function main()
{
Hello();
}
function Hello()
{
// How do you find out the caller function is 'main'?
}
有办法找到调用堆栈吗?
function main()
{
Hello();
}
function Hello()
{
// How do you find out the caller function is 'main'?
}
有办法找到调用堆栈吗?
当前回答
我认为下面的代码段可能会有帮助:
window.fnPureLog = function(sStatement, anyVariable) {
if (arguments.length < 1) {
throw new Error('Arguments sStatement and anyVariable are expected');
}
if (typeof sStatement !== 'string') {
throw new Error('The type of sStatement is not match, please use string');
}
var oCallStackTrack = new Error();
console.log(oCallStackTrack.stack.replace('Error', 'Call Stack:'), '\n' + sStatement + ':', anyVariable);
}
执行以下代码:
window.fnPureLog = function(sStatement, anyVariable) {
if (arguments.length < 1) {
throw new Error('Arguments sStatement and anyVariable are expected');
}
if (typeof sStatement !== 'string') {
throw new Error('The type of sStatement is not match, please use string');
}
var oCallStackTrack = new Error();
console.log(oCallStackTrack.stack.replace('Error', 'Call Stack:'), '\n' + sStatement + ':', anyVariable);
}
function fnBsnCallStack1() {
fnPureLog('Stock Count', 100)
}
function fnBsnCallStack2() {
fnBsnCallStack1()
}
fnBsnCallStack2();
日志是这样的:
Call Stack:
at window.fnPureLog (<anonymous>:8:27)
at fnBsnCallStack1 (<anonymous>:13:5)
at fnBsnCallStack2 (<anonymous>:17:5)
at <anonymous>:20:1
Stock Count: 100
其他回答
我可以在2021年使用这些,并获得从调用者函数开始的堆栈:
1. console.trace();
2. console.log((new Error).stack)
// do the same as #2 just with better view
3. console.log((new Error).stack.split("\n"))
你可以得到完整的堆栈跟踪:
arguments.callee.caller
arguments.callee.caller.caller
arguments.callee.caller.caller.caller
直到调用者为空。
注意:它会导致递归函数的无限循环。
使用*arguments.callee更安全。调用自参数。不赞成调用…
概括一下(并说得更清楚)…
这段代码:
function Hello() {
alert("caller is " + arguments.callee.caller.toString());
}
等价于:
function Hello() {
alert("caller is " + Hello.caller.toString());
}
显然,第一部分更容易移植,因为你可以改变函数的名字,从“Hello”说成“Ciao”,但仍然可以让整个程序正常工作。
在后一种情况下,如果你决定重构被调用函数的名称(Hello),你将不得不改变它的所有出现:
试着访问这个:
arguments.callee.caller.name