带有const修饰的指针解读
2017-09-20 14:03
246 查看
通过实例来看看const所修饰的指针使用方法:
1、
先看一个普通的指针使用。
形式:int *pTmp
含义:一个普通的指向int类型的指针
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int *pTmp = &i;
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
(*pTmp)++;
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
输出:
[plain] view
plaincopyprint?
pTmp = 1
pTmp = 2
pTmp = 3
请按任意键继续. . .
没有任何问题,pTmp指针你可以任意操作。
2、
形式:const int *pTmp
含义:表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
const int *pTmp = &i; /*正确:表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变*/
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*正确:将pTmp指向变量j的地址*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*错误,pTmp指向的对象是只读的,编译出错提示:assignment of read-only location */
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
3、
形式:int const *pTmp
含义:与2相同, 表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
const int *pTmp = &i; /*正确:表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变*/
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*正确:将pTmp指向变量j的地址*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*错误,pTmp指向的对象是只读的,编译出错提示:assignment of read-only location */
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
4、
形式:int * const pTmp
含义:pTmp不可修改,但pTmp所指向的对象可以修改。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
int * const pTmp = &i;
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*错误,pTmp不可修改,编译提示:error:assignment of read-only variable `pTmp'*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*正确,pTmp所指向的对象可以修改 */
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
5、
形式:const int * const pTmp
含义:pTmp不可修改,pTmp所指对象也不能修改。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
const int * const pTmp = &i;
printf("pTmp = %d\n", *pTmp);
i = 2; /*正确*/
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*错误:error: assignment of read-only variable `pTmp'*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*错误:error: increment of read-only location*/
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
1、
先看一个普通的指针使用。
形式:int *pTmp
含义:一个普通的指向int类型的指针
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int *pTmp = &i;
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
(*pTmp)++;
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
输出:
[plain] view
plaincopyprint?
pTmp = 1
pTmp = 2
pTmp = 3
请按任意键继续. . .
没有任何问题,pTmp指针你可以任意操作。
2、
形式:const int *pTmp
含义:表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
const int *pTmp = &i; /*正确:表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变*/
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*正确:将pTmp指向变量j的地址*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*错误,pTmp指向的对象是只读的,编译出错提示:assignment of read-only location */
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
3、
形式:int const *pTmp
含义:与2相同, 表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
const int *pTmp = &i; /*正确:表示pTmp所指向的对象是只读的,但pTmp可指向其他地址,即pTmp可变*/
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*正确:将pTmp指向变量j的地址*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*错误,pTmp指向的对象是只读的,编译出错提示:assignment of read-only location */
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
4、
形式:int * const pTmp
含义:pTmp不可修改,但pTmp所指向的对象可以修改。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
int * const pTmp = &i;
printf("pTmp = %d\n", *pTmp);
i = 2;
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*错误,pTmp不可修改,编译提示:error:assignment of read-only variable `pTmp'*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*正确,pTmp所指向的对象可以修改 */
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
5、
形式:const int * const pTmp
含义:pTmp不可修改,pTmp所指对象也不能修改。
[cpp] view
plaincopyprint?
int main(int argc, char *argv[])
{
int i = 1;
int j = 100;
const int * const pTmp = &i;
printf("pTmp = %d\n", *pTmp);
i = 2; /*正确*/
printf("pTmp = %d\n", *pTmp);
pTmp = &j; /*错误:error: assignment of read-only variable `pTmp'*/
printf("pTmp = %d\n", *pTmp);
(*pTmp)++; /*错误:error: increment of read-only location*/
printf("pTmp = %d\n", *pTmp);
system("PAUSE");
return 0;
}
相关文章推荐
- 带有const修饰的指针解读
- C:带有const修饰的指针解读
- C:带有const修饰的指针解读
- C++:带有const修饰的指针解读
- const在修饰指针时的应用
- const 修饰指针
- [c/c++]const修饰指针
- const关键字修饰指针的疑惑点
- const修饰指针和引用的用法【转贴】
- const修饰的指针
- const用法之修饰指向常量的指针
- C++中Const 修饰指针就近原则
- 关于const修饰指针
- const修饰指针的情况分析
- const修饰词在指针中作用
- 做题笔记(一)const修饰的一阶二阶指针
- 整理:#define和typedefine的区别 以及const修饰指针的用法
- 关于const修饰指针
- const修饰指针和引用的用法
- const修饰的指针的用法解析