blocks|key|2061212|text|您必须更改为以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2061213|def+increment(num):
++++try:
++++++++return+int(num)+%2B+1

++++except+Exception+as+e:
++++++++print("ufff+error+occured+:",+e)

n+=+input("enter+your+num:+")
a+=+increment(n)
print(a)|code-block|syntax|javascript|2061214|原因是int()函数引发了异常，因为它不能将您的输入转换为int。对int()的调用在增量函数之前。因此，该异常未被处理。|2061215|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You would have to change to the following:
<pre><code>def increment(num):
 try:
 return int(num) + 1

 except Exception as e:
 print(&quot;ufff error occured :&quot;, e)

n = input(&quot;enter your num: &quot;)
a = increment(n)
print(a)
</code></pre>
The reason for that is that the exception was raised by the int() functions, because it couldn't convert your input to int. The call to int() was before the increment function. Therefore the exception was unhandled.

blocks|key|2061248|text|在进入try/except块之前，您尝试将输入转换为int。如果你只是把它放在try块中，它就会工作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2061249|def+increment(num):
++++try:
++++++++num+=+int(num)
++++++++return+num+%2B+1

++++except+Exception+as+e:
++++++++print("ufff+error+occured+:",+e)


n+=+input("enter+your+num:+")
a+=+increment(n)
print(a)|code-block|syntax|javascript|2061250|entityMap^0|3|A|Q|3|13|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

You tried converting the input to <code>int</code> before going into the <code>try/except</code> block. If you just put it in the <code>try</code> block it will work.
<pre><code>def increment(num):
 try:
 num = int(num)
 return num + 1

 except Exception as e:
 print(&quot;ufff error occured :&quot;, e)


n = input(&quot;enter your num: &quot;)
a = increment(n)
print(a)
</code></pre>

blocks|key|4377889|text|我可以推荐以下内容吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4377890|def+increment(num):
++++try:
++++++++return+int(num)+%2B+1
++++except+Exception+as+e:
++++++++print("ufff+error+occurred:",+e)

n+=+input("enter+your+num:+")
a+=+increment(n)
print(a)|code-block|syntax|javascript|4377891|这里唯一的区别是，将用户输入转换为int的尝试发生在try-except块内部，而不是外部。这允许捕获类型转换错误和您的消息("ufff+error+occurred：...")要显示的。|offset|length|style|CODE|4377892|4377893|编辑:关于您的评论：|4377894|4377895|是的，谢谢大家，但谁能告诉我，如果我在输入行中使用int，我应该做什么，我需要做哪些更改|blockquote|4377896|4377897|4377898|或者，您可以执行以下操作：|4377899|def+increment(num):
++++return+num+%2B+1


try:
++++n+=+int(input("enter+your+num:+"))
++++a+=+increment(n)
++++print(a)
except+Exception+as+e:
++++print("ufff+error+occurred:",+e)|4377900|在这种情况下，我建议删除increment函数中的try-except块，因为您可以安全地假设传递给该函数的所有值都是数值。|4377901|entityMap^0|0|0|H|3|Q|A|0|0|0|0|0|0|0|0|0|C|9|P|A|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|15|8|@$I|16|J|17|K|L]|$I|18|J|19|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1B|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1D|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1G|8|@]|9|@]|A|$]]|$1|X|3|Y|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|Z|3|10|5|6|7|1I|8|@$I|1J|J|1K|K|L]|$I|1L|J|1M|K|L]]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|12|$]]

Might I recommend the following?
<pre class="lang-py prettyprint-override"><code>def increment(num):
 try:
 return int(num) + 1
 except Exception as e:
 print(&quot;ufff error occurred:&quot;, e)

n = input(&quot;enter your num: &quot;)
a = increment(n)
print(a)
</code></pre>
The only difference here is that the attempt to convert the user's input to an <code>int</code> occurs within the <code>try-except</code> block, rather than outside. This allows the type conversion error to be caught and your message (&quot;ufff error occurred: ...&quot;) to be displayed.
<hr />
Edit: regarding your comment:
<blockquote>
yes it worked out thanks everyone, but can anyone tell what should i have to do if i use int with input line what changes i've to make
</blockquote>
Optionally, you could do the following:
<pre class="lang-py prettyprint-override"><code>def increment(num):
 return num + 1


try:
 n = int(input(&quot;enter your num: &quot;))
 a = increment(n)
 print(a)
except Exception as e:
 print(&quot;ufff error occurred:&quot;, e)

</code></pre>
In this case, I would recommend removing the <code>try-except</code> block in the <code>increment</code> function, because you can safely assume that all values passed to that function will be numeric.

blocks|key|2061352|text|你必须在块中使用类型转换，这样如果num不是一个数字，控制器将移动到expect部分。这里的问题是您在try-catch块之外使用类型转换。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2061353|2061354|另一点需要注意的是，将e转换为string是一种很好的做法。|ordered-list-item|2061355|2061356|2061357|第三点需要注意的是，最后一行print(a)本身会在出现异常的情况下造成麻烦，因为您没有从异常部分返回任何内容，而是直接打印了错误。在那里使用return语句，否则它将返回None类型，这本身就会使流程变得混乱。|2061358|2061359|def+increment(num):
++++try:
++++++++return+int(num)+%2B+1++++++++#Updated+Line

++++except+Exception+as+e:
++++++++return+"ufff+error+occurred+:"+%2B+str(e)+++#Updated+Line

n+=+input("enter+your+num:+")
a+=+increment(n)
print(a)|code-block|syntax|javascript|2061360|entityMap^0|H|3|1F|9|0|0|B|1|0|0|0|E|8|1Z|6|2E|4|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|10|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|14|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|I|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1E|8|@]|D|@]|E|$R|S]]|$1|T|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|U|$]]

<ol>
<li>You have to use the type conversion within the try block so that if <code>num</code> is not a number, the controller would move to expect portion.
The problem here is you are using type conversion outside of <code>try-catch</code> block.
</li>
<li>Another point to note, convert <code>e</code> to string as a good practice.
</li>
<li>Third point to note,the last line <code>print(a)</code> will itself cause trouble in cases of exception as you have returned nothing from the exception part and directly printed the error. Use a <code>return</code> statement in there else it would return <code>None</code> type and which itself will mess up the flow.
</li>
</ol>
<pre><code>def increment(num):
 try:
 return int(num) + 1 #Updated Line

 except Exception as e:
 return &quot;ufff error occurred :&quot; + str(e) #Updated Line

n = input(&quot;enter your num: &quot;)
a = increment(n)
print(a)
</code></pre>

blocks|key|4377964|text|唯一会引发异常的代码是int(...)转换--它不受try/except块的保护。你应该把它放在try下|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4377965|def+increment():
++++try:
++++++++return+int(input("enter+your+num:+"))+%2B+1
++++except+ValueError+as+e:
++++++++return+f"ufff+error+occured:+{e}"

print(increment())|code-block|syntax|javascript|4377966|您应该始终尝试捕获最窄的异常。在本例中，为ValueError。|4377967|entityMap^0|B|8|Q|A|1C|3|0|0|L|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

The only code that would raise an exception is the <code>int(...)</code> conversion - and it is not protected under a <code>try/except</code> block. You should put it under the <code>try</code>:
<pre class="lang-py prettyprint-override"><code>def increment():
 try:
 return int(input(&quot;enter your num: &quot;)) + 1
 except ValueError as e:
 return f&quot;ufff error occured: {e}&quot;

print(increment())
</code></pre>
You should always try to catch the narrowest exception. In this case, a <code>ValueError</code>.

I am beginner to python and I have doubt in one program when using try except part
I try to take input from user and add that number by 1 like if you enter 5 then 6 will be printed but if you enter string then except would come in role and print enter only number
I've written code but except is not working there, code is working like if I enter 5 then 6 gets printed but except statement is not working
Here is the code that I've written
<pre><code>def increment(num):
 try:
 return num + 1

 except Exception as e:
 print(&quot;ufff error occured :&quot;, e)

n = int(input(&quot;enter your num: &quot;))
a = increment(n)
print(a)
</code></pre>

Exception In Python Doesn't Do Anything

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我是python初学者，在使用try时，我对一个程序有疑问。除了部分I尝试从用户输入并将数字加1，如果您输入5，则将打印6，但如果您输入字符串，则将打印6，但如果您输入字符串，则除了在角色中输入和打印外，仅输入数字我已经编写了代码，但除了在那里不起作用外，代码是否正常工作，如果我输入5，则打印6，但除了语句不起作用下面...

问Python中的异常不做任何事情
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