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STL 源码剖析 stl_numeric.h
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描述、源码、示例
version 1:普通操作版本
version 2: 泛化操作版本
1.accumulate
描述:计算 init 和 [first, last) 内所有元素的总和
源码:
//version 1 template <class InputIterator, class T> T accumulate(InputIterator first, InputIterator last, T init) { for ( ; first != last; ++first) init = init + *first; return init; } //version 2 template <class InputIterator, class T, class BinaryOperation> T accumulate(InputIterator first, InputIterator last, T init, BinaryOperation binary_op) { for ( ; first != last; ++first) init = binary_op(init, *first); return init; }
示例:
int main() { int A[] = {1, 2, 3, 4, 5}; const int N = sizeof(A) / sizeof(int); cout << "The sum of all elements in A is " << accumulate(A, A + N, 0) // 15 << endl; cout << "The product of all elements in A is " << accumulate(A, A + N, 1, multiplies<int>()) //120 << endl; }
2.inner_product
描述:计算[first1, last1) 和 [first2, first2 + (last1 - first1)) 的内积
源码:
//version 1 template <class InputIterator1, class InputIterator2, class T> T inner_product(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, T init) { for ( ; first1 != last1; ++first1, ++first2) init = init + (*first1 * *first2); return init; } //version 2 template <class InputIterator1, class InputIterator2, class T, class BinaryOperation1, class BinaryOperation2> T inner_product(InputIterator1 first1, InputIterator1 last1, InputIterator2 first2, T init, BinaryOperation1 binary_op1, BinaryOperation2 binary_op2) { for ( ; first1 != last1; ++first1, ++first2) init = binary_op1(init, binary_op2(*first1, *first2)); return init; }
示例:
int main() { int A1[] = {1, 2, 3}; int A2[] = {4, 1, -2}; const int N1 = sizeof(A1) / sizeof(int); cout << "The inner product of A1 and A2 is " << inner_product(A1, A1 + N1, A2, 0) //0 << endl; }
3.partial_sum
描述: 计算局部总和。
源码:
//version 1 template <class InputIterator, class OutputIterator, class T> OutputIterator __partial_sum(InputIterator first, InputIterator last, OutputIterator result, T*) { T value = http://www.mamicode.com/*first;>
示例:int main() { const int N = 10; int A[N]; fill(A, A+N, 1); cout << "A: "; copy(A, A+N, ostream_iterator<int>(cout, " ")); // 1 2 3 4 5 6 7 8 9 10 cout << endl; cout << "Partial sums of A: "; partial_sum(A, A+N, ostream_iterator<int>(cout, " ")); // 1 3 6 10 15 21 28 36 45 55 cout << endl; }
4.adjacent_difference
描述:计算[first, last) 相邻元素的差额
源码://version 1 template <class InputIterator, class OutputIterator, class T> OutputIterator __adjacent_difference(InputIterator first, InputIterator last, OutputIterator result, T*) { T value = http://www.mamicode.com/*first;>
示例:int main() { int A[] = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}; const int N = sizeof(A) / sizeof(int); int B[N]; cout << "A[]: "; copy(A, A + N, ostream_iterator<int>(cout, " ")); //1 4 9 16 25 36 49 64 81 100 cout << endl; adjacent_difference(A, A + N, B); cout << "Differences: "; copy(B, B + N, ostream_iterator<int>(cout, " ")); // 3 5 7 9 11 13 15 17 19 cout << endl; cout << "Reconstruct: "; partial_sum(B, B + N, ostream_iterator<int>(cout, " ")); //1 4 9 16 25 36 49 64 81 100 cout << endl; }
5.power
描述:计算某数的 n 幂次方。 SGI 专属,不在STL标准中
源码:// Returns x ** n, where n >= 0. Note that "multiplication" // is required to be associative, but not necessarily commutative. template <class T, class Integer, class MonoidOperation> T power(T x, Integer n, MonoidOperation op) { //这里用的是 Russian Peasant Algorithm if (n == 0) return identity_element(op); else { while ((n & 1) == 0) { n >>= 1; x = op(x, x); } T result = x; n >>= 1; while (n != 0) { x = op(x, x); if ((n & 1) != 0) result = op(result, x); n >>= 1; } return result; } } template <class T, class Integer> inline T power(T x, Integer n) { return power(x, n, multiplies<T>()); }
示例:int main() { cout << "2 ** 30 = " << power(2, 30) << endl; // --> 我编译不通过。说找不到标识符 power ,我已经 include 了 <numeric> }
6.iota
描述:设定某个区间的内容,使其内的每一个元素从指定的value 值 开始,呈现递增状态
源码:template <class ForwardIterator, class T> void iota(ForwardIterator first, ForwardIterator last, T value) { while (first != last) *first++ = value++; }
示例:int main() { vector<int> V(10); iota(V.begin(), V.end(), 7); copy(V.begin(), V.end(), ostream_iterator<int>(cout, " ")); // 7 8 9 10 11 12 13 14 15 16 cout << endl; }
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