Signed to unsigned conversion in C - is it always safe?

When one unsigned and one signed variable are added (or any binary operation) both are implicitly converted to unsigned, which would in this case result in a huge result.

So it is safe in the sense of that the result might be huge and wrong, but it will never crash.


Short Answer

Your i will be converted to an unsigned integer by adding UINT_MAX + 1, then the addition will be carried out with the unsigned values, resulting in a large result (depending on the values of u and i).

Long Answer

According to the C99 Standard:

6.3.1.8 Usual arithmetic conversions

  1. If both operands have the same type, then no further conversion is needed.
  2. Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
  3. Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
  4. Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
  5. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

In your case, we have one unsigned int (u) and signed int (i). Referring to (3) above, since both operands have the same rank, your i will need to be converted to an unsigned integer.

6.3.1.3 Signed and unsigned integers

  1. When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.
  2. Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.
  3. Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

Now we need to refer to (2) above. Your i will be converted to an unsigned value by adding UINT_MAX + 1. So the result will depend on how UINT_MAX is defined on your implementation. It will be large, but it will not overflow, because:

6.2.5 (9)

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.

Bonus: Arithmetic Conversion Semi-WTF

#include <stdio.h>

int main(void)
{
  unsigned int plus_one = 1;
  int minus_one = -1;

  if(plus_one < minus_one)
    printf("1 < -1");
  else
    printf("boring");

  return 0;
}

You can use this link to try this online: https://repl.it/repls/QuickWhimsicalBytes

Bonus: Arithmetic Conversion Side Effect

Arithmetic conversion rules can be used to get the value of UINT_MAX by initializing an unsigned value to -1, ie:

unsigned int umax = -1; // umax set to UINT_MAX

This is guaranteed to be portable regardless of the signed number representation of the system because of the conversion rules described above. See this SO question for more information: Is it safe to use -1 to set all bits to true?


Conversion from signed to unsigned does not necessarily just copy or reinterpret the representation of the signed value. Quoting the C standard (C99 6.3.1.3):

When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.

Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.

Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

For the two's complement representation that's nearly universal these days, the rules do correspond to reinterpreting the bits. But for other representations (sign-and-magnitude or ones' complement), the C implementation must still arrange for the same result, which means that the conversion can't just copy the bits. For example, (unsigned)-1 == UINT_MAX, regardless of the representation.

In general, conversions in C are defined to operate on values, not on representations.

To answer the original question:

unsigned int u = 1234;
int i = -5678;

unsigned int result = u + i;

The value of i is converted to unsigned int, yielding UINT_MAX + 1 - 5678. This value is then added to the unsigned value 1234, yielding UINT_MAX + 1 - 4444.

(Unlike unsigned overflow, signed overflow invokes undefined behavior. Wraparound is common, but is not guaranteed by the C standard -- and compiler optimizations can wreak havoc on code that makes unwarranted assumptions.)


Referring to The C Programming Language, Second Edition (ISBN 0131103628),

  • Your addition operation causes the int to be converted to an unsigned int.
  • Assuming two's complement representation and equally sized types, the bit pattern does not change.
  • Conversion from unsigned int to signed int is implementation dependent. (But it probably works the way you expect on most platforms these days.)
  • The rules are a little more complicated in the case of combining signed and unsigned of differing sizes.

Tags:

Type Conversion

C

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