32bit架构下各种指针大小的面试题
char str[] = " http://www.ibegroup.com/";
char *p = str ;
void Foo ( char str[100]){
}
void *p = malloc( 100 );
上述4种情况中的sizeof str,p,str,p是多少反过来？
我已经在我的机器（似乎是64位）下测试了它，结果如下：
25 8 8 8
但还不明白原因。
char str[] = " http://www.ibegroup.com/";
char *p = str ;
void Foo ( char str[100]){
}
void *p = malloc( 100 );
What's the sizeof str,p,str,p in the above 4 case in turn?
I've tested it under my machine(which seems to be 64bit) with these results:
25 8 8 8
But don't understand the reason yet.
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sizeof(char[])返回字符串中的字节数，即strlen()+1（对于填充整个数组的以 null 结尾的 C 字符串）。 数组不会衰减为sizeof中的指针。str是一个数组，字符串有 25 个字符加上一个空字节，因此sizeof(str)应该是 26。您是否在该值中添加了空格？当然，指针的大小始终仅由机器体系结构决定，因此
p的两个实例在 64 位体系结构上都是 8 字节，在 32 位体系结构上都是 4 字节。在函数参数中，数组 do 衰减为指针，因此您得到的结果与指针得到的结果相同。因此，以下定义是等效的：
sizeof(char[])returns the number of bytes in the string, i.e.strlen()+1for null-terminated C strings filling the entire array. Arrays don't decay to pointers insizeof.stris an array, and the string has 25 characters plus a null byte, sosizeof(str)should be 26. Did you add a space to the value?The size of a pointer is of course always determined just by the machine architecture, so both instances of
pare 8 bytes on 64-bit architectures and 4 bytes on 32-bit architectures.In function arguments, arrays do decay to pointers, so you're getting the same result that you get for a pointer. Therefore, the following definitions are equivalent:
第一个是内置数组的
sizeof，它是元素的数量（24 + 字符串末尾的 null）。第二个是指针的
sizeof，它是系统的本机字大小，在您的情况下为 64 位或 8 字节。第三个是指向数组第一个元素的指针的 sizeof ，该元素的大小与任何其他指针相同，即系统的本机字大小。为什么要使用指向数组第一个元素的指针？因为数组的大小信息在传递给函数时会丢失，并且会隐式转换为指向第一个元素的指针。
第四个是指针的
sizeof，它与任何其他指针的大小相同。The first is the
sizeofof an built-in array, which is the amount of elements (24 + null on the end of the string).The second is the
sizeofof a pointer which is the native word size of your system, in your case 64 bit or 8 bytes.The third is the
sizeofof a pointer to the first element of an array which has the same size as any other pointer, the native word size of your system. Why a pointer to the first element of an array? Because size information of an array goes lost when passed to a function and it gets implicitly converted to a pointer to the first element instead.The fourth is the
sizeofof a pointer which has the same size as any other pointer.str是一个 8 位字符的数组，包括 null 终止符。p是一个指针，通常是机器本机字大小（32 位或 64 位）的大小。无论指针指向的内存大小如何，指针占用的大小都保持不变。
编辑
在 C++ 中，数组参数是通过引用传递的（内部是指针类型），这就是为什么
str的第二个实例具有sizeof8.stris an array of 8-bit characters, including null terminator.pis a pointer, which is typically the size of the machine's native word size (32 bit or 64 bit).The size taken up by a pointer stays constant, regardless of the size of the memory to which it points.
EDIT
In c++, arguments that are arrays are passed by reference (which internally is a pointer type), that's why the second instance of
strhassizeof8.在这种情况下，已知 的大小
为 25 (24+1)，因为实际分配了那么多内存。
的情况下
在没有分配内存
in the cases the size of
is known to be 25 (24+1), because that much memory is actually allocated.
In the case of
no memory is allocated