数组与& array [0]之间的区别 [英] Difference between array and &array[0]

问题描述

int array[] = {1,2,3,4};

据我了解，array只是指向&array[0]

但是，为什么sizeof(array);知道数组的大小，而不仅仅是知道它只有4个字节的数字?

But how come then sizeof(array); knows size of an array and not just that it's only 4 byte number?

推荐答案

除非它是sizeof或一元&运算符的操作数，或者是用于在声明中初始化另一个数组的字符串文字，类型为"T的N个元素的数组"的表达式将被转换(衰变")为类型为指向T的指针"的表达式，该表达式的值为数组第一个元素的地址.

Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T, and the value of the expression will be the address of the first element of the array.

C中的数组不存储任何有关其大小的元数据或其他任何内容，也不为任何类型的指针留出任何存储空间.它们的布局大致如下:

Arrays in C don't store any metadata about their size or anything else, nor is any storage set aside for any sort of pointer. They're laid out pretty much as follows:

     +---+
arr: | 1 | arr[0]
     +---+
     | 2 | arr[1]
     +---+
     | 3 | arr[2]
     +---+
     | 4 | arr[3]
     +---+

除了数组元素本身之外，没有单独的变量arr存储.如您所见，arr的地址和arr[0]的地址是相同的.这就是为什么即使类型，表达式arr，&arr和&arr[0]都为您提供相同的 value (第一个元素的地址)的原因这些表达方式是不同的.

There's no separate storage for a variable arr apart from the array elements themselves. As you can see, the address of arr and the address of arr[0] are the same. This is why the expressions arr, &arr, and &arr[0] all give you the same value (the address of the first element), even though the types of those expressions are different.

除非操作数是可变长度数组，否则sizeof的结果是在编译时计算的，在这种情况下，编译器会将数组操作数视为数组.否则，它将数组表达式视为指向第一个元素的指针.

Except when the operand is a variable-length array, the result of sizeof is computed at compile time, and the compiler treats array operands as arrays in those circumstances; otherwise, it treats the array expression as a pointer to the first element.

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