如何空指针运算在GCC发生 [英] How void pointer arithmetic is happening in GCC

问题描述

int main()  
{  
    int a;  
    void *p;  
    p = &a;  
    printf("%ld\n",(long)p);  
    p = p+1;  
    printf("%ld\n",(long)p);  
}  

在这个程序中， P + 1 1。只是递增p值，我知道无效指针运算不是ç可能，所以 GCC 被隐式这样做。如果是，则它以它为字符指针。另外，为什么取消引用是不可能的空指针，如果是隐式做指针运算。

In this program, p+1 is just incrementing the value of p by 1. I know void pointer arithmetic is not possible in C, so GCC is doing it implicitly. And if yes, then is it taking it as char pointer. Also, why dereferencing is not possible for void pointer, if it is implicitly doing pointer arithmetic.

推荐答案

C不会允许与指针运算无效* 指针类型。

C does not allow pointer arithmetic with void * pointer type.

GNU C允许它通过考虑无效的大小为 1 。

GNU C allows it by considering the size of void is 1.

从的 6.23算术上无空隙和函数指针的：

在GNU C，加法和减法运算支持指针为void和函数指针。这是通过处理空隙的函数或大小作为1进行。

In GNU C, addition and subtraction operations are supported on pointers to void and on pointers to functions. This is done by treating the size of a void or of a function as 1.

http://gcc.gnu.org/onlinedocs/gcc/Pointer- Arith.html

现在来回答这个问题：

此外，为什么非关联化是不可能的空指针，如果是隐式做指针运算。

Also, why dereferencing is not possible for void pointer, if it is implicitly doing pointer arithmetic.

GNU C允许指针算法与无效* ，但仍不允许类型的对象无效来声明。

GNU C allows pointer arithmetic with void * but still does not allow an object of type void to be declared.

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