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yasuo_ohgaki at hotmail dot com
17 years ago
Manual defines "expression is anything that has value", Therefore, parser will give error for following code.

($val) ? echo('true') : echo('false');
Note: "? : " operator has this syntax  "expr ? expr : expr;"

since echo does not have(return) value and ?: expects expression(value).

However, if function/language constructs that have/return value, such as include(), parser compiles code.

Note: User defined functions always have/return value without explicit return statement (returns NULL if there is no return statement). Therefore, user defined functions are always valid expressions.
[It may be useful to have VOID as new type to prevent programmer to use function as RVALUE by mistake]

For example,

($val) ? include('') : include('');

is valid, since "include" returns value.

The fact "echo" does not return value(="echo" is not a expression), is less obvious to me.

Print() and Echo() is NOT identical since print() has/returns value and can be a valid expression.

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