[英]C++ vector.begin() and vector[0]
why this two line prints different addresses? 为什么这两行打印不同的地址?
vector<int> v1{ 12,2,34 };
printf_s("%d - 0x%p\n", v1[0], &v1[0]);
printf_s("%d - 0x%p\n",*v1.begin(), v1.begin());
Values in this addresses are same but address itself is different. 该地址中的值相同,但地址本身不同。 Does it mean that there is two copy of the same array? 这是否意味着同一数组有两个副本?
EDIT : In debug mode it prints different addresses, in release mode there are same addresses :) 编辑 :在调试模式下它打印不同的地址,在发布模式下有相同的地址:)
v1.begin()
returns an std::vector<int>::iterator
, which is not necessarily an address to v1[0]
. v1.begin()
返回一个std::vector<int>::iterator
,它不一定是v1[0]
的地址。 In fact, attempting to print it out using printf
gives me a warning: 实际上,尝试使用printf
将其打印出来会给我一个警告:
warning: format '%p' expects argument of type 'void*' , but argument 3 has type 'std::vector::iterator` 警告:格式'%p'期望类型为'void *'的参数,但是参数3的类型为'std :: vector :: iterator`
Unless you're sure that in your particular Standard Library implementation and with your current compilation options std::vector<int>::iterator
is an alias for int*
, the comparison is meaningless. 除非您确定在特定的标准库实现中并且使用当前的编译选项std::vector<int>::iterator
是int*
的别名,否则比较是没有意义的。
for a vector object v1: 对于矢量对象v1:
&V[0]
is the the address of first element of the object v1. &V[0]
是对象v1的第一个元素的地址。
when we create an iterator eg vector<>::iterator iter
the iter
itself is an other object as explain in this document . 当我们创建一个迭代例如vector<>::iterator iter
的iter
本身是一个其他对象,因为这在解释文件 。
Iterator: a pointer-like object that can be incremented with ++, dereferenced with *, and compared against another iterator with !=. 迭代器:类似指针的对象,可以用++递增,用*取消引用,然后与!=的另一个迭代器进行比较。
I hope this reference manual answer your question in detail. 希望本参考手册能详细回答您的问题。
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