CAN＆QUOT;的sizeof（ARR [0]）QUOT;导致未定义行为？ [英] Can "sizeof(arr[0])" lead to undefined behavior?

问题描述

有就是要弄清楚数组长度的一个众所周知的模式：

There is a well known pattern of figuring out array length:

int arr[10]; 
size_t len = sizeof(arr) / sizeof(arr[0]); 
assert(len == 10); 

此模式适用于静态数组和恒定的尺寸自动阵列。它也适用于在C99中可变长度阵列。

This pattern applies to static arrays and auto arrays of constant size. It also applies to variable length arrays in C99.

我想申请类似的想法以字节为单位计算出的动态数组的大小：

I want to apply similar idea for figuring out dynamic array size in bytes:

size_t known_len = 10; 
int *ptr = malloc(known_len * sizeof(int)); 
size_t size = known_len * sizeof(ptr[0]); 
assert(size == known_len * sizeof(int)); 

这是不是 known_len *的sizeof（INT），因为的sizeof（PTR [0]）不漂亮参考实际的数组元素类型。因此，它不需要$ C $的c读者了解的类型。

This is nicer than known_len * sizeof(int) because sizeof(ptr[0]) doesn't refer to actual array element type. Hence it doesn't require reader of the code to know the type.

不过，目前尚不清楚也好，我前pression 的sizeof（PTR [0]）可能会导致不确定的行为。由于扩展：

However it is unclear to me whether expression sizeof(ptr[0]) can lead to undefined behavior. As it is expanded:

sizeof(ptr[0]) -> sizeof(*((ptr) + (0))) -> sizeof(*ptr) 

结果前pression是值得怀疑的情况下，如果 PTR 是 0 ：

sizeof(*((int*) 0)) 

根据C99标准：

According to a C99 standard:

（C99，6.3.2.3p3）：整型常量前pression与价值 0 ，
  或者这样一个前pression投键入无效* ，被称为空指针
  不变的。提领一空指针是未定义的行为。

(C99, 6.3.2.3p3 ): "An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant." Dereferencing a null pointer is undefined behavior.

（C99，6.5.3.2.p4）如果无效值已被分配给该
  指针，一元的行为* 运算符是undefined.87）

(C99, 6.5.3.2.p4) "If an invalid value has been assigned to the pointer, the behavior of the unary * operator is undefined.87)"

87）：在由解引用指针无效值
  一元 * 操作符是一个空指针，地址不适当
  为对象的类型对齐指出，和一个地址
  其生命周期结束后的对象。

87): "Among the invalid values for dereferencing a pointer by the unary * operator are a null pointer, an address inappropriately aligned for the type of object pointed to, and the address of an object after the end of its lifetime."

但是它从来没有指明的前pression的sizeof的是否会导致不确定的行为。其实这样的sizeof应该在编译时进行评估。

But it's never specified whether sizeof of such expression can lead to undefined behavior. In fact such sizeof should be evaluated at compile time.

我的问题是：

• 可以除权pression 的sizeof（PTR [0]）在code时使用类型的 PTR 是已知的， PTR 不知道？


• 可以这样使用可根据C99标准是否合理？ GNU GCC规范呢？

• Can the expression sizeof(ptr[0]) be used in the code when type of ptr is known and the value of ptr is not known?
• Can such use be justified according to C99 standard? GNU GCC spec?

推荐答案

这位前pression PTR [0] 不会被评估的sizeof（PTR [0]）。大小将仅使用 PTR [0] 的类型在编译时确定。

The expression ptr[0] will not be evaluated in sizeof(ptr[0]). Size will be determined by just using the type of ptr[0] at compile time.

的的sizeof 运营商产生的操作数的大小（以字节为单位），这可能是一个前pression或类型的括号名称。 的大小是从操作数的类型确定。结果是整数。 如果操作数的类型是可变长度数组类型，操作数被评估;否则，操作数不计算，结果是aninteger不变。

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is aninteger constant.

这意味着，没有任何不确定的行为。

That means, there is no undefined behavior.

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