如何递增指针地址和指针的值？

Question

我们假设，

int *p;
int a = 100;
p = &a;

以下代码将实际执行什么以及如何执行？

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会发生什么。

注意：假设a=5120300的地址，它存储在地址为3560200指针p 。 现在，执行每个声明后， p & a的价值是多少？

Answer 1

首先，++运算符优先于*运算符，而（）运算符优先于其他所有运算符。

其次，如果您没有将它们分配给任何东西，则++数字运算符与数字++运算符相同。 差异是数字++返回数字然后递增数字，并且++数字首先递增然后返回它。

第三，通过增加指针的值，您可以通过其内容的大小来递增它，也就是说，您正在递增它，就好像您在数组中进行迭代一样。

所以，总结一下：

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 of ptr is incremented
++(*ptr); // The value of ptr is incremented
++*(ptr); // The value of 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 of 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

由于这里有很多案例，我可能犯了一些错误，如果我错了，请纠正我。

编辑：

所以我错了，优先级比我写的要复杂一点，在这里查看： http ： //en.cppreference.com/w/cpp/language/operator_precedence

Answer 2

检查程序，结果为，

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

Answer 3

关于“如何递增指针地址和指针的值？” 我认为++(*p++); 实际上是很好的定义，并做你要求的，例如：

#include <stdio.h>

int main() {
  int a = 100;
  int *p = &a;
  printf("%p\n",(void*)p);
  ++(*p++);
  printf("%p\n",(void*)p);
  printf("%d\n",a);
  return 0;
}

它不是在序列点之前修改两次相同的东西。 虽然对于大多数用途我并不认为它是好的风格 - 但是我觉得它有点太神秘了。

Answer 4

以下是各种“只是打印它”建议的实例化。 我发现它很有启发性。

#include "stdio.h"

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

它回来了

(int) p   => 256688152
(int) p++ => 256688152
(int) ++p => 256688156
++*p      => 6
++(*p)    => 6
++*(p)    => 6
*p++      => 5
(*p)++    => 5
*(p)++    => 5
*++p      => 0
*(++p)    => 0

我将指针地址转换为int s，以便可以轻松比较它们。

我用GCC编译了它。

Answer 5

        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)  <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++;             // incrementing the address(pointer)
                                // check above about the prcedence and associativity
        }
        free(str[0]);
        free(str[1]);
        free(str);