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I read that decimal literals are signed by default.
For simplicity, assume that the values that int can hold are the integers [-128,127], and long can hold the integer 128. Now, what happens if I code the literal -128? What I know is that the literal here is simply ‘128’, which can't fit into an int but rather would go into a long! or does the unary minus operator do something else?
So, how does the unary minus sign work with integer literals?

Mason
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  • The unary `-` is used in an expression. So `-128` applies the unary `-` to the decimal literal `128`, and gives a negative result. Since `128` has type `int`, so does `-128`. And since the `int` type is guaranteed to be able to represent any value in the range `-32767` to `32767`, there is no need for conversions. An `int` can represent `128` or `-128` - guaranteed. – Peter Nov 04 '19 at 06:06
  • @Peter Does it mean, that `int i = -2147483648;` requires a conversion with 32-bit `int`? UPDATE: It's discussed here: [How do I define a constant equal to -2147483648?](https://stackoverflow.com/q/27612996/580083). – Daniel Langr Nov 04 '19 at 06:27

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From cppreference.com:

The type of the integer literal is the first type in which the value can fit, from the list of types which depends on which numeric base and which integer-suffix was used.

When using a decimal base and no suffix, as in your example, the possible types are int, long int, and long long int. If the value (ignoring the minus sign) fits in a long but not in an int, then the type of the value is long.

After the type is determined, the unary minus operator is applied as normal. Applying unary minus to a long results in a long (even if the result could fit in an int).

JaMiT
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