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上会发生什么。
注意:假设
a=5120300
的地址存储在指针p
中,其地址为3560200
。现在,每条语句执行后p & a
的值是多少?
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 ++number 运算符分配给任何内容,则 ++number 运算符与 number++ 运算符相同。区别是 number++ 返回 number 然后将 number 递增,而 ++number 先递增然后返回。
第三,通过增加指针的值,您可以按其内容的大小来递增它,也就是说,您可以像在数组中迭代一样来递增它。
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 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 gets accessed, with your code, segfault
*(++ptr); // Pointer moves to the next int position, and then gets accessed, with your code, segfault
由于案例较多,我可能有一些错误,如有错误请指正。
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
检查了程序,结果如下,
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
以下是各种“仅打印”建议的实例。我发现它很有启发。
#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
,以便可以轻松比较它们。
我用GCC编译的。
关于“如何增加指针地址和指针的值?”我认为
++(*p++);
实际上定义良好并且可以满足您的要求,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;
}
它不会在序列点之前两次修改同一事物。虽然对于大多数用途来说,我认为这不是一个好的风格 - 对于我的喜好来说,它有点太神秘了。
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);