C:带有const修饰的指针解读 (一)

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]
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;
}
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]
pTmp = 1
pTmp = 2
pTmp = 3
请按任意键继续. . .
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]
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;
}
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]
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;
}
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]
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;
}
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