can include header <cstdint> and use type std::uintmax_t instead of long.
This question already has answers here:
Difference between unsigned and unsigned int in C
(5 answers)
Closed 9 years ago.
I saw in some C++ code the keyword "unsigned" in the following form:
const int HASH_MASK = unsigned(-1) >> 1;
and later:
unsigned hash = HASH_SEED;
(it is taken from the CS106B/X reader - of Stanford - by Eric S. Roberts - on the topic of "implementation of the hash code function for strings").
Can someone tell me please what does that keyword mean and when do I use it anyway?
Thanks!
Take a look: https://stackoverflow.com/a/7176690/1758762
unsigned is a modifier which can apply to any integral type (char,
short, int, long, etc.) but on its own it is identical to unsigned
int.
It's a short version of unsigned int. Syntactically, you can use it anywhere you would use any other datatype like float or short.
Unsigned types are types that can't represent negative numbers; only zero and positive numbers. In C++, they use modular arithmetic; the modulus for an N-bit type is 2^N. It's a good idea to use unsigned rather than signed types when messing around with bit patterns (for example, when calculating hash codes), since C++ allows several different representations of negative numbers which could lead to portability issues.
unsigned can be used as a qualifier for any integer type (e.g. unsigned int or unsigned long long); or on its own as shorthand for unsigned int.
So the first converts -1 into unsigned int. Due to modular arithmetic, this gives the largest representable value. This could also be written (more clearly, in my opinion) as std::numeric_limits<unsigned>::max().
The second declares and initialises a variable of type unsigned int.
Values are signed by default, which means they can be positive or negative. The unsigned keyword is used to specify that a value must be positive.
Signed variables use 1 bit to specify whether the value is positive or not. The unsigned keyword actualy makes this bit part of the value (thus allowing bigger numbers to be stored).
Lastly, unsigned hash is interpreted by compilers as unsigned int hash (int being the default type in C programming).
To get a good idea what unsigned means, one has to understand signed and unsigned integers. For a full explanation of twos-compliment, search Wikipedia, but in a nutshell, a computer stores negative numbers by subtracting negative numbers from 2^32 (for a 32-bit integer). In this way, -1 is stored as 2^32-1. This does mean that you only have 2^31 positive numbers, but that is by the by. This is known as signed integers (as it can have positive or negative sign)
Unsigned tells the compiler that you don't want twos compliment and are dealing only in positive numbers. When -1 is typecast (as it is in the code) to an unsigned int it becomes
2^32-1 = 0b111111111...
Thus that is an easy way of getting a whole lot of 1s in binary.
Use unsigned rarely. If you need to do bit operations, or for some reason need only positive integers bigger than 2^31. Otherwise, if you leave it out, c++ assumes signed integers.
C allows chars to be signed or unsigned, depending on which is more efficient for the host computer. if you want to be sure your char is unsigned, you can declare your variable to be unsigned char. You can use signed char if you want the ensure signed interpretation.
Incidentally, the C and C++ compilers treatd char, signed char, and unsigned char as three distinct types, even though char is compiled into one of the other two.
I want signed integers to overflow when they become too big. How do I achieve that without using the next biggest datatype (or when I am already at int128_t)?
For example using 8bit integers 19*12 is commonly 260, but I want the result 1 11 10 01 00 with the 9th bit cut off, thus -27.
Signed overflow is undefined in C, and that's for real.
One solution follows:
signed_result = (unsigned int)one_argument + (unsigned int)other_argument;
The above solution involves implementation-defined behavior in the final conversion from unsigned to int but do not invoke undefined behavior. With most compilation platforms' implementation-defined choices, the result is exactly the two's complement result that you expect.
Finally, an optimizing compiler for one of the numerous platforms on which implementation-defined choices force the compiler to give you the behavior you expect will compile the above code to the obvious assembly instruction.
Alternately, if you are using gcc, then the options -fwrapv/-fno-strict-overflow may be exactly what you want. They provide an additional guarantee with respect to the standard that signed overflows wrap around. I'm not sure about the difference between the two.
Signed integer overflow is undefined according to both C and C++ standards. There's no way to accomplish what you want without a specific platform in mind.
It is possible to do this in a correct standard C manner, so long as you have access to an unsigned type that is of the same width as your signed type (that is, has one more value bit). To demonstrate with int64_t:
int64_t mult_wrap_2scomp(int64_t a, int64_t b)
{
uint64_t result = (uint64_t)a * (uint64_t)b;
if (result > INT64_MAX)
return (int64_t)(result - INT64_MAX - 1) - INT64_MAX - 1;
else
return (int64_t)result;
}
This does not produce any problematic intermediate results.
You could create an objective wrapper around int, but that would involve quite a lot of overhead code.
Assuming two's complement signed integer arithmetic (which is a reasonable assumption these days), for addition and subtraction, just cast to unsigned to do the calculation. For multiplication and division, ensure the operands are positive, cast to unsigned, calculate and adjust the signs.
It sounds like you want to do unsinged integer arithmetic, then stuff the result into a signed integer:
unsigned char a = 19;
unsigned char b = 12;
signed char c = (signed char)(a*b);
should give you what you're looking for. Let us know if it doesn't.
Use bigger datatypes. With GMP you will have all the space you probably need.