blocks|key|4402330|text|m+=+n%2B%2B+-j%2B10;与|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4402331|m+=+n+-j%2B10;
n+=+n+%2B+1;+//+m+is+11.|code-block|syntax|javascript|4402332|如果它是%2B%2Bn，它将是|4402333|n+=+n+%2B+1;
m+=+n+-j%2B10;+//m+is+12.|4402334|4402335|，那么表达式就求解了，即j%2B10+=+25|blockquote|4402336|4402337|4402338|不是的。应该是-j%2B10+=+-5|4402339|我的建议是，不要写复杂的表达式，除非你完全确定你在写什么。|BOLD|4402340|entityMap^0|0|E|0|0|4|3|0|0|0|0|0|0|7|5|0|6|N|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1A|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|W|3|X|5|6|7|1G|8|@$9|1H|A|1I|B|Y]]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|10|$]]

<code>m = n++ -j+10;</code> is same as 

<pre><code>m = n -j+10;
n = n + 1; // m is 11.
</code></pre>

If it was <code>++n</code> It would be 

<pre><code>n = n + 1;
m = n -j+10; //m is 12.
</code></pre>

<blockquote>
 then the expression is solved i.e. j+10 = 25
</blockquote>

No. It would be <code>-j+10</code> = -5

My suggestion is, dont write complex expression unless you are completely sure what you are writing.

blocks|key|363129|text|你在这方面有一些地方错了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|363130|363131|n%2B%2B将递增n并返回原始结果，因此您将获得m+=+16+....|ordered-list-item|offset|length|style|CODE|363132|-j，因此您将获得m+=+16+-+15+....|363133|%2B10，因此您将获得m+=+16+-+15+%2B+10.|363134|363135|上一次我做数学运算的时候，就像你们看到的，结果是m+=+11。|363136|如果你希望它是m+=+17+-+(15+%2B+10)，那么你想：|363137|int+m,+n,+j;
n=16;
j=15;
m+=+%2B%2Bn+-(j%2B10);
printf("%25d",+m);|code-block|syntax|javascript|363138|entityMap^0|0|0|0|3|6|1|L|A|0|0|2|9|F|0|0|3|A|G|0|0|O|6|0|7|I|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|11|8|@$F|12|G|13|H|I]|$F|14|G|15|H|I]|$F|16|G|17|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|18|8|@$F|19|G|1A|H|I]|$F|1B|G|1C|H|I]]|9|@]|A|$]]|$1|L|3|M|5|E|7|1D|8|@$F|1E|G|1F|H|I]|$F|1G|G|1H|H|I]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1J|8|@$F|1K|G|1L|H|I]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1M|8|@$F|1N|G|1O|H|I]]|9|@]|A|$]]|$1|S|3|T|5|U|7|1P|8|@]|9|@]|A|$V|W]]|$1|X|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|Y|$]]

You've got a few things wrong there.

<ol>
<li><code>n++</code> will increment <code>n</code> and return the original result, so you've then got <code>m = 16 ...</code>.</li>
<li><code>-j</code> so you've got <code>m = 16 - 15 ...</code>.</li>
<li><code>+10</code> so you've got <code>m = 16 - 15 + 10</code>.</li>
</ol>

Now the last time I did maths that would come out as <code>m = 11</code> like you're seeing.

If you wanted it to be <code>m = 17 - (15 + 10)</code> then you wanted:

<pre><code>int m, n, j;
n=16;
j=15;
m = ++n -(j+10);
printf("%d", m);
</code></pre>

blocks|key|363247|text|实际上，后增量操作是在操作之后对n执行的…你有16-15%2B10+=+11，但如果你打印n，你应该有17。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|363248|首先，你可以阅读一些关于基础知识的书籍，但这个例子并不简单；它包含了操作符的优先顺序，这可能会很棘手。|363249|开始简单..。用c.+http://www.cs.cf.ac.uk/Dave/C/node4.html编写不可读的代码非常简单。|offset|length|363250|希望能有所帮助|363251|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cs.cf.ac.uk/Dave/C/node4.html^0|0|0|B|14|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$F|U|G|V|1|W]]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

in fact the post increment operation is done on n after the operation...
you have 16-15+10 = 11
but if you print n you should have 17.

to begin, you can read some book on basics but this example is not simple; it include the precedence of operator which can be tricky.

begin simple... it's quite simple to write unreadable code in c.
<a href="http://www.cs.cf.ac.uk/Dave/C/node4.html" rel="nofollow">http://www.cs.cf.ac.uk/Dave/C/node4.html</a>

hope it helps

blocks|key|4402242|text|n%2B%2B首先返回n的值，然后递增该值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4402243|因此，实际发生的计算是m+=+16+-+15+%2B+10是11|4402244|我想你想要的是：|4402245|m+=+(n%2B1)+-+(j%2B10);|code-block|syntax|javascript|4402246|%2B%2B运算符的用法是在使用n的当前值计算m之后，递增其值以供将来使用。|4402247|entityMap^0|0|3|7|1|0|B|G|0|0|0|0|2|C|1|J|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|10|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|11|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|R|$]]

<code>n++</code> first returns the value of <code>n</code> and then increments it. 
 so, the actual computation that takes place is <code>m = 16 - 15 + 10</code> which is 11

i think what you want is:

<pre><code>m = (n+1) - (j+10);
</code></pre>

the use of the <code>++</code> operator is to increment the value of <code>n</code> for future use after you use it's current value to compute <code>m</code>.

blocks|key|363076|text|对于m，首先计算n+-+j+%2B+10并将其分配给m。在此之后，将执行n%2B%2B。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|363077|在n+=+17的末尾，m+=+11|363078|entityMap^0|2|1|8|A|O|1|Y|3|0|1|6|B|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|I|$]]

for <code>m</code>, first of all calculate <code>n - j + 10</code> and assign it to <code>m</code>. After that <code>n++</code> is executed.

at the end <code>n = 17</code>, <code>m = 11</code>

blocks|key|4402409|text|你做了两个错误的解释。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4402410|首先，正如在其他答案中指出的那样，n%2B%2B仅在整个表达式计算完毕后才会递增n。|offset|length|style|CODE|4402411|其次，你有-j%2B10。这不等于-(j%2B10)，所以说j%2B10是25，而您正在寻找something+-+25，这是错误的。另一种看待-j%2B10的方式是10-j。|4402412|entityMap^0|0|H|3|10|1|0|5|5|F|7|Q|4|V|2|14|E|1U|5|23|4|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|K|$]]

You're making two incorrect interpretations.

Firstly, as indicated in other answers, <code>n++</code> only increments <code>n</code> after the entire expression has been evaluated.

Secondly, you have <code>-j+10</code>. This is not equal to <code>-(j+10)</code>, so it is wrong to say that <code>j+10</code> is <code>25</code> and you are looking a <code>something - 25</code>. Another way of view <code>-j+10</code> is <code>10-j</code>.

blocks|key|414741|text|回答你关于一本好书的问题--你可能想要考虑学习C%2B%2B而不仅仅是普通的老式C语言，因为C%2B%2B是C的超集。对于C%2B%2B，你需要得到Bjarne+Stroustrup的“C%2B%2B编程语言”。这本书很容易阅读，并且会在你的书架上保存很长一段时间，作为一个很好的参考。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|414742|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

In answer to your question about a good book - you probably want to consider learning C++ instead of just plain old C, since C++ is a superset of C. And for C++ you need to get Bjarne Stroustrup's 'The C++ Programming Language'. It's easy enough to read and will last a long time on your bookshelf as a good reference.

blocks|key|4402549|text|在expression+m+=+n%2B%2B+-j%2B10;中|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4402550|编译器将表达式视为m=++n%2B%2B+((-j)%2B10)|4402551|因为n和j的初始化值是n+=+16和j+=+15。我们有m+=+16%2B%2B+((-15)%2B10)。我们得到的输出是11。|4402552|执行完表达式后，n将递增。|4402553|entityMap^0|C|E|0|9|H|0|B|6|I|6|S|J|1K|2|0|8|1|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|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|M|$]]

In the expression <code>m = n++ -j+10;</code> 
The compiler treats the expression as <code>m= n++ ((-j)+10)</code> 
As the intialized values of n and j are <code>n = 16</code> and <code>j = 15</code>. We have <code>m = 16++ ((-15)+10)</code>. We get output as <code>11</code>. 
After the expression is executed <code>n</code> will be incremented.

blocks|key|4402267|text|n在之后递增，它的值在表达式中使用，最后以m结束。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|4402268|entityMap^0|0|1|L|1|6|J|0^^$0|@$1|2|3|4|5|6|7|I|8|@$9|J|A|K|B|C]|$9|L|A|M|B|C]|$9|N|A|O|B|D]]|E|@]|F|$]]|$1|G|3|-4|5|6|7|P|8|@]|E|@]|F|$]]]|H|$]]

<code>n</code> is incremented after its value is used in the expression which ends up in <code>m</code>.

blocks|key|363165|text|n%2B%2B是后增量。它只会在执行以下操作后递增n的值：m+=+n%2B%2B+-j%2B10;|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|363166|%2B%2Bn是预增量。它在计算m.+m+=+%2B%2Bn+-j%2B10;之前递增n的值。|363167|entityMap^0|0|3|L|1|P|E|4|3|0|0|3|F|E|X|1|0^^$0|@$1|2|3|4|5|6|7|K|8|@$9|L|A|M|B|C]|$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|E|@]|F|$]]|$1|I|3|-4|5|6|7|10|8|@]|E|@]|F|$]]]|J|$]]

<code>n++</code> is post incrementation. It only increments the value of <code>n</code> after doing: <code>m = n++ -j+10;</code>

<code>++n</code> is pre incrementation. It increments the value of <code>n</code> before calculing m. <code>m = ++n -j+10;</code>

<pre><code>int m, n, j;
n=16;
j=15;
m = n++ -j+10;
printf("%d", m);
</code></pre>

Output: 11.

Here, first, the old value of <code>n</code> is given to <code>m</code> and then it is incremented so the new value i get is 17 and then the expression is solved i.e. <code>j+10</code> = 25 then the new value of n is subtracted by 25 i.e 17-25.
Am i right ? but the answer doesn't match the output <code>11</code>.
Then how does this work ?
And also, i am new to programming and started learning C. Which book will you suggest is the best for me ? As I've no programming experience.
Thank you.

Can someone tell me how does this work ? (C programming)

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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int m, n, j;n=16;j=15;m = n++ -j+10;printf("%d", m);输出: 11。这里，首先，n的旧值被赋予m，然后它被递增，因此我得到的新值是17，然后求解表达式，即j+10 = 25，然后将n的新值减去25，即17-25。我说的对吗？但是答案与输出的11不匹配。那么这是如何工作的呢？此外，我是编程新手，并开始学习C语言。你会推荐哪本书最适合我？因为我没有编程

问有人能告诉我这是怎么工作的吗？(C编程)
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回答 10

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