问题
Suppose i have something like
int x = 2 + 3;
Is x considered to be a literal?
回答1:
x is a symbol. 2 + 3 is an expression. 2 and 3 are literals.
回答2:
No, it's two literals in a compile-time constant expression. Depending on how it gets compiled, you may not be able to tell the difference in the resulting binary.
回答3:
- int - type of the variable, identifier
- x - name of the variable, identifier
- = - assignment
- 2 - literal
- + - operation between two literals
- 3 - literal
- ; - end of the statement
回答4:
- int is a type
- x is an identifier
- =,+ are operators
- 2,3 are literals as "StringLiteral" would be
To learn how all these syntacically elements are named you could browse i.g.
http://java.sun.com/docs/books/jls/second_edition/html/lexical.doc.html
http://java.sun.com/docs/books/jls/second_edition/html/typesValues.doc.html
回答5:
yeah it is a literal. the expression will be reduced to literal, i.e. even before the program is run, so it is literal
other types of literal you might see in real-world (the programmer can't be bothered too compute it by himself, let the compiler do the grunt work of determining the literal):
const unsigned int NEGATIVE_TESTER_FOR_32_BIT = 1 << 31;
const char ALPHABET_PIVOT = 'A' + ( ('Z' - 'A') / 2);
[EDIT: regarding depending on the compiler] yeah it depends on the compiler. but i guess most of compiler writers do their homework, if they can generate the necessarily assembly or virtual machine instruction of the language, computing things in compile time is just a walk in the park for them. in fact, even the character literal 'H' for example, doesn't even get stored in file as verbatim 'H', it gets stored as number literal (72, or in hex view 0x48 or 48h) in final machine code. i can hazard a guess that all compilers will reduce the two literals to a literal, not expression. it's a walk in the park for them(compiler writers)
来源：https://stackoverflow.com/questions/2800696/is-23-considered-as-a-literal