If your goal is to use a profiler, use one of the suggested ones.
(如果您的目标是使用探查器，请使用建议的探查器之一。)
However, if you're in a hurry and you can manually interrupt your program under the debugger while it's being subjectively slow, there's a simple way to find performance problems.
(但是，如果您急于在主观上很慢的情况下在调试器下手动中断程序，则有一种简单的方法可以查找性能问题。)
Just halt it several times, and each time look at the call stack.
(暂停几次，每次查看调用堆栈。)
If there is some code that is wasting some percentage of the time, 20% or 50% or whatever, that is the probability that you will catch it in the act on each sample. (如果有一些代码浪费了一定比例的时间（20％或50％或其他），那么这就是您在每次采样时都将其捕获的概率。)
So that is roughly the percentage of samples on which you will see it. (因此，这大约是您将看到样品的百分比。)
There is no educated guesswork required. (不需要有根据的猜测。)
If you do have a guess as to what the problem is, this will prove or disprove it. (如果您确实怀疑问题出在哪里，这将证明或不证明它。)
You may have multiple performance problems of different sizes.
(您可能会遇到多个不同大小的性能问题。)
If you clean out any one of them, the remaining ones will take a larger percentage, and be easier to spot, on subsequent passes. (如果您清除其中任何一个，其余的将在以后的传递中占更大的比例，并且更容易发现。)
This magnification effect , when compounded over multiple problems, can lead to truly massive speedup factors. (当放大多个问题时，这种放大效果会导致真正巨大的加速因素。)
Caveat: Programmers tend to be skeptical of this technique unless they've used it themselves.
(警告：除非程序员自己使用过，否则程序员往往会对这种技术持怀疑态度。)
They will say that profilers give you this information, but that is only true if they sample the entire call stack, and then let you examine a random set of samples. (他们会说探查器会为您提供此信息，但是只有当他们对整个调用堆栈进行采样，然后让您检查随机的一组采样时，这才是正确的。)
(The summaries are where the insight is lost.) Call graphs don't give you the same information, because (（摘要是丢失洞察力的地方。）调用图不会为您提供相同的信息，因为)
- they don't summarize at the instruction level, and
(他们没有在教学水平上进行总结，并且)
- they give confusing summaries in the presence of recursion.
(在递归存在的情况下，它们给出了令人困惑的摘要。)
They will also say it only works on toy programs, when actually it works on any program, and it seems to work better on bigger programs, because they tend to have more problems to find.
(他们还会说，它实际上仅对玩具程序有效，而实际上对任何程序都有效，并且似乎在较大的程序上效果更好，因为它们往往会发现更多的问题。)
They will say it sometimes finds things that aren't problems, but that is only true if you see something once . (他们会说有时发现没有问题的东西，但这只有在您看到一次之后才是真的。)
If you see a problem on more than one sample, it is real. (如果您在多个样本上发现问题，那是真实的。)
PS This can also be done on multi-thread programs if there is a way to collect call-stack samples of the thread pool at a point in time, as there is in Java.
(PS如果可以像Java中那样在某个时间点收集线程池的调用堆栈样本，也可以在多线程程序上完成。)
PPS As a rough generality, the more layers of abstraction you have in your software, the more likely you are to find that that is the cause of performance problems (and the opportunity to get speedup).
(PPS大致来说，软件中的抽象层越多，您越有可能发现这是性能问题的原因（并且有提高速度的机会）。)
Added: It might not be obvious, but the stack sampling technique works equally well in the presence of recursion.
(补充：可能并不明显，但是在存在递归的情况下，堆栈采样技术同样有效。)
The reason is that the time that would be saved by removal of an instruction is approximated by the fraction of samples containing it, regardless of the number of times it may occur within a sample. (原因是通过删除一条指令可以节省的时间大约等于包含该指令的样本所占的比例，而不管该指令在一个样本中可能发生的次数。)
Another objection I often hear is: " It will stop someplace random, and it will miss the real problem ".
(我经常听到的另一个反对意见是：“ 它将在某个地方随机停止，并且将错过真正的问题 ”。)
This comes from having a prior concept of what the real problem is. (这源于对实际问题有一个先验的概念。)
A key property of performance problems is that they defy expectations. (性能问题的一个关键特性是它们无法兑现预期。)
Sampling tells you something is a problem, and your first reaction is disbelief. (抽样告诉您某些问题，而您的第一个反应是难以置信。)
That is natural, but you can be sure if it finds a problem it is real, and vice-versa. (那是很自然的，但是您可以确定它是否发现了真正的问题，反之亦然。)
ADDED: Let me make a Bayesian explanation of how it works.
(添加：让我对它的工作方式进行贝叶斯解释。)
Suppose there is some instruction I (call or otherwise) which is on the call stack some fraction f of the time (and thus costs that much). (假设有一条指令I （调用或其他方式）在调用堆栈上占时间的比例为f （因此花费了很多）。)
For simplicity, suppose we don't know what f is, but assume it is either 0.1, 0.2, 0.3, ... 0.9, 1.0, and the prior probability of each of these possibilities is 0.1, so all of these costs are equally likely a-priori. (为简单起见，假设我们不知道f是什么，但假设它是0.1、0.2、0.3，... 0.9、1.0，并且每种可能性的先验概率为0.1，因此所有这些成本均相等可能是先验的。)
Then suppose we take just 2 stack samples, and we see instruction I on both samples, designated observation o=2/2 .
(然后假设我们仅取2个堆栈样本，并且在两个样本上都看到指令I ，将其指定为观察值o=2/2 。)
This gives us new estimates of the frequency f of I , according to this: (根据这一点，这为我们提供了I的频率f的新估计：)
Prior
P(f=x) x P(o=2/2|f=x) P(o=2/2&&f=x) P(o=2/2&&f >= x) P(f >= x | o=2/2)
0.1 1 1 0.1 0.1 0.25974026
0.1 0.9 0.81 0.081 0.181 0.47012987
0.1 0.8 0.64 0.064 0.245 0.636363636
0.1 0.7 0.49 0.049 0.294 0.763636364
0.1 0.6 0.36 0.036 0.33 0.857142857
0.1 0.5 0.25 0.025 0.355 0.922077922
0.1 0.4 0.16 0.016 0.371 0.963636364
0.1 0.3 0.09 0.009 0.38 0.987012987
0.1 0.2 0.04 0.004 0.384 0.997402597
0.1 0.1 0.01 0.001 0.385 1
P(o=2/2) 0.385
The last column says that, for example, the probability that f >= 0.5 is 92%, up from the prior assumption of 60%.
(最后一栏说，例如， f > = 0.5的概率为92％，高于先前假设的60％。)
Suppose the prior assumptions are different.
(假设先前的假设是不同的。)
Suppose we assume P(f=0.1) is .991 (nearly certain), and all the other possibilities are almost impossible (0.001). (假设我们假设P（f = 0.1）为.991（几乎可以肯定），而其他所有可能性几乎都是不可能的（0.001）。)
In other words, our prior certainty is that I is cheap. (换句话说，我们的先验是I很便宜。)
Then we get: (然后我们得到：)
Prior
P(f=x) x P(o=2/2|f=x) P(o=2/2&& f=x) P(o=2/2&&f >= x) P(f >= x | o=2/2)
0.001 1 1 0.001 0.001 0.072727273
0.001 0.9 0.81 0.00081 0.00181 0.131636364
0.001 0.8 0.64 0.00064 0.00245 0.178181818
0.001 0.7 0.49 0.00049 0.00294 0.213818182
0.001 0.6 0.36 0.00036 0.0033 0.24
0.001 0.5 0.25 0.00025 0.00355 0.258181818
0.001 0.4 0.16 0.00016 0.00371 0.269818182
0.001 0.3 0.09 0.00009 0.0038 0.276363636
0.001 0.2 0.04 0.00004 0.00384 0.279272727
0.991 0.1 0.01 0.00991 0.01375 1
P(o=2/2) 0.01375
Now it says P(f >= 0.5) is 26%, up from the prior assumption of 0.6%.
(现在它说P（f> = 0.5）为26％，高于先前假设的0.6％。)
So Bayes allows us to update our estimate of the probable cost of I . (因此，贝叶斯允许我们更新对I的可能成本的估计。)
If the amount of data is small, it doesn't tell us accurately what the cost is, only that it is big enough to be worth fixing. (如果数据量很小，它并不能准确地告诉我们成本是多少，而只是告诉我们它足够大才能值得解决。)
Yet another way to look at it is called the Rule Of Succession .
(另一种看待它的方法称为继承规则 。)
If you flip a coin 2 times, and it comes up heads both times, what does that tell you about the probable weighting of the coin? (如果您掷硬币两次，并且两次都出现正面，那么该硬币可能的权重又告诉您什么？)
The respected way to answer is to say that it's a Beta distribution, with average value (number of hits + 1) / (number of tries + 2) = (2+1)/(2+2) = 75%. (尊重的回答方式是说它是Beta分布，平均值（命中数+ 1）/（尝试数+ 2）=（2 + 1）/（2 + 2）= 75％。)
(The key is that we see I more than once. If we only see it once, that doesn't tell us much except that f > 0.)
(（关键是我们见到I不止一次。如果我们只看到一次，那么除了f > 0之外，这对我们没有多大意义。）)
So, even a very small number of samples can tell us a lot about the cost of instructions that it sees.
(因此，即使是非常少量的样本也可以告诉我们有关它所看到的指令成本的很多信息。)
(And it will see them with a frequency, on average, proportional to their cost. If n samples are taken, and f is the cost, then I will appear on nf+/-sqrt(nf(1-f)) samples. Example, n=10 , f=0.3 , that is 3+/-1.4 samples.) (（平均来看，它们的频率与成本成正比。如果抽取n样本，而f为成本，那么I将出现在nf+/-sqrt(nf(1-f))样本上。 ， n=10 ， f=0.3 ，即3+/-1.4样本。）)
ADDED, to give an intuitive feel for the difference between measuring and random stack sampling:
(添加，以便直观地了解测量和随机堆栈采样之间的区别：)
There are profilers now that sample the stack, even on wall-clock time, but what comes out is measurements (or hot path, or hot spot, from which a "bottleneck" can easily hide).
(现在有分析器可以对堆栈进行采样，即使是在墙上时钟的时间，但结果就是测量值（或热路径或热点，“瓶颈”可以从中轻松隐藏）。)
What they don't show you (and they easily could) is the actual samples themselves. (他们没有向您显示（并且很容易做到）是实际样本本身。)
And if your goal is to find the bottleneck, the number of them you need to see is, on average , 2 divided by the fraction of time it takes. (而且，如果您的目标是找到瓶颈，那么平均而言 ，您需要查看的瓶颈数量为2除以所需的时间。)
So if it takes 30% of time, 2/.3 = 6.7 samples, on average, will show it, and the chance that 20 samples will show it is 99.2%. (因此，如果花费30％的时间，平均将显示2 / .3 = 6.7个样本，而20个样本将显示99.2％的机会。)
Here is an off-the-cuff illustration of the difference between examining measurements and examining stack samples.
(这是检验测量值和检验烟囱样品之间差异的现成插图。)
The bottleneck could be one big blob like this, or numerous small ones, it makes no difference. (瓶颈可能是这样的一个大斑点，也可能是很多小的斑点，这没有什么区别。)
