将整数文字分配给指针? [英] Assign integer literal to pointer?

问题描述

这个问题可能太糟糕了，但是我冒着风险在此处发布这个问题来解决我的困惑.

This question might be too bad but I can take risk to post this question here to address my confusion.

实际上我的问题是我们只能将地址分配给指针，如:-

Actually my question is that we can only assign address to pointer like :-

int *p,a;
p = &a; // OK
p = 1; // Error because you cannot assign integer literal to p*  

但是我们可以为p赋NULL:

But we can assign NULL to p like :

p = NULL;

实际上，NULL是一个值为0的宏，在由编译器编译此代码之前，它会由前置处理程序替换为0.因此，更换后的样子

Indeed, NULL is a macro which is value is 0 and before compiling this code by compiler it get replaced with 0 by prepocessor. So after replacement its look like

 p = 0;

我知道这意味着p不指向任何东西，但是根据规则，我们只能将地址分配给指针，而0是整数. 所以这不是违反规则吗?

I know it means p is point to nothing but according to rule we can only assign address to pointer but 0 is an integer. So this isn't break the rule ?

谢谢.

推荐答案

barak manos已经在他的评论中指出了这一点:

barak manos already pointed it out in his comment:

如果要将指针设置为文字值，则需要首先将文字值转换为相应的指针类型.

If you want to set a pointer to a literal value, you need to cast the literal value to the corresponding pointer type first.

NULL也可以定义为(void *) 0 ...，它可以隐式转换为任何指针类型.

NULL could just as well be defined as (void *) 0... which is implicitly convertible to any pointer type.

在任何一种情况下，您最终都有一个指向文字地址的指针.

In either case, you end up with a pointer pointing to a literal address.

但是，在任何情况下，您的指针都不会指向包含文字4的内存 .据我所知，如果不先将字面值分配给int，则不可能:

In no case, however, does your pointer point to memory containing a literal 4, though. This is, to my knowledge, not possible without assigning that literal value to an int first:

int i = 4;
int * p = &i;

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