C++ – Signed/unsigned comparisons

ccomparisonsignedunsignedvisual-studio-2005

I'm trying to understand why the following code doesn't issue a warning at the indicated place.

//from limits.h
#define UINT_MAX 0xffffffff /* maximum unsigned int value */
#define INT_MAX  2147483647 /* maximum (signed) int value */
            /* = 0x7fffffff */

int a = INT_MAX;
//_int64 a = INT_MAX; // makes all warnings go away
unsigned int b = UINT_MAX;
bool c = false;

if(a < b) // warning C4018: '<' : signed/unsigned mismatch
    c = true;
if(a > b) // warning C4018: '<' : signed/unsigned mismatch
    c = true;
if(a <= b) // warning C4018: '<' : signed/unsigned mismatch
    c = true;
if(a >= b) // warning C4018: '<' : signed/unsigned mismatch
    c = true;
if(a == b) // no warning <--- warning expected here
    c = true;
if(((unsigned int)a) == b) // no warning (as expected)
    c = true;
if(a == ((int)b)) // no warning (as expected)
    c = true;

I thought it was to do with background promotion, but the last two seem to say otherwise.

To my mind, the first == comparison is just as much a signed/unsigned mismatch as the others?

Best Answer

When comparing signed with unsigned, the compiler converts the signed value to unsigned. For equality, this doesn't matter, -1 == (unsigned) -1. For other comparisons it matters, e.g. the following is true: -1 > 2U.

EDIT: References:

5/9: (Expressions)

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

If either operand is of type long double, the other shall be converted to long double.

Otherwise, if either operand is double, the other shall be converted to double.

Otherwise, if either operand is float, the other shall be converted to float.

Otherwise, the integral promotions (4.5) shall be performed on both operands.54)

Then, if either operand is unsigned long the other shall be converted to unsigned long.

Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int.

Otherwise, if either operand is long, the other shall be converted to long.

Otherwise, if either operand is unsigned, the other shall be converted to unsigned.

4.7/2: (Integral conversions)

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2ⁿ where n is the number of bits used to represent the unsigned type). [Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). ]

EDIT2: MSVC warning levels

What is warned about on the different warning levels of MSVC is, of course, choices made by the developers. As I see it, their choices in relation to signed/unsigned equality vs greater/less comparisons make sense, this is entirely subjective of course:

-1 == -1 means the same as -1 == (unsigned) -1 - I find that an intuitive result.

-1 < 2 does not mean the same as -1 < (unsigned) 2 - This is less intuitive at first glance, and IMO deserves an "earlier" warning.

Related Solutions

Signed versus Unsigned Integers

Unsigned can hold a larger positive value and no negative value.

Yes.

Unsigned uses the leading bit as a part of the value, while the signed version uses the left-most-bit to identify if the number is positive or negative.

There are different ways of representing signed integers. The easiest to visualise is to use the leftmost bit as a flag (sign and magnitude), but more common is two's complement. Both are in use in most modern microprocessors — floating point uses sign and magnitude, while integer arithmetic uses two's complement.

Signed integers can hold both positive and negative numbers.

Yes.

C++ – Iteration over std::vector: unsigned vs signed index variable

For iterating backwards see this answer.

Iterating forwards is almost identical. Just change the iterators / swap decrement by increment. You should prefer iterators. Some people tell you to use std::size_t as the index variable type. However, that is not portable. Always use the size_type typedef of the container (While you could get away with only a conversion in the forward iterating case, it could actually go wrong all the way in the backward iterating case when using std::size_t, in case std::size_t is wider than what is the typedef of size_type):

Using std::vector

Using iterators

for(std::vector<T>::iterator it = v.begin(); it != v.end(); ++it) {
    /* std::cout << *it; ... */
}

Important is, always use the prefix increment form for iterators whose definitions you don't know. That will ensure your code runs as generic as possible.

Using Range C++11

for(auto const& value: a) {
     /* std::cout << value; ... */

Using indices

for(std::vector<int>::size_type i = 0; i != v.size(); i++) {
    /* std::cout << v[i]; ... */
}

Using arrays

Using iterators

for(element_type* it = a; it != (a + (sizeof a / sizeof *a)); it++) {
    /* std::cout << *it; ... */
}

Using Range C++11

for(auto const& value: a) {
     /* std::cout << value; ... */

Using indices

for(std::size_t i = 0; i != (sizeof a / sizeof *a); i++) {
    /* std::cout << a[i]; ... */
}

Read in the backward iterating answer what problem the sizeof approach can yield to, though.

Best Answer

Related Solutions

Signed versus Unsigned Integers

C++ – Iteration over std::vector: unsigned vs signed index variable

Using std::vector

Using iterators

Using Range C++11

Using indices

Using arrays

Using iterators

Using Range C++11

Using indices

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