根据C89的基本原理，标准的作者不想要求编译器给出如下代码:

int x;
int test(double *p)
{
  x=5;
  *p = 1.0;
  return x;
}

应该要求在赋值和返回语句之间重新加载x的值，以便允许p指向x的可能性，而对*p的赋值可能因此改变x的值。编译器应该有权假定在上述情况下不会出现混叠的概念是没有争议的。

不幸的是，C89的作者写规则的方式，如果从字面上读，甚至会使下面的函数调用未定义行为:

void test(void)
{
  struct S {int x;} s;
  s.x = 1;
}

because it uses an lvalue of type int to access an object of type struct S, and int is not among the types that may be used accessing a struct S. Because it would be absurd to treat all use of non-character-type members of structs and unions as Undefined Behavior, almost everyone recognizes that there are at least some circumstances where an lvalue of one type may be used to access an object of another type. Unfortunately, the C Standards Committee has failed to define what those circumstances are.

大部分问题是缺陷报告#028的结果，它询问了程序的行为，如:

int test(int *ip, double *dp)
{
  *ip = 1;
  *dp = 1.23;
  return *ip;
}
int test2(void)
{
  union U { int i; double d; } u;
  return test(&u.i, &u.d);
}

缺陷报告#28指出，程序调用了未定义行为，因为写入类型为“double”的联合成员并读取类型为“int”的联合成员的操作调用了实现定义的行为。这样的推理是毫无意义的，但却形成了有效类型规则的基础，这些规则不必要地使语言复杂化，而对解决原始问题毫无帮助。

解决原始问题的最好办法可能是治疗 关于规则目的的脚注，就像它是规范的一样，并作出 除非实际涉及使用别名的冲突访问，否则该规则不可执行。假设是这样的:

 void inc_int(int *p) { *p = 3; }
 int test(void)
 {
   int *p;
   struct S { int x; } s;
   s.x = 1;
   p = &s.x;
   inc_int(p);
   return s.x;
 }

在inc_int中没有冲突，因为所有通过*p访问的存储都是使用int类型的左值完成的，在test中也没有冲突，因为p明显地派生于结构体S，并且在下次使用S时，所有通过p访问的存储都已经发生了。

如果代码稍微改变一下……

 void inc_int(int *p) { *p = 3; }
 int test(void)
 {
   int *p;
   struct S { int x; } s;
   p = &s.x;
   s.x = 1;  //  !!*!!
   *p += 1;
   return s.x;
 }

这里，p和对s.x的访问在被标记的行上存在别名冲突，因为在执行时存在另一个引用，该引用将用于访问相同的存储。

如果缺陷报告028说原始示例调用UB是因为两个指针的创建和使用之间有重叠，那么事情就会变得更清楚，而不必添加“有效类型”或其他类似的复杂性。

严格的混叠是不允许不同的指针类型指向相同的数据。

本文将帮助您全面详细地理解这个问题。

根据C89的基本原理，标准的作者不想要求编译器给出如下代码:

int x;
int test(double *p)
{
  x=5;
  *p = 1.0;
  return x;
}

应该要求在赋值和返回语句之间重新加载x的值，以便允许p指向x的可能性，而对*p的赋值可能因此改变x的值。编译器应该有权假定在上述情况下不会出现混叠的概念是没有争议的。

不幸的是，C89的作者写规则的方式，如果从字面上读，甚至会使下面的函数调用未定义行为:

void test(void)
{
  struct S {int x;} s;
  s.x = 1;
}

because it uses an lvalue of type int to access an object of type struct S, and int is not among the types that may be used accessing a struct S. Because it would be absurd to treat all use of non-character-type members of structs and unions as Undefined Behavior, almost everyone recognizes that there are at least some circumstances where an lvalue of one type may be used to access an object of another type. Unfortunately, the C Standards Committee has failed to define what those circumstances are.

大部分问题是缺陷报告#028的结果，它询问了程序的行为，如:

int test(int *ip, double *dp)
{
  *ip = 1;
  *dp = 1.23;
  return *ip;
}
int test2(void)
{
  union U { int i; double d; } u;
  return test(&u.i, &u.d);
}

缺陷报告#28指出，程序调用了未定义行为，因为写入类型为“double”的联合成员并读取类型为“int”的联合成员的操作调用了实现定义的行为。这样的推理是毫无意义的，但却形成了有效类型规则的基础，这些规则不必要地使语言复杂化，而对解决原始问题毫无帮助。

解决原始问题的最好办法可能是治疗 关于规则目的的脚注，就像它是规范的一样，并作出 除非实际涉及使用别名的冲突访问，否则该规则不可执行。假设是这样的:

 void inc_int(int *p) { *p = 3; }
 int test(void)
 {
   int *p;
   struct S { int x; } s;
   s.x = 1;
   p = &s.x;
   inc_int(p);
   return s.x;
 }

在inc_int中没有冲突，因为所有通过*p访问的存储都是使用int类型的左值完成的，在test中也没有冲突，因为p明显地派生于结构体S，并且在下次使用S时，所有通过p访问的存储都已经发生了。

如果代码稍微改变一下……

 void inc_int(int *p) { *p = 3; }
 int test(void)
 {
   int *p;
   struct S { int x; } s;
   p = &s.x;
   s.x = 1;  //  !!*!!
   *p += 1;
   return s.x;
 }

这里，p和对s.x的访问在被标记的行上存在别名冲突，因为在执行时存在另一个引用，该引用将用于访问相同的存储。

如果缺陷报告028说原始示例调用UB是因为两个指针的创建和使用之间有重叠，那么事情就会变得更清楚，而不必添加“有效类型”或其他类似的复杂性。

通过指针强制转换(而不是使用联合)的类型双关是打破严格混叠的一个主要例子。

这是严格的混叠规则，可以在c++ 03标准的3.10节中找到(其他答案提供了很好的解释，但没有一个提供了规则本身):

If a program attempts to access the stored value of an object through an lvalue of other than one of the following types the behavior is undefined: the dynamic type of the object, a cv-qualified version of the dynamic type of the object, a type that is the signed or unsigned type corresponding to the dynamic type of the object, a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic 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), a type that is a (possibly cv-qualified) base class type of the dynamic type of the object, a char or unsigned char type.

c++ 11和c++ 14的措辞(强调更改):

If a program attempts to access the stored value of an object through a glvalue of other than one of the following types the behavior is undefined: the dynamic type of the object, a cv-qualified version of the dynamic type of the object, a type similar (as defined in 4.4) to the dynamic type of the object, a type that is the signed or unsigned type corresponding to the dynamic type of the object, a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic type of the object, an aggregate or union type that includes one of the aforementioned types among its elements or non-static data members (including, recursively, an element or non-static data member of a subaggregate or contained union), a type that is a (possibly cv-qualified) base class type of the dynamic type of the object, a char or unsigned char type.

有两个变化很小:glvalue代替了lvalue，并澄清了聚合/并集的情况。

第三个变化提供了更强的保证(放宽强混叠规则):类似类型的新概念现在可以安全地进行混叠。

还有C的措辞(C99;Iso / iec 9899:1999 6.5/7;在ISO/IEC 9899:2011§6.5¶7中使用了完全相同的措辞:

An object shall have its stored value accessed only by an lvalue expression that has one of the following types 73) or 88): 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. 73) or 88) The intent of this list is to specify those circumstances in which an object may or may not be aliased.

我找到的最好的解释是Mike Acton的《Understanding Strict Aliasing》。本文主要关注PS3的开发，但这基本上只是GCC的工作。

摘自文章:

严格混叠是C(或c++)编译器的一个假设，即指向不同类型对象的指针的解引用永远不会指向相同的内存位置(即相互混叠)。

所以基本上，如果你有一个int*指向一些包含int型的内存，然后你把一个float*指向那个内存，并把它用作浮点数，你就违反了规则。如果你的代码不尊重这一点，那么编译器的优化器很可能会破坏你的代码。

该规则的例外是一个char*，它被允许指向任何类型。