var x = 1;
if (function f(){}) {
x += typeof f;
}
console.log(x);
"1undefined"
var x = 1;
if (function f(){}) {
x += typeof f;
}
console.log(x);
"1undefined"
if(/*expression*/)
expression, if (function f () {}) is a function expression with a name , not a function declaration statement function f () {} in
f
function f(){}()
() (),()
because return is also followed by an expression, function f () {} is considered an expression, so no error is reported when executed directly.
if , so the variable f
to sum up, there are two test points:
1 expression if (/ expression /) receives expressions instead of function declarations.
2, the function name cannot be taken externally in the function expression (because it is optional). The function name is required in the function declaration and optional in the function expression.
so the question just now can't get the f. / p > in if ().
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