我为什么不能做算术上的一个空指针的投？ [英] Why can I not do arithmetic on a cast of a void pointer?

问题描述

void foo(void *ptr, int numBytes)
{
    (char*)ptr += numBytes;
}

这并不在C编译我所知道的替代品。但为什么这个不行？有什么问题？

This doesn't compile in C. I know the alternative. But why does this not work? What is the problem?

推荐答案

问题

问题是，（字符*）PTR 不会产生一个的左值的，这意味着价值不能被修改 - 一眼就能看出来作为演员的临时结果，演员将产生的右值的类型的char * 。

The problem is that (char*)ptr will not yield an lvalue, meaning that the value cannot be modified - one can see it as a temporary result of the cast, the cast will yield a rvalue of type char*.

这是语义上一样，如果你不得不在下面的例子中，铸造产生一个临时的价值，这样的价值不能被赋予一个新的值。

It's semantically the same as if you'd have the below example, a cast yields a temporary value, such a value cannot be assigned a new value.

int x = 123;

(float)x += 0.12f; /* (1), illegal                                     */
                   /*  ^-- sementically equivalent to `123.f += 0.12f` */

---

解决方案

在你的问题，你都表示你已经知道一种解决方法这个问题，但我想明确写入该解决方案显示一个如何修改 PTR 即使蒙上产生不可修改的值。

In your question you have stated that you already know a workaround to this problem, but I'd like to explicitly write the solution to show how one can modify the value of ptr even when casts yields non-modifiable values.

1. 取地址你的指向void 的


2. 施放此地址到的指针指向char 的


3. 解引用该指针，产生的字符指针的


4. 修改此收益率的左值的，如果原来的无效* 是类型的char *

1. Take the address of your pointer to void,
2. cast this address to a pointer to pointer to char,
3. dereference that pointer, yielding a pointer to char,
4. modify this yield lvalue as if the original void* was of type char*

*((char**)&ptr) += numbytes; // make `ptr` move forward `numbytes`

（注意：当解引用指针，你得到的左值的，否则这将是不可能改变的值的指出的值位于存储在指针中的地址。）

( Note: When dereferencing a pointer you get an lvalue, otherwise it would be impossible to change the value of the pointed to value located at an address stored in a pointer. )

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