Why does the most negative int value cause an error about ambiguous function overloads?

Question

I'm learning about function overloading in C++ and came across this:

void display(int a)
{
    cout << "int" << endl;
}

void display(unsigned a)
{
    cout << "unsigned" << endl;
}

int main()
{
    int i = -2147483648;
    cout << i << endl; //will display -2147483648
    display(-2147483648);
}

From what I understood, any value given in the int range (in my case int is 4 byte) will call display(int) and any value outside this range will be ambiguous (since the compiler cannot decide which function to call). It is valid for the complete range of int values except its min value i.e. -2147483648 where compilation fails with the error

call of overloaded display(long int) is ambiguous

But taking the same value to an int and printing the value gives 2147483648. I'm literally confused with this behavior.

Why is this behavior observed only when the most negative number is passed? (The behavior is the same if a short is used with -32768 - in fact, in any case where the negative number and positive number have the same binary representation)

Compiler used: g++ (GCC) 4.8.5

Meshed

Int's min value is "throwing a compiler error". What error? You should include it in the question – Willson

I see call of overloaded ‘display(long int)’ is ambiguous. – Visitant

@Visitant Yes, the error should be in the question – Willson

Not related, but you should update the compiler. There is already GCC 7.1. – Theoretics

Yeah the same what @Visitant has mentioned! – Meshed

Here's my guess: typeof(-2147483648) != int. The literal is 2147483648, which is too big for an int, so it's a long, and it's being negated – Willson

@HolyBlackCat, the behaviour is same with MSVC++ 14.0 compiler – Meshed

perhaps somewhat related: #8771909 – Fornax

Possible duplicate of large negative integer literals – Fornax

@infiniteloop I'm not saying that's because of the compiler. (That's why I said 'not related'.) I just suggest getting an ungrade. – Theoretics

Casting minimum 32-bit integer (-2147483648) to float gives positive number (2147483648.0), (-2147483648> 0) returns true in C++?, Warning: this decimal constant is unsigned only in ISO C90 – Ovariectomy

Why does MSVC pick a long long as the type for -2147483648? – Ovariectomy

Interestingly, g++ (6.4 and 7.1, at least) don't complain that int j{-2147483648}; is a narrowing conversion. Almost worth a question in itself, that. It's probably related to allowing (e.g.) long long constexpr values such as 2147483647LL to be narrowed in initialization. – Valeda

Answer 1

This is a very subtle error. What you are seeing is a consequence of there being no negative integer literals in C++. If we look at [lex.icon] we get that a integer-literal,

integer-literal
        decimal-literal integer-suffix_opt
        [...]

can be a decimal-literal,

decimal-literal:
        nonzero-digit
        decimal-literal ’ _opt digit

where digit is [0-9] and nonzero-digit is [1-9] and the suffix par can be one of u, U, l, L, ll, or LL. Nowhere in here does it include - as being part of the decimal literal.

In §2.13.2, we also have:

An integer literal is a sequence of digits that has no period or exponent part, with optional separating single quotes that are ignored when determining its value. An integer literal may have a prefix that specifies its base and a suffix that specifies its type. The lexically first digit of the sequence of digits is the most significant. A decimal integer literal (base ten) begins with a digit other than 0 and consists of a sequence of decimal digits.

^{(emphasis mine)}

Which means the - in -2147483648 is the unary operator -. That means -2147483648 is actually treated as -1 * (2147483648). Since 2147483648 is one too many for your int it is promoted to a long int and the ambiguity comes from that not matching.

If you want to get the minimum or maximum value for a type in a portable manner you can use:

std::numeric_limits<type>::min();  // or max()

Borden

-2147483647 - 1 would also work without warning as a negative literal expression – Inconsistency

Or INT_MIN for the least verbose option. Less generic, though. – Lomalomas

@NathanOliver, Can you kindly explain me this case display(2147483649);. Why can't it call the unsigned int func in this case? and why it treats the arg 2147483649 as long int instead of unsigned int? – Meshed

@infiniteloop Decimal integer literals go from int to long int to long long int. You wlll never get an unsigned type for a decimal literal unless you use the u/U suffix. – Borden

@Borden So, until explicitly given an unsigned int variable or unsigned cast, the unsigned function cannot be called? – Meshed

In this example yes. To call display(unsigned a) you need either display(1234u); or display(static_cast<unsigned>(1234)); or unsigned foo = 1234; display(foo); – Borden

Answer 2

The expression -2147483648 is actually applying the - operator to the constant 2147483648. On your platform, int can't store 2147483648, it must be represented by a larger type. Therefore, the expression -2147483648 is not deduced to be signed int but a larger signed type, signed long int.

Since you do not provide an overload for long the compiler is forced to choose between two overloads that are both equally valid. Your compiler should issue a compiler error about ambiguous overloads.

Shell

Answer 3

Expanding on others' answers

---

To clarify why the OP is confused, first: consider the signed int binary representation of 2147483647, below.

---

Next, add one to this number: giving another signed int of -2147483648 (which the OP wishes to use)

---

Finally: we can see why the OP is confused when -2147483648 compiles to a long int instead of a signed int, since it clearly fits in 32 bits.

But, as the current answers mention, the unary operator (-) is applied after resolving 2147483648 which is a long int and does NOT fit in 32 bits.

Mylesmylitta