c - 关于指向 char 的通用指针和严格别名

Question

我不知道为什么下面的代码工作正常，没有 gcc错误( -fstrict-aliasing -Wstrict-aliasing=1 )。

#include <stdio.h>

int 
main(void) 
{
    char  n = 42;
    char *p = &n;
    int  *q = (int *)p;

    *q = 10;

    printf("%d|%d\n", *p, *q);

    return 0;
}

如果我遵循严格的别名规则:

n1570, § 6.5 Expressions

An object shall have its stored value accessed only by an lvalue expression that has one of the following types:

— a type compatible with the effective type of the object,

— a qualified version of a type compatible with the effective type of the object,

— a type that is the signed or unsigned type corresponding to the effective type of the object,

— a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,

— an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or

— a character type.

但是 *q没有与 *p 兼容的类型，要么是限定版本，要么是对应的有符号类型，要么是无符号类型，要么是字符类型。

那么，为什么允许呢？

Answer 1

这是不允许的。您发布的标准部分显示了您允许为对象设置别名的类型。

编译的代码并不意味着它是正确的。您的代码存在三个问题，导致程序表现出未定义的行为。

首先是您将 char 指针分配给 int 指针。
标准不强制它们的对齐和表示，因此结果指针无效。

int  *q = (int *)p;

然后你解释了 char n对象作为整数，违反了严格的别名。(请注意问题中标准的引用)。

*q;

最后你写了一个 int进入 char 对象的内存( char n )，这会导致溢出，因为 int 的大小总是大于 char 的大小.

*q = 10;