在C ++ / C中获得错误的绝对值

Question

I am trying to find the absolute value in C++ by the following method:

#include <cmath>
unsigned value1=4;
unsigned value2=10;

unsigned absoluteValue=abs(value1-value2);
int absValue=abs(value1-value2);

Answer: absoluteValue=4294967294
        absValue=-2147483648

Desired Answer=6

The value that I am getting for absoluteValue is 4294967294...which is wrong. The answer is also wrong for int. I need to compute the absolute value several times in code..is there a more efficient way to achieve this?

Is there an efficient way to find absolute value for integers?

Answer 1

By subtract unsigned values like this, it might wrap.

If you want to get the absolute value of a unsigned do something like this:

unsigned absoluteValue = (value1>value2)?(value1-value2):(value2-value1);undeflow

Answer 2

First of all in C++ such call is ambiguous. If you compiled the code in C then your result shall be equal to 6 in the both cases. The result you got can be get if you will cast the argument to double. For example

    absValue = std::abs( double( value1 -value2 ) );
    std::cout << "absoluteValue = " << absoluteValue << ", absValue = " << absValue << std::endl;

Compare two results of the function call in the following code snippet

    unsigned value1 = 4;
    unsigned value2 = 10;

    unsigned absoluteValue = std::abs( int( value1-value2 ) );
    int absValue = std::abs( int( value1 -value2 ) );
    std::cout << "absoluteValue = " << absoluteValue << ", absValue = " << absValue << std::endl;
    std::cout << std::endl;

    absoluteValue = std::abs( double( value1-value2 ) );
    absValue = std::abs( double( value1 -value2 ) );
    std::cout << "absoluteValue = " << absoluteValue << ", absValue = " << absValue << std::endl;
    std::cout << std::endl;

The output will be

absoluteValue = 6, absValue = 6

absoluteValue = 4294967290, absValue = -2147483648

Why is there such a difference?

Expression value1 - valu2 has type unsigned int. Here is two-complement arithmetic. Then function abs is applied in the first case the expression is converted to signed int and will be equal to -6 according to the two-complement arithmetic. So function will return 6 for the both variable.

In the second case the (positive due to the type unsigned int) expression is converted to double and the function returns a double. It will be a positive value equal to 4294967290. Now this value for the second variable is converted to signed int and you get -2147483648

Answer 3

实际上，在您的C ++实现中， value1-value2等于4294967294，因为value1和value2是无符号的，因此是模类型。

Answer 4

Well, first off, you should be using signed integers, not unsigned. Then it's basically

 
 
 
  
  (ab)<0 ? -(ab) : (ab)
 
  

eg

 
 
 
  
  int main() { int a=24; int b=10; unsigned ab=(ab<0) ? -(ab) : (ab); printf("%lu\\n", ab); }
 
  

KoKuToru's answer does this right for unsigned.

Answer 5

You have to cast these values to integer first. Try:

#include <cmath>
unsigned value1=4;
unsigned value2=10;

unsigned absoluteValue=abs(static_cast<int>(value1)-static_cast<int>(value2));
int absValue=abs(static_cast<int>(value1)-static_cast<int>(value2));

Answer 6

What about this?

unsigned int abs( int a )
 {
   if( a < 0 )
      return a * (-1);
   return a;
 }

Answer 7

Cast value1 and value2 to long long

take absolute value of difference

then cast answer back to unsigned - it will always fit