std::bsearch在排序數組中搜索元素。在ptr指向的數組中查找與key指向的元素相等的元素。
如果數組包含comp表示與搜索到的元素相等的多個元素，則未指定函數將作為結果返回的元素。
用法:
void* bsearch( const void* key, const void* ptr, std::size_t count,
std::size_t size, * comp );
參數：
key - element to be found
ptr - pointer to the array to examine
count - number of element in the array
size - size of each element in the array in bytes
comp - comparison function which returns ?a negative integer value if
the first argument is less than the second,
a positive integer value if the first argument is greater than the second
and zero if the arguments are equal.
返回值：
Pointer to the found element or null pointer if the element has not been found.
實現二進製謂詞comp：
// Binary predicate which returns 0 if numbers found equal
int comp(int* a, int* b)
{
if (*a < *b)
return -1;
else if (*a > *b)
return 1;
// elements found equal
else
return 0;
}Implementation
// CPP program to implement
// std::bsearch
#include <bits/stdc++.h>
// Binary predicate
int compare(const void* ap, const void* bp)
{
// Typecasting
const int* a = (int*)ap;
const int* b = (int*)bp;
if (*a < *b)
return -1;
else if (*a > *b)
return 1;
else
return 0;
}
// Driver code
int main()
{
// Given array
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
// Size of array
int ARR_SIZE = sizeof(arr) / sizeof(arr[0]);
// Element to be found
int key1 = 4;
// Calling std::bsearch
// Typecasting the returned pointer to int
int* p1 = (int*)std::bsearch(&key1, arr, ARR_SIZE, sizeof(arr[0]), compare);
// If non-zero value is returned, key is found
if (p1)
std::cout << key1 << " found at position " << (p1 - arr) << '\n';
else
std::cout << key1 << " not found\n";
// Element to be found
int key2 = 9;
// Calling std::bsearch
// Typecasting the returned pointer to int
int* p2 = (int*)std::bsearch(&key2, arr, ARR_SIZE, sizeof(arr[0]), compare);
// If non-zero value is returned, key is found
if (p2)
std::cout << key2 << " found at position " << (p2 - arr) << '\n';
else
std::cout << key2 << " not found\n";
}輸出：
4 found at position 3 9 not found
使用位置：二進製搜索可用於要查找關鍵字的已排序數據。它可以用於諸如排序列表中鍵的計算頻率之類的情況。
為什麽選擇二進製搜索？
二進製搜索比線性搜索更有效，因為它使每一步的搜索空間減半。這對於長度為9的數組而言並不重要。這裏，線性搜索最多需要9個步驟，而二進製搜索最多需要4個步驟。但是考慮具有1000個元素的數組，此處線性搜索最多需要1000步，而二進製搜索最多需要10步。
對於10億個元素，二進製搜索最多可以在30個步驟中找到我們的關鍵字。
相關用法
注：本文由純淨天空篩選整理自 std::bsearch in C++。非經特殊聲明，原始代碼版權歸原作者所有，本譯文未經允許或授權，請勿轉載或複製。