blocks|key|1017338|text|您所写的内容相当于|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1017339|def+anon():
++++return+print+"x"|code-block|syntax|javascript|1017340|这也会导致一个SyntaxError，python不允许在2.xx中为print赋值；在python3中您可以这样说|1017341|lambda:+print('hi')|1017342|它可以工作，因为他们已经将print更改为一个函数，而不是语句。|1017343|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

what you've written is equivalent to

<pre><code>def anon():
 return print "x"
</code></pre>

which also results in a SyntaxError, python doesn't let you assign a value to print in 2.xx; in python3 you could say

<pre><code>lambda: print('hi')
</code></pre>

and it would work because they've changed print to be a function instead of a statement.

blocks|key|1017407|text|lambda的主体必须是返回值的表达式。作为一条语句，print不会返回任何内容，甚至None也不会。同样，您不能将print的结果赋给一个变量：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1017408|>>>+x+=+print+"hello"
++File+"<stdin>",+line+1
++++x+=+print+"hello"
++++++++++++%5E
SyntaxError:+invalid+syntax|code-block|syntax|javascript|1017409|您也不能将变量赋值放在lambda中，因为赋值是语句：|1017410|>>>+lambda+y:+(x+=+y)
++File+"<stdin>",+line+1
++++lambda+y:+(x+=+y)
+++++++++++++++++%5E
SyntaxError:+invalid+syntax|1017411|entityMap^0|R|5|17|4|1M|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

The body of a lambda has to be an expression that returns a value. <code>print</code>, being a statement, doesn't return anything, not even <code>None</code>. Similarly, you can't assign the result of <code>print</code> to a variable:

<pre><code>&gt;&gt;&gt; x = print "hello"
 File "&lt;stdin&gt;", line 1
 x = print "hello"
 ^
SyntaxError: invalid syntax
</code></pre>

You also can't put a variable assignment in a lambda, since assignments are statements:

<pre><code>&gt;&gt;&gt; lambda y: (x = y)
 File "&lt;stdin&gt;", line 1
 lambda y: (x = y)
 ^
SyntaxError: invalid syntax
</code></pre>

blocks|key|1232068|text|Here，你看到了你问题的答案。它说，print不是Python中的表达式。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1232069|entityMap|0|LINK|mutability|MUTABLE|url|http://mail.python.org/pipermail/tutor/2007-July/055260.html^0|J|5|0|4|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<a href="http://mail.python.org/pipermail/tutor/2007-July/055260.html" rel="nofollow noreferrer">Here</a>, you see an answer for your question. <code>print</code> is not expression in Python, it says.

blocks|key|1232119|text|在我使用它进行简单的存根输出的情况下，我使用以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1232120|fn+=+lambda+x:+sys.stdout.write(str(x)+%2B+"\n")|code-block|syntax|javascript|1232121|它工作得很完美。|1232122|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|$]]

In cases where I am using this for simple stubbing out I use this:

<pre><code>fn = lambda x: sys.stdout.write(str(x) + "\n")
</code></pre>

which works perfectly.

blocks|key|1232171|text|使用Python3.x，print可以在lambda中工作，而不会改变lambda的语义。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1232172|以一种特殊的方式使用，这对于调试非常方便。我贴出了这个“延迟回复”，因为这是我经常使用的一个实用技巧。|1232173|假设您的“未插入指令的”lambda是：|1232174|lambda:+4|code-block|syntax|javascript|1232175|那么你的“仪表化”的lambda是：|1232176|lambda:+(print+(3),+4)+[1]|1232177|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

With Python 3.x, print CAN work in a lambda, without changing the semantics of the lambda.

Used in a special way this is very handy for debugging.
I post this 'late answer', because it's a practical trick that I often use.

Suppose your 'uninstrumented' lambda is:

<pre><code>lambda: 4
</code></pre>

Then your 'instrumented' lambda is:

<pre><code>lambda: (print (3), 4) [1]
</code></pre>

blocks|key|1017608|text|在python3中，打印是一个函数，您可以像Jacques+de+Hooge建议的那样打印和返回一些东西，但我喜欢其他方法：lambda+x:+print("Message")+or+x|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1017609|print函数不返回任何内容，因此None+or+x代码返回x：|1017610|只有当x为false-ish时，lambda+x:+x+or+print("Message")才会打印消息|1017611|这在lua中被广泛使用，在python中，您也可以用cond+and+a+or+b代替a+if+cond+else+b编写lua。|1017612|entityMap^0|1Q|V|0|0|5|H|9|U|1|0|G|V|0|Q|F|17|G|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|M|$]]

in python3 print is a function, and you can print and return something as Jacques de Hooge suggests, but i like other approach: <code>lambda x: print(&quot;Message&quot;) or x</code>
<code>print</code> function returns nothing, so <code>None or x</code> code returns <code>x</code>
other way around: <code>lambda x: x or print(&quot;Message&quot;)</code> would print message only if x is false-ish
this is widely used in lua, and in python you can too instead of <code>a if cond else b</code> write <code>cond and a or b</code>

Why doesn't this work?

<pre><code>lambda: print "x"
</code></pre>

Is this not a single statement, or is it something else?
The documentation seems a little sparse on what is allowed in a lambda...

Why doesn't print work in a lambda?

Python

为什么这个不起作用？lambda: print "x"这不是一个单独的陈述，还是其他的东西？文档似乎对lambda中允许的内容有点稀疏……

问为什么打印不能在lambda中工作？
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