std::fill_n

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C++
 
算法库
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Functions
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Non-modifying sequence operations
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all_of
any_of
none_of		（C++11）
（C++11）
（C++11）
for_each
count
count_if
mismatch
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Partitioning operations
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is_partitioned（C++11）
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Sorting operations (on sorted ranges)
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Binary search operations (on sorted ranges)
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Defined in header <algorithm>
template< class OutputIt, class Size, class T >

void fill_n( OutputIt first, Size count, const T& value );
template< class OutputIt, class Size, class T >

OutputIt fill_n( OutputIt first, Size count, const T& value );
 （至 C++11）

（C++11 起）
分配给给定的值valuecount如果first开始的范围内，在第一count>0元素。什么都不做，否则.....
Original:
Assigns the given value value to the first count elements in the range beginning at first if count>0. Does nothing otherwise.
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目录

[编辑] 参数

first -
修改的范围内的元素开始
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the beginning of the range of elements to modify
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count -
要修改的元素的数量
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number of elements to modify
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value -
被分配的值
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the value to be assigned
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Type requirements
-
OutputIt must meet the requirements of OutputIterator.

[编辑] 返回值

（无）（至 C++11）
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(none) （至 C++11）
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如果count>0，first否则分配的最后一个元素的迭代器一个过去。 （C++11 起）
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Iterator one past the last element assigned if count>0, first otherwise. （C++11 起）
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[编辑] 复杂性

究竟count任务，count>0.
Original:
Exactly count assignments, for count>0.
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[编辑] 可能的实现

template<class OutputIt, class Size, class T>
OutputIt fill_n(OutputIt first, Size count, const T& value)
{
    for (Size i = 0; i < count; i++) {
        *first++ = value;
    }
    return first;
}

[编辑] 为例

下面的代码使用fill_n()分配-1，上半年的向量整数
Original:
The following code uses fill_n() to assign -1 to the first half of a vector of integers:
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#include <algorithm>
#include <vector>
#include <iostream>
 
int main()
{
    std::vector<int> v1{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
 
    std::fill_n(v1.begin(), 5, -1);
 
    for (vector<int>::iterator it = v1.begin(); it != v1.end(); ++it) {
        std::cout << *it << " ";
    }
    std::cout << "\n";
}

Output: 

-1 -1 -1 -1 -1 5 6 7 8 9

[编辑] 另请参阅
分配一个一定值范围内的元素
Original:
assigns a range of elements a certain value
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（函数模板） [edit]
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