我有一个Perl项目,我通过打一个循环包调用问题.下面的代码演示了这个问题.
执行此操作时,每个程序包将调用另一个程序包,直到计算机的所有内存都被占用并锁定.我同意这是一个糟糕的设计,这样的循环调用不应该在设计中进行,但我的项目足够大,我想在运行时检测到这一点.
我已经阅读了弱化函数和Data :: Structure :: Util,但我还没有找到一种方法来检测是否存在循环包加载(我假设,因为在每次迭代时都会生成一个新副本并存储在$ this hash的每个副本中).有任何想法吗？
use system::one;
my $one = new system::one();
package system::one;
use strict;
use system::two;
sub new {
my ($class) = @_;
my $this = {};
bless($this,$class);
# attributes
$this->{two} = new system::two();
return $this;
}
package system::two;
use strict;
use system::one;
sub new {
my ($class) = @_;
my $this = {};
bless($this,$class);
# attributes
$this->{one} = new system::one();
return $this;
}
有没有人知道更好/更快的方式获得调用堆栈比"StackWalk"？我也认为对于有很多变量的方法,stackwalk也可能会变慢...(我想知道商业分析器的用途是什么？)我在Windows上使用C++.:) 谢谢 :)
我打开了一个进程(使用C++/Windows)
if( CreateProcessA( NULL, // No module name (use command line)
(LPSTR)path, //argv[1], // Command line
NULL, // Process handle not inheritable
NULL, // Thread handle not inheritable
FALSE, // Set handle inheritance to FALSE
creationFlags, // No creation flags
NULL, // Use parent's environment block
NULL, // Use parent's starting directory
&startInfo, // Pointer to STARTUPINFO structure
&processInfo ) // Pointer to PROCESS_INFORMATION structure
哪里
DWORD creationFlags = DEBUG_PROCESS | DEBUG_ONLY_THIS_PROCESS;
然后我试着用它来堆积它
bool ok = StackWalk64(IMAGE_FILE_MACHINE_I386,m_ps.Handle ,m_th.Handle,
&m_stackframe, &m_threadContext, …是否可以检索方法/构造函数的调用者实例？
这个问题已经发布,但每次答案都是关于调用者类(使用堆栈跟踪)而不是调用者实例.如果存在解决方案,那么构建对象图(具有公共超类型)并使用默认构造函数处理父子导航可能非常方便.
public class TestCallStack {
public static class BaseClass {
BaseClass owner;
// //ok, this is the correct way to do it
// public BaseClass(BaseClass owner) {
// this.owner = owner;
// }
public BaseClass() {
//this.owner = ???????;
}
}
public static class Parent extends BaseClass {
Child child = new Child();
}
public static class Child extends BaseClass {
}
public static void main(String[] args) {
Parent parent = new Parent();
System.out.println(parent.child.owner==parent); // must be true
} …我知道您可能已经厌倦了再次回答相同的问题,但我仍然在其他几个 问题中讨论了错误:
承诺已经在评估中:递归默认参数引用还是早期问题？
即使我确实遵循了前期的"繁琐"建议".":
show.large.objects.threshold <- 100000
show.large.objects.exclude <- c("closure")
show.large.objects <- function (.envir = sys.frame(),
threshold = show.large.objects.threshold,
exclude = show.large.objects.exclude) {
for (n in print(ls(.envir, all.names = TRUE))) tryCatch({
o <- get(n,envir = .envir)
s <- object.size(o)
if (s > threshold && !(typeof(o) %in% exclude)) {
cat(n,": ")
print(s,units="auto")
}
}, error = function(e) { cat("n=",n,"\n"); print(e) })
}
show.large.objects.stack <- function (.threshold = show.large.objects.threshold,
skip.levels = 1,# do not examine the last …我很好奇CasperJS如何处理与调用堆栈有关的事件.
假设我们有一些代码:
casper.on('foo', function() {
this.wait(60000);
this.echo('foo');
});
casper.start('http://www.stackoverflow.com', function() {
this.echo('start');
this.emit('foo');
});
casper.then(function() {
this.echo('done');
});
casper.run();
我知道then()将等待检查3个标志:pendingWait,loadInProgress和navigationRequested.打印出调用堆栈会将emit调用显示在函数start()中,因此在事件结束之前start()不会被视为已完成？即,然后()等待事件结束
我等了60秒测试了这个,我确实得到了输出:
start
foo
done
虽然我不确定超过某个超时是否会触发下一个().
我们正在围绕LLVM库进行研究,我们发现IR库有时会达到最多29个方法调用的调用堆栈.
有时当我在iOS框架中看到一些崩溃时,我也会观察到相当深的调用堆栈.
我的问题是,我们是否可以推断一个代码的设计是否存在问题,而这个代码的设计自称是如此之大.
这是一个例子:
/usr/local/LLVM/llvm/unittests/IR/AttributesTest.cpp:54
/usr/local/LLVM/llvm/lib/IR/LLVMContext.cpp:162
/usr/local/LLVM/llvm/lib/IR/LLVMContext.cpp:162
/usr/local/LLVM/llvm/lib/IR/LLVMContextImpl.cpp:54
/usr/local/LLVM/llvm/lib/IR/LLVMContextImpl.cpp:59
/usr/local/LLVM/llvm/lib/IR/Module.cpp:60
/usr/local/LLVM/llvm/lib/IR/Module.cpp:62
/usr/local/LLVM/llvm/lib/IR/Module.cpp:456
/usr/local/LLVM/llvm/lib/IR/Function.cpp:350
/usr/local/LLVM/llvm/lib/IR/BasicBlock.cpp:98
/usr/local/LLVM/llvm/include/llvm/ADT/ilist.h:282
/usr/local/LLVM/llvm/include/llvm/ADT/ilist.h:267
/usr/local/LLVM/llvm/lib/IR/SymbolTableListTraitsImpl.h:76
/usr/local/LLVM/llvm/lib/IR/BasicBlock.cpp:90
/usr/local/LLVM/llvm/lib/IR/SymbolTableListTraitsImpl.h:58
/usr/local/LLVM/llvm/lib/IR/ValueSymbolTable.cpp:75
/usr/local/LLVM/llvm/lib/IR/ValueSymbolTable.cpp:47
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:132
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:112
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:122
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:96
/usr/local/LLVM/llvm/include/llvm/IR/Value.h:777
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:132
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:122
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:75
/usr/local/LLVM/llvm/include/llvm/IR/Value.h:771
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:132
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:122
/usr/local/LLVM/llvm/include/llvm/Support/Casting.h:75
/usr/local/LLVM/llvm/include/llvm/IR/Value.h:759
PS示例调用堆栈实际上是由LLVMContext类的析构函数生成的:LLVMContext::~LLVMContext().这是来自Java世界的一篇非常古老的帖子的另一个例子:Java调用栈 - 从HTTP到JDBC作为图片.
在x86程序集中,是否可以从堆栈中删除值而不存储它？有什么东西沿着pop word null？我显然可以使用add esp,4,但也许有一个很好的和干净的cisc助记符我不见了？
AFAIK，有两种方法可以获取调用堆栈以在Haskell中进行调试：
HasCallStack在代码中添加约束ghc -prof -fprof-auto-top我的测试代码：
import GHC.Stack
-- | a wrapper function to make "last" from base traceable
last' :: HasCallStack => [a] -> a
last' xs = case xs of [] -> error "abuse last"; _ -> last xs
-- | a untraceable partial function
foo :: [Int] -> Int
foo xs = last' xs + 1
-- | a untraceable partial function
-- , which looks like traceable, but it's call stack is cut off …