在 C++ 中切片向量
在 C++ 中，是否有等效的列表切片 [1:] 来自 Python 的向量?我只想从向量中获取除第一个元素之外的所有元素.
Is there an equivalent of list slicing [1:] from Python in C++ with vectors? I simply want to get all but the first element from a vector.
Python 的列表切片操作符:
Python's list slicing operator:
list1 = [1, 2, 3]
list2 = list1[1:]
print(list2) # [2, 3]
C++ 期望的结果:
std::vector<int> v1 = {1, 2, 3};
std::vector<int> v2;
v2 = v1[1:];
std::cout << v2 << std::endl; //{2, 3}
推荐答案
这可以使用 std::vector 的复制构造函数轻松完成:
This can easily be done using std::vector's copy constructor:
v2 = std::vector<int>(v1.begin() + 1, v1.end());
正如@AlessandroFlati 建议的那样，我应该澄清这将包括 v1.end()
As @AlessandroFlati suggested I should clarify that this will include v1.end()
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