var c = 1;
function myFunction(){
c = 2;
var c = 4;
console.log(c);
}
console.log(c);
myFunction();
console.log(c);
为什么最后一个console.log吐出1?这是应该在我的大脑中起作用的方式:
var c = 1; // create global variable called 'c'
function myFunction(){
c = 2; // assign value of global 'c' to 2
var c = 4; // create a new variable called 'c' which has a new address in memory (right?) with a value of 4
console.log(c); // log the newly created 'c' variable (which has the value of 4)
}
console.log(c); //log the global c
myFunction(); //log the internal c
console.log(c); //log the updated global c
最佳答案
在console.log运行的范围内,仅存在全局变量。而且该函数永远不会触及全局变量。
细节:
var c = 1; // create global variable called 'c'
function myFunction(){
// the global is unavailable here, as you declared a local c.
// Its declaration is hoisted to the top of the function, so
// c exists here, and its value is undefined.
c = 2; // assign value of local 'c' to 2
var c = 4; // change local c to 4
console.log(c); // log local c (4)
}
console.log(c); //log the global c
myFunction(); //log the internal c
console.log(c); //log the global c (which never changed)
由于上述提升,您的函数的行为类似于以下代码:
function myFunction(){
var c; // declares a local c that shadows the global one
c = 2;
c = 4;
console.log(c);
}
有关起重的一些参考
MDN page for the
var
statementMDN scope cheatsheet
How are javascript variables "hoisted" in these examples from MDN
Javascript hoisting explained
JavaScript Scoping and Hoisting