blocks|key|725364|text|首先，%2B%2B运算符的优先级高于*运算符，而()运算符的优先级高于其他所有运算符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|725365|其次，如果没有将%2B%2Bnumber运算符分配给任何对象，那么它们与number%2B%2B运算符是相同的。不同之处在于number%2B%2B先返回number，然后递增number，而%2B%2Bnumber先递增，然后再返回number。|725366|第三，通过增加指针的值，你增加了它的内容的大小，也就是说，你增加了它，就像你在数组中迭代一样。|725367|所以，总结一下：|725368|ptr%2B%2B;++++//+Pointer+moves+to+the+next+int+position+(as+if+it+was+an+array)
%2B%2Bptr;++++//+Pointer+moves+to+the+next+int+position+(as+if+it+was+an+array)
%2B%2B*ptr;+++//+The+value+pointed+at+by+ptr+is+incremented
%2B%2B(*ptr);+//+The+value+pointed+at+by+ptr+is+incremented
%2B%2B*(ptr);+//+The+value+pointed+at+by+ptr+is+incremented
*ptr%2B%2B;+++//+Pointer+moves+to+the+next+int+position+(as+if+it+was+an+array).+But+returns+the+old+content
(*ptr)%2B%2B;+//+The+value+pointed+at+by+ptr+is+incremented
*(ptr)%2B%2B;+//+Pointer+moves+to+the+next+int+position+(as+if+it+was+an+array).+But+returns+the+old+content
*%2B%2Bptr;+++//+Pointer+moves+to+the+next+int+position,+and+then+get's+accessed,+with+your+code,+segfault
*(%2B%2Bptr);+//+Pointer+moves+to+the+next+int+position,+and+then+get's+accessed,+with+your+code,+segfault|code-block|syntax|javascript|725369|因为这里有很多案例，我可能犯了一些错误，如果我错了，请纠正我。|725370|编辑：|725371|所以我错了，优先级比我写的要复杂一点，请在这里查看：http://en.cppreference.com/w/cpp/language/operator_precedence|offset|length|725372|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/language/operator_precedence^0|0|0|0|0|0|0|0|Q|1P|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|14|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|15|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@$S|1A|T|1B|1|1C]]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

First, the ++ operator takes precedence over the * operator, and the () operators take precedence over everything else.
Second, the ++number operator is the same as the number++ operator if you're not assigning them to anything. The difference is number++ returns number and then increments number, and ++number increments first and then returns it.
Third, by increasing the value of a pointer, you're incrementing it by the sizeof its contents, that is you're incrementing it as if you were iterating in an array.
So, to sum it all up:
<pre><code>ptr++; // Pointer moves to the next int position (as if it was an array)
++ptr; // Pointer moves to the next int position (as if it was an array)
++*ptr; // The value pointed at by ptr is incremented
++(*ptr); // The value pointed at by ptr is incremented
++*(ptr); // The value pointed at by ptr is incremented
*ptr++; // Pointer moves to the next int position (as if it was an array). But returns the old content
(*ptr)++; // The value pointed at by ptr is incremented
*(ptr)++; // Pointer moves to the next int position (as if it was an array). But returns the old content
*++ptr; // Pointer moves to the next int position, and then get's accessed, with your code, segfault
*(++ptr); // Pointer moves to the next int position, and then get's accessed, with your code, segfault
</code></pre>
As there are a lot of cases in here, I might have made some mistake, please correct me if I'm wrong.
EDIT:
So I was wrong, the precedence is a little more complicated than what I wrote, view it here:
<a href="http://en.cppreference.com/w/cpp/language/operator_precedence" rel="noreferrer">http://en.cppreference.com/w/cpp/language/operator_precedence</a>

blocks|key|725419|text|检查了程序，结果是，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|725420|p%2B%2B;++++//+use+it+then+move+to+next+int+position
%2B%2Bp;++++//+move+to+next+int+and+then+use+it
%2B%2B*p;+++//+increments+the+value+by+1+then+use+it+
%2B%2B(*p);+//+increments+the+value+by+1+then+use+it
%2B%2B*(p);+//+increments+the+value+by+1+then+use+it
*p%2B%2B;+++//+use+the+value+of+p+then+moves+to+next+position
(*p)%2B%2B;+//+use+the+value+of+p+then+increment+the+value
*(p)%2B%2B;+//+use+the+value+of+p+then+moves+to+next+position
*%2B%2Bp;+++//+moves+to+the+next+int+location+then+use+that+value
*(%2B%2Bp);+//+moves+to+next+location+then+use+that+value|code-block|syntax|javascript|725421|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

checked the program and the results are as,

<pre><code>p++; // use it then move to next int position
++p; // move to next int and then use it
++*p; // increments the value by 1 then use it 
++(*p); // increments the value by 1 then use it
++*(p); // increments the value by 1 then use it
*p++; // use the value of p then moves to next position
(*p)++; // use the value of p then increment the value
*(p)++; // use the value of p then moves to next position
*++p; // moves to the next int location then use that value
*(++p); // moves to next location then use that value
</code></pre>

blocks|key|509011|text|以下是各种“只需打印”建议的实例化。我发现它很有启发性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|509012|#include+"stdio.h"

int+main()+{
++++static+int+x+=+5;
++++static+int+*p+=+&x;
++++printf("(int)+p+++=>+%25d\n",(int)+p);
++++printf("(int)+p%2B%2B+=>+%25d\n",(int)+p%2B%2B);
++++x+=+5;+p+=+&x;
++++printf("(int)+%2B%2Bp+=>+%25d\n",(int)+%2B%2Bp);
++++x+=+5;+p+=+&x;
++++printf("%2B%2B*p++++++=>+%25d\n",%2B%2B*p);
++++x+=+5;+p+=+&x;
++++printf("%2B%2B(*p)++++=>+%25d\n",%2B%2B(*p));
++++x+=+5;+p+=+&x;
++++printf("%2B%2B*(p)++++=>+%25d\n",%2B%2B*(p));
++++x+=+5;+p+=+&x;
++++printf("*p%2B%2B++++++=>+%25d\n",*p%2B%2B);
++++x+=+5;+p+=+&x;
++++printf("(*p)%2B%2B++++=>+%25d\n",(*p)%2B%2B);
++++x+=+5;+p+=+&x;
++++printf("*(p)%2B%2B++++=>+%25d\n",*(p)%2B%2B);
++++x+=+5;+p+=+&x;
++++printf("*%2B%2Bp++++++=>+%25d\n",*%2B%2Bp);
++++x+=+5;+p+=+&x;
++++printf("*(%2B%2Bp)++++=>+%25d\n",*(%2B%2Bp));
++++return+0;
}|code-block|syntax|javascript|509013|它返回|509014|(int)+p+++=>+256688152
(int)+p%2B%2B+=>+256688152
(int)+%2B%2Bp+=>+256688156
%2B%2B*p++++++=>+6
%2B%2B(*p)++++=>+6
%2B%2B*(p)++++=>+6
*p%2B%2B++++++=>+5
(*p)%2B%2B++++=>+5
*(p)%2B%2B++++=>+5
*%2B%2Bp++++++=>+0
*(%2B%2Bp)++++=>+0|509015|我将指针地址转换为int，以便比较它们。|offset|length|style|CODE|509016|我和GCC一起编的。|509017|entityMap^0|0|0|0|0|9|3|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

The following is an instantiation of the various "just print it" suggestions. I found it instructive.

<pre><code>#include "stdio.h"

int main() {
 static int x = 5;
 static int *p = &amp;x;
 printf("(int) p =&gt; %d\n",(int) p);
 printf("(int) p++ =&gt; %d\n",(int) p++);
 x = 5; p = &amp;x;
 printf("(int) ++p =&gt; %d\n",(int) ++p);
 x = 5; p = &amp;x;
 printf("++*p =&gt; %d\n",++*p);
 x = 5; p = &amp;x;
 printf("++(*p) =&gt; %d\n",++(*p));
 x = 5; p = &amp;x;
 printf("++*(p) =&gt; %d\n",++*(p));
 x = 5; p = &amp;x;
 printf("*p++ =&gt; %d\n",*p++);
 x = 5; p = &amp;x;
 printf("(*p)++ =&gt; %d\n",(*p)++);
 x = 5; p = &amp;x;
 printf("*(p)++ =&gt; %d\n",*(p)++);
 x = 5; p = &amp;x;
 printf("*++p =&gt; %d\n",*++p);
 x = 5; p = &amp;x;
 printf("*(++p) =&gt; %d\n",*(++p));
 return 0;
}
</code></pre>

It returns

<pre><code>(int) p =&gt; 256688152
(int) p++ =&gt; 256688152
(int) ++p =&gt; 256688156
++*p =&gt; 6
++(*p) =&gt; 6
++*(p) =&gt; 6
*p++ =&gt; 5
(*p)++ =&gt; 5
*(p)++ =&gt; 5
*++p =&gt; 0
*(++p) =&gt; 0
</code></pre>

I cast the pointer addresses to <code>int</code>s so they could be easily compared.

I compiled it with GCC.

blocks|key|725507|text|++++++++Note:
++++++++1)+Both+%2B%2B+and+*+have+same+precedence(priority),+so+the+associativity+comes+into+picture.
++++++++2)+in+this+case+Associativity+is+from+**Right-Left**

++++++++important+table+to+remember+in+case+of+pointers+and+arrays:+

++++++++operators+++++++++++precedence++++++++associativity

++++1)++()+,+[]++++++++++++++++1+++++++++++++++left-right
++++2)++*++,+identifier++++++++2+++++++++++++++right-left
++++3)++<data+type>++++++++++++3+++++++++++++++----------

++++++++let+me+give+an+example,+this+might+help;

++++++++char+**str;
++++++++str+=+(char+**)malloc(sizeof(char*)*2);+//+allocate+mem+for+2+char*
++++++++str[0]=(char+*)malloc(sizeof(char)*10);+//+allocate+mem+for+10+char
++++++++str[1]=(char+*)malloc(sizeof(char)*10);+//+allocate+mem+for+10+char

++++++++strcpy(str[0],"abcd");++//+assigning+value
++++++++strcpy(str[1],"efgh");++//+assigning+value

++++++++while(*str)
++++++++{
++++++++++++cout<<*str<<endl;+++//+printing+the+string
++++++++++++*str%2B%2B;+++++++++++++//+incrementing+the+address(pointer)
++++++++++++++++++++++++++++++++//+check+above+about+the+prcedence+and+associativity
++++++++}
++++++++free(str[0]);
++++++++free(str[1]);
++++++++free(str);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|725508|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> Note:
 1) Both ++ and * have same precedence(priority), so the associativity comes into picture.
 2) in this case Associativity is from **Right-Left**

 important table to remember in case of pointers and arrays: 

 operators precedence associativity

 1) () , [] 1 left-right
 2) * , identifier 2 right-left
 3) &lt;data type&gt; 3 ----------

 let me give an example, this might help;

 char **str;
 str = (char **)malloc(sizeof(char*)*2); // allocate mem for 2 char*
 str[0]=(char *)malloc(sizeof(char)*10); // allocate mem for 10 char
 str[1]=(char *)malloc(sizeof(char)*10); // allocate mem for 10 char

 strcpy(str[0],"abcd"); // assigning value
 strcpy(str[1],"efgh"); // assigning value

 while(*str)
 {
 cout&lt;&lt;*str&lt;&lt;endl; // printing the string
 *str++; // incrementing the address(pointer)
 // check above about the prcedence and associativity
 }
 free(str[0]);
 free(str[1]);
 free(str);
</code></pre>

Let us assume,
<pre><code>int *p;
int a = 100;
p = &amp;a;
</code></pre>
What will the following code actually do and how?
<pre><code>p++;
++p;
++*p;
++(*p);
++*(p);
*p++;
(*p)++;
*(p)++;
*++p;
*(++p);
</code></pre>
I know, this is kind of messy in terms of coding, but I want to know what will actually happen when we code like this.
Note : Lets assume that the address of <code>a=5120300</code>, it is stored in pointer <code>p</code> whose address is <code>3560200</code>. Now, what will be the value of <code>p &amp; a</code> after the execution of each statement?

How to increment a pointer address and pointer's value?

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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让我们假设，int *p;int a = 100;p = &a;下面的代码实际上会做什么以及如何做？p++;++p;++*p;++(*p);++*(p);*p++;(*p)++;*(p)++;*++p;*(++p);我知道，这在编码方面有点混乱，但我想知道当我们这样编码时，实际会发生什么。注意:假设a=5120300的...

问如何递增指针地址和指针的值？
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问如何递增指针地址和指针的值？
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