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         


        

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::min();  // or max()