使用 std::fill 填充带有递增数字的向量
我想用 std::fill 填充一个 vector，但不是一个值，该向量应该包含后面按升序排列的数字.
I would like to fill a vector<int> using std::fill, but instead of one value, the vector should contain numbers in increasing order after.
我尝试通过将函数的第三个参数迭代 1 来实现这一点，但这只会给我填充 1 或 2 的向量(取决于 ++ 运算符的位置).
I tried achieving this by iterating the third parameter of the function by one, but this would only give me either vectors filled with 1 or 2 (depending of the position of the ++ operator).
示例:
vector<int> ivec;
int i = 0;
std::fill(ivec.begin(), ivec.end(), i++); // elements are set to 1
std::fill(ivec.begin(), ivec.end(), ++i); // elements are set to 2
推荐答案
最好使用 std::iota 像这样:
Preferably use std::iota like this:
std::vector<int> v(100) ; // vector with 100 ints.
std::iota (std::begin(v), std::end(v), 0); // Fill with 0, 1, ..., 99.
也就是说，如果您没有任何 c++11 支持(仍然是我工作的真正问题)，请使用 std::generate 像这样:
That said, if you don't have any c++11 support (still a real problem where I work), use std::generate like this:
struct IncGenerator {
int current_;
IncGenerator (int start) : current_(start) {}
int operator() () { return current_++; }
};
// ...
std::vector<int> v(100) ; // vector with 100 ints.
IncGenerator g (0);
std::generate( v.begin(), v.end(), g); // Fill with the result of calling g() repeatedly.
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