blocks|key|897548|text|*运算符在%2B%2B之后工作，因此它返回错误地址的值。在c%2B%2B中，需要知道运算符的优先顺序。看看这个：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|897549|http://en.cppreference.com/w/cpp/language/operator_precedence|897550|添加括号以更改运算符的优先顺序，例如：|897551|#include+<iostream>
using+namespace+std;
void+__print(int+x,+int+*px)+{
+cout+<<+"(x,+*px)+=+("+<<+x+<<+",+"+<<+*px+<<+")"+<<+endl;
}

int+main()+{
+int+x+=+99;
+int+*px+=+&x;
+__print(x,+px);
+x%2B%2B;+__print(x,+px);
+x+=+x+%2B+1;+__print(x,+px);
+*px+=+*px+%2B+1;+__print(x,+px);
+(*px)%2B%2B;+__print(x,+px);
+return+0;
}|code-block|syntax|javascript|897552|结果：|897553|(x,+*px)+=+(99,+99)
(x,+*px)+=+(100,+100)
(x,+*px)+=+(101,+101)
(x,+*px)+=+(102,+102)
(x,+*px)+=+(103,+103)|897554|entityMap|0|LINK|mutability|MUTABLE|url^0|0|1|5|2|0|0|1P|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|H|3|I|5|6|7|18|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|19|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1B|8|@]|D|@]|E|$M|N]]|$1|S|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|G]]]]

the <code>*</code> operator works after the <code>++</code> so it returns the value of a wrong address. the operator precedence is important to know in c++. take a look at this :
<a href="http://en.cppreference.com/w/cpp/language/operator_precedence" rel="nofollow noreferrer">http://en.cppreference.com/w/cpp/language/operator_precedence</a>
Add parentheses to change operator precedence, for example:
<pre><code>#include &lt;iostream&gt;
using namespace std;
void __print(int x, int *px) {
 cout &lt;&lt; &quot;(x, *px) = (&quot; &lt;&lt; x &lt;&lt; &quot;, &quot; &lt;&lt; *px &lt;&lt; &quot;)&quot; &lt;&lt; endl;
}

int main() {
 int x = 99;
 int *px = &amp;x;
 __print(x, px);
 x++; __print(x, px);
 x = x + 1; __print(x, px);
 *px = *px + 1; __print(x, px);
 (*px)++; __print(x, px);
 return 0;
}
</code></pre>
result:
<pre><code>(x, *px) = (99, 99)
(x, *px) = (100, 100)
(x, *px) = (101, 101)
(x, *px) = (102, 102)
(x, *px) = (103, 103)
</code></pre>

blocks|key|897556|text|问题出在运算符的优先级上。试试(*px)%2B%2B；|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|897557|entityMap^0|F|7|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

The problem is with operator precedence. Try <code>(*px)++</code>;

blocks|key|1331180|text|当你执行*px%2B%2B时，你实际上是在地址上加1，然后得到值。你很可能指的是(*px)%2B%2B。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1331181|entityMap^0|4|5|10|7|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

When you do <code>*px++</code> you actually add 1 to the address and then get the value. You most likely meant <code>(*px)++</code>.

blocks|key|1919537|text|*px+%2B1取px指向的值并将其加1|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1919538|*px%2B%2B首先递增地址，然后取px指向的值|1919539|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

*px + 1 takes the value that px points to and add it by 1

*px++ increments the address first and then takes the value px points to

blocks|key|1914466|text|这是一个优先级的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1914467|*px+=+*px+%2B1读取为(*px)+=+(*px)+%2B+1，因为px+=+&x，所以它类似于x=x%2B1。|offset|length|style|CODE|1914468|*px%2B%2B读取为*(px%2B%2B)，因此您实际上是将指针向前移动，并获取内存作为相对于x变量的下一个位置的值(最有可能是垃圾)。|1914469|entityMap^0|0|0|C|F|H|1D|5|0|0|5|8|7|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]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$]]

It's a problem of precedence:

<code>*px = *px +1</code> reads as <code>(*px) = (*px) + 1</code>, and since px = &amp;x, it is like <code>x=x+1</code>.

<code>*px++</code> reads as <code>*(px++)</code> hence you are actually moving the pointer forward and getting the value of what the memory holds as the next position respect to the x variable (most likely garbage).

blocks|key|897637|text|%2B%2B的优先级大于*。因此计算(px%2B%2B)。现在*(px%2B%2B)是一个浪费的声明。您既没有从该位置赋值，也没有从该位置读取值。此外，(px%2B%2B)具有可能尚未初始化的位置的地址。所以你得到的是垃圾值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|897638|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|$]]

Precedence for ++ is more than *. Hence (px++) is calculated. Now *(px++) is a waste statement. You are neither assigning nor reading the value from the location. Moreover (px++) has a address of a location which may not have initialized. So you are getting the garbage value.

blocks|key|1914520|text|哇，这么多人不知道C%2B%2B运算符是如何工作的。:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1914521|每个人都正确地指出了这是一个优先顺序的问题，然而，问题到底是什么，似乎每个人都回避了。|1914522|*p%2B%2B；作为一条语句只做一件事。它会递增指针。在对其求值并取消引用其原始值之后(在本例中将忽略该原始值)。|1914523|int++a[+2+]+=+{+10,+20+};
int*+b+=+&a[+0+];
int++c+=+*b%2B%2B;|code-block|syntax|javascript|1914524|在上面的例子中，c将等于10，b将指向a+(20)的第二个元素。因为指针b将在增量之前求值。|offset|length|style|CODE|1914525|entityMap^0|0|0|0|0|10|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Wow, so many people don't know how C++ operators work. :)

Everyone is correct in pointing out it's a problem with the order of precedence, exactly what the problem is, however, seems to be eluding everyone.

*p++; as a statement does exactly one thing. It increments the pointer. After evaluating it and dereferencing its original value (which is then ignored in this case).

<pre><code>int a[ 2 ] = { 10, 20 };
int* b = &amp;a[ 0 ];
int c = *b++;
</code></pre>

In the above example, c will equal 10, and b will point to the second element of a (20). Because the pointer <code>b</code> will be evaluated before the increment.

I am taking my first steps in C++ having a good background in Java. I need to clear out some peculiarities of the ++ operator in C++. Consider the following program:

<pre><code>#include &lt;iostream&gt;
using namespace std;
void __print(int x, int *px) {
 cout &lt;&lt; "(x, *px) = (" &lt;&lt; x &lt;&lt; ", " &lt;&lt; *px &lt;&lt; ")" &lt;&lt; endl;
}

int main() {
 int x = 99;
 int *px = &amp;x;
 __print(x, px);
 x++; __print(x, px);
 x = x + 1; __print(x, px);
 *px = *px + 1; __print(x, px);
 *px++; __print(x, px);
 return 0;
}
</code></pre>

Surprisingly to me, the program prints:

<pre><code>(x, *px) = (99, 99)
(x, *px) = (100, 100)
(x, *px) = (101, 101)
(x, *px) = (102, 102)
(x, *px) = (102, 134514848)
</code></pre>

It seems that *px = *px + 1 does not have the same effect on *px as on x. But aren't these things the same??? Isn't it *px == x?

C++ operator ++ and pointers

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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我在C++上迈出了我的第一步，有很好的Java背景。我需要清除C++中++运算符的一些特性。考虑以下程序：#include <iostream>using namespace std;void __print(int x, int *px) { cout << "(x, *px) = (" << x << ", " <...

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