我正在尝试实现功能// once at he beginning of your program (before any use of default SSLContext)
// or instead can be done on the command line with -Dprop=value
System.setProperty ("javax.net.ssl.keyStore", filepath);
System.setProperty ("javax.net.ssl.keyStorePassword", password);
System.setProperty ("javax.net.ssl.keyStoreType", "PKCS12" or "JKS"); -- if not default
System.setProperty ("javax.net.ssl.trustStore", filepath);
System.setProperty ("javax.net.ssl.trustStorePassword", password);
System.setProperty ("javax.net.ssl.trustStoreType", "PKCS12" or "JKS"); -- if not default
// when (each time) you want to make a request:
URL url = new URL ("https://hello-signs.api-dev.myname.com/v2/oauth2/client_credential/accesstoken?grant_type=client_credentials");
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
// we aren't actually using S-specific items, so could instead use
// HttpURLConnection especially if you want to mix HTTPS and HTTP;
// or you could use the Https version to tweak some SSL/TLS parameters
conn.setRequestMethod ("POST");
conn.setRequestHeader ("Content-Type","application/x-www-form-urlencoded");
conn.setDoOutput (true);
OutputStream os = conn.getOutputStream();
os.write ("client_id=xxccvvbbnniioopp&client_secret=zoopopopopopppp".getBytes());
os.close();
os.connect(); // optional, can help localize exceptions
// depending on what you want to do with the result(s):
... conn.getResponseCode() and/or conn.getResponseMessage() ...
... conn.getHeaderField{,Int,Long,s}("id") ...
... conn.getContent{Type,Length[Long],Encoding,}() ...
... conn.getInputStream() then read from it, including decoding various formats or charsets as needed/desired ...
conn.disconnect(); // when done
，该功能试图查看单词是否与预排序词典中的任何单词匹配。下面是我当前的实现方式：
findMatchesInDict void findMatchesInDict(string word, int start, const string dict[], int end, string results[], int& totalResults)
{
// initial start = 0 index
// initial end = last index of dict array
int middle = start + (end - start) / 2;
if (end < start)
return;
if (word == dict[middle]) // if we found a match
storeUniqueMatches(word, 0, results, totalResults);
else if (word < dict[middle])
findMatchesInDict(word, start, dict, middle - 1, results, totalResults);
else
findMatchesInDict(word, middle + 1, dict, end, results, totalResults);
}
函数正常工作（这只是将匹配的单词存储在storeUniqueMatches数组中，确保没有重复的单词被存储。
该功能将只匹配词典中的选定单词，而不匹配其他单词。
关于为什么可能无法正常工作的任何想法？
作为参考，此实现有效，但效率极低，并且会导致堆栈溢出错误。
results答案 0 :(得分：0)
我仍然相信OP犯错了1次。
我强烈怀疑
findMatchesInDict(word, start, dict, middle - 1, results, totalResults);
应该是
findMatchesInDict(word, start, dict, middle, results, totalResults);
我做了自己的小样本。 （因此，我对OP的运行方式感到不走运，因此对代码进行了一些重新设计。）
#include <iostream>
#include <string>
size_t find(const std::string &word, const std::string dict[], size_t i0, size_t size)
{
if (!size) return (size_t)-1; // bail out with invalid index
const size_t i = i0 + size / 2;
return word == dict[i]
? i
: word < dict[i]
? find(word, dict, i0, i - i0)
: find(word, dict, i + 1, i0 + size - (i + 1));
}
int main()
{
const std::string dict[] = {
"Ada", "BASIC", "C", "C++",
"D", "Haskell", "INTERCAL", "Modula2",
"Oberon", "Pascal", "Scala", "Scratch",
"Vala"
};
const size_t sizeDict = sizeof dict / sizeof *dict;
unsigned nErrors = 0;
// brute force tests to find something what is in
for (size_t n = 1; n <= sizeDict; ++n) {
for (size_t i = 0; i < n; ++i) {
if (find(dict[i], dict, 0, n) >= n) {
std::cerr << "ALERT! Unable to find entry " << i << " in " << n << " entries!\n";
++nErrors;
}
}
}
// brute force tests to find something what is not in
for (size_t n = 1; n <= sizeDict; ++n) {
if (find("", dict, 0, n) < n) {
std::cerr << "ALERT! Able to find entry '' in " << n << " entries!\n";
++nErrors;
}
for (size_t i = 0; i < n; ++i) {
if (find(dict[i] + " + Assembler", dict, 0, n) < n) {
std::cerr << "ALERT! Able to find entry '" << dict[i] << " + Assembler' in " << n << " entries!\n";
++nErrors;
}
}
}
// report
if (!nErrors) std::cout << "All tests passed OK.\n";
else std::cerr << nErrors << " tests failed!\n";
// done
return nErrors > 0;
}
此代码大部分是蛮力测试代码：
从1到dict大小的每个长度都经过测试。对于每个长度，将搜索dict的任何条目。
从1到dict大小的每个长度都经过测试。对于每个长度，都会测试空字符串（在任何其他条目之前）以及任何经过修改的条目。 （修改后，该修改将介于未修改的条目与其后继条目之间，或者位于最后一个条目之后。）
输出：
All tests passed OK.
一切顺利。
然后我替换了
find(word, dict, i0, i - i0)
使用
find(word, dict, i0, i - i0 > 0 ? i - i0 - 1 : 0)
类似于OP的代码在我看来是什么错误。
输出：
ALERT! Unable to find entry 0 in 2 entries!
ALERT! Unable to find entry 0 in 3 entries!
ALERT! Unable to find entry 1 in 4 entries!
ALERT! Unable to find entry 1 in 5 entries!
ALERT! Unable to find entry 3 in 5 entries!
ALERT! Unable to find entry 0 in 6 entries!
ALERT! Unable to find entry 2 in 6 entries!
ALERT! Unable to find entry 4 in 6 entries!
ALERT! Unable to find entry 0 in 7 entries!
ALERT! Unable to find entry 2 in 7 entries!
ALERT! Unable to find entry 4 in 7 entries!
ALERT! Unable to find entry 0 in 8 entries!
ALERT! Unable to find entry 3 in 8 entries!
ALERT! Unable to find entry 5 in 8 entries!
ALERT! Unable to find entry 0 in 9 entries!
ALERT! Unable to find entry 3 in 9 entries!
ALERT! Unable to find entry 6 in 9 entries!
ALERT! Unable to find entry 1 in 10 entries!
ALERT! Unable to find entry 4 in 10 entries!
ALERT! Unable to find entry 7 in 10 entries!
ALERT! Unable to find entry 1 in 11 entries!
ALERT! Unable to find entry 4 in 11 entries!
ALERT! Unable to find entry 7 in 11 entries!
ALERT! Unable to find entry 9 in 11 entries!
ALERT! Unable to find entry 1 in 12 entries!
ALERT! Unable to find entry 3 in 12 entries!
ALERT! Unable to find entry 5 in 12 entries!
ALERT! Unable to find entry 8 in 12 entries!
ALERT! Unable to find entry 10 in 12 entries!
ALERT! Unable to find entry 1 in 13 entries!
ALERT! Unable to find entry 3 in 13 entries!
ALERT! Unable to find entry 5 in 13 entries!
ALERT! Unable to find entry 7 in 13 entries!
ALERT! Unable to find entry 9 in 13 entries!
ALERT! Unable to find entry 11 in 13 entries!
35 tests failed!
好吧。实际上，这并没有证明与OP的代码有关。
但是，这显示了
“减1”可以从本质上破坏二进制搜索。
如何设计暴力测试以发现此类错误。
因此，这有望帮助OP自己发现算法中的错误（实际上对他来说更有价值）。