Is unsigned integer subtraction defined behavior

Question

I have come across code from someone who appears to believe there is a problem subtracting an unsigned integer from another integer of the same type when the result would be negative  So that code like this would be incorrect even if it happens to work on most architectures   unsigned int To  Tf   To   getcounter    while  1        Tf   getcounter        if   Tf-To   gt   TIME LIMIT            break             This is the only vaguely relevant quote from the C standard I could find      A computation involving unsigned operands can never over ow  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    I suppose one could take that quote to mean that when the right operand is larger the operation is adjusted to be meaningful in the context of modulo truncated numbers   i e   0x0000 - 0x0001    0x 1 0000 - 0x0001    0xFFFF  as opposed to using the implementation dependent signed semantics   0x0000 - 0x0001     unsigned  0   -1      0xFFFF but also 0xFFFE or 0x8001   Which or what interpretation is right  Is it defined at all

User · Answer

Well  the first interpretation is correct  However  your reasoning about the  signed semantics  in this context is wrong   Again  your first interpretation is correct  Unsigned arithmetic follow the rules of modulo arithmetic  meaning that 0x0000 - 0x0001 evaluates to 0xFFFF for 32-bit unsigned types   However  the second interpretation  the one based on  signed semantics   is also required to produce the same result  I e  even if you evaluate 0 - 1 in the domain of signed type and obtain -1 as the intermediate result  this -1 is still required to produce 0xFFFF when later it gets converted to unsigned type  Even if some platform uses an exotic representation for signed integers  1 s complement  signed magnitude   this platform is still required to apply rules of modulo arithmetic when converting signed integer values to unsigned ones   For example  this evaluation  signed int a   0  b   1  unsigned int c   a - b    is still guaranteed to produce UINT MAX in c  even if the platform is using an exotic representation for signed integers

User · Answer

When you work with unsigned types  modular arithmetic  also known as  wrap around  behavior  is taking place  To understand this modular arithmetic  just have a look at these clocks      9   4   1  13 mod 12   so to the other direction it is  1 - 4   9  -3 mod 12   The same principle is applied while working with unsigned types  If the result type is unsigned  then modular arithmetic takes place     Now look at the following operations storing the result as an unsigned int   unsigned int five   5  seven   7  unsigned int a   five - seven          a    -2   2 32    4294967294   int one   1  six   6  unsigned int b   one - six             b    -5   2 32    4294967291   When you want to make sure that the result is signed  then stored it into signed variable or cast it to signed  When you want to get the difference between numbers and make sure that the modular arithmetic will not be applied  then you should consider using abs   function defined in stdlib h   int c   five - seven           c   -2 int d   abs five - seven       d    2   Be very careful  especially while writing conditions  because   if  abs five - seven   lt  seven        if  2  lt  7              if  five - seven  lt  -1                if  -2  lt  -1              if  one - six  lt  1                    if  -5  lt  1              if   int  five - seven   lt  1          if  -2  lt  1               but  if  five - seven  lt  1         if   unsigned int -2  lt  1    if  4294967294  lt  1              if  one - six  lt  five         if   unsigned int -5  lt  5    if  4294967291  lt  5

User · Answer

With unsigned numbers of type unsigned int or larger  in the absence of type conversions  a-b is defined as yielding the unsigned number which  when added to b  will yield a   Conversion of a negative number to unsigned is defined as yielding the number which  when added to the sign-reversed original number  will yield zero  so converting -5 to unsigned will yield a value which  when added to 5  will yield zero    Note that unsigned numbers smaller than unsigned int may get promoted to type int before the subtraction  the behavior of a-b will depend upon the size of int

User · Answer

The result of a subtraction generating a negative number in an unsigned type is well-defined               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     ISO IEC 9899 1999  E    6 2 5 9       As you can see   unsigned 0 -  unsigned 1 equals -1 modulo UINT MAX 1  or in other words  UINT MAX   Note that although it does say  A computation involving unsigned operands can never overflow   which might lead you to believe that it applies only for exceeding the upper limit  this is presented as a motivation for the actual binding part of the sentence   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   This phrase is not restricted to overflow of the upper bound of the type  and applies equally to values too low to be represented

[c] Is unsigned integer subtraction defined behavior?

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