When Do Integer Literals Become 64-Bit?

64-Bit Integer Literals in Default Int Contexts

In programming, integer literals represent whole numbers. Typically, these literals are assigned to the default integer data type, which is usually 32 bits wide. However, certain circumstances can result in larger integer literals being created by default.

One such case is when the integer literal exceeds the range of the default integer type. For example, in C and C , literals without an appended "L" have a type that can represent their value. If the value is too large for an int type, it will be promoted to a long int or long long int type.

This promotion resolves the overflow issue that would otherwise occur if the literal were cast as a 32-bit integer. The C 11 standard specifies this behavior in [lex.icon] ¶2, stating that the literal's type will be the first in the given list where the value can be represented:

int
long int
long long int

Similarly, C99 defines this promotion mechanism in §6.4.4.1. This ensures that even if an integer literal exceeds the default integer range, it will be implicitly converted to a larger type, allowing it to be represented correctly.

In rare cases, the integer literal may still be too large for the available integer types. In that case, both C99 and C 11 mandate a compilation error, indicating that the literal cannot be represented by any supported type. This prevents runtime issues caused by integer overflow.

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