Question

这是我的代码：

int *p;
p=4;
printf("p is %p\n",p);
free(p);
//need p=NULL but I don't
int *q;
q=5:
printf("q is %i",*q);

然后出现错误。我只需要解释。

Answer 1

int *p;

是指向int的指针。

p = 4;

使其指向地址0x4

free(p);

尝试取消分配地址0x4

基本上，您正在尝试释放无法释放的资源。

int *q;
q = 5;

将q指向地址'0x5;

*q;

从0x5地址读取，很可能会崩溃。（此地址也未对齐）。

指针不是整数...您编写的程序显示出对什么是指针以及为什么/应该如何使用它们缺乏了解。

Answer 2

Int *p; // You had better use int* p = NULL; we don't like wild pointers
//p = new int;
p=4; // I guess you want put the value of 4 to a variable, so *p = 4;
printf("p is %p\n",p);//    printf("p is %d\n",*p);
free(p);//just correct in theory, no one can tell what happens in reality.
/* 
if ( NULL != p )
{
    delete p;
    p = NULL;
}
*/

我只是不明白这种指针情况

2 个答案: