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c++ - 使用 map 将多行 if 语句转换为单行

转载 作者:行者123 更新时间:2023-12-02 10:01:38 27 4
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注:c++98

我对 C++ 有点陌生,我想清理我的代码。我有一个 if 语句检查数组中的数据类型,如果匹配,则执行相应的语句。

我想将此多行 if 语句转换为单行,以检查 map 中是否存在任何这些类型,以及它们是否执行它。

我的代码:

if (boost::iequals(sqlBufferTypes[i][j], "INTEGER")                 ||
boost::iequals(sqlBufferTypes[i][j], "INT") ||
boost::iequals(sqlBufferTypes[i][j], "BIGINT") ||
boost::iequals(sqlBufferTypes[i][j], "uint8_t") ||
boost::iequals(sqlBufferTypes[i][j], "uint16_t") ||
boost::iequals(sqlBufferTypes[i][j], "LONG"))
{
// do stuff
}

并希望将其转换为类似于:

map<int, string> dataTypes;

dataTypes[1,"INT"];
dataTypes[2,"BIGINT"];
dataTypes[3,"uint8_t"];
dataTypes[4,"uint16_t"];
dataTypes[5,"LONG"];

if (boost::iequals(dataTypes.begin(), dataTypes.end())
{
// do stuff
}

最佳答案

我想 真实 挑战是拥有一个map<>不区分大小写地比较键。

您可以通过使用比较谓词来做到这一点:

struct ci_less {
bool operator()(std::string_view a, std::string_view b) const {
return boost::lexicographical_compare(a, b, boost::is_iless{});
}
};

您声明 map 以使用该谓词:
std::map<std::string, int, ci_less> const dataTypes {
{ "INT", 1 },
{ "BIGINT", 2 },
{ "uint8_t", 3 },
{ "uint16_t", 4 },
{ "LONG", 5 },
};

Note that it is now const, and I flipped the key/value pairs. See below



一些测试: Live On Coliru
// your sqlBufferTypes[i][j] e.g.:
for (std::string const key : { "uint32_t", "long", "lONg" }) {
if (auto match = dataTypes.find(key); match != dataTypes.end()) {
std::cout << std::quoted(key) << " maps to " << match->second;

// more readable repeats lookup:
std::cout << " or the same: " << dataTypes.at(key) << "\n"; // throws unless found
} else {
std::cout << std::quoted(key) << " not found\n";
}
}

打印
"uint32_t" not found
"long" maps to 5 or the same: 5
"lONg" maps to 5 or the same: 5

翻转键/值

字典有一个用于在所有语言/库中查找的关键字段。因此,要反向查找,您最终会进行线性搜索(只需查看每个元素)。

在助推你 可以通过定义一个多索引容器来吃你的蛋糕。

多指标

这有助于通过多个索引(包括复合键)进行查找。 (搜索我的答案以获取更多真实示例)

Live On Coliru
#include <boost/algorithm/string.hpp> // for is_iless et al.
#include <string_view>
#include <boost/multi_index_container.hpp>
#include <boost/multi_index/member.hpp>
#include <boost/multi_index/ordered_index.hpp>
#include <iostream> // for std::cout
#include <iomanip> // for std::quoted
#include <boost/locale.hpp>

namespace bmi = boost::multi_index;

struct ci_less {
bool operator()(std::string_view a, std::string_view b) const {
return boost::lexicographical_compare(a, b, boost::is_iless{});
}
};

struct DbType {
std::string_view name;
int type_id;

friend std::ostream& operator<<(std::ostream& os, DbType const& t) {
return os << "DbType{" << std::quoted(t.name) << ", " << t.type_id << "}";
}
};

using Map = bmi::multi_index_container<
DbType,
bmi::indexed_by<
bmi::ordered_unique<
bmi::tag<struct by_id>,
bmi::member<DbType, int, &DbType::type_id> >,
bmi::ordered_unique<
bmi::tag<struct by_name>,
bmi::member<DbType, std::string_view, &DbType::name>, ci_less>
>
>;

int main() {

Map dataTypes {
{ "INT", 1 },
{ "BIGINT", 2 },
{ "uint8_t", 3 },
{ "uint16_t", 4 },
{ "LONG", 5 },
};

auto& idx = dataTypes.get<by_name>();

// your sqlBufferTypes[i][j] e.g.:
for (std::string_view const key : { "uint32_t", "long", "lONg" }) {
if (auto match = idx.find(key); match != idx.end()) {
std::cout << std::quoted(key) << " -> " << *match << std::endl;
} else {
std::cout << std::quoted(key) << " not found\n";
}
}
}

打印
"uint32_t" not found
"long" -> DbType{"LONG", 5}
"lONg" -> DbType{"LONG", 5}

双图

Boost Bimap 是 map 的专门化。它的选项更少,尤其是增加了 operator[]风格界面回来了。
using Map = boost::bimap<
int,
boost::bimaps::set_of<std::string_view, ci_less>>;

遗憾的是构造函数不支持初始化列表,但我们可以使用迭代器接口(interface),然后我们使用 right按名称进行查找的 bimap View :

Live On Coliru
static const Map::relation s_mappings[] = {
{ 1, "INT" },
{ 2, "BIGINT" },
{ 3, "uint8_t" },
{ 4, "uint16_t" },
{ 5, "LONG" },
};

Map const dataTypes { std::begin(s_mappings), std::end(s_mappings) };

// your sqlBufferTypes[i][j] e.g.:
auto& vw = dataTypes.right;
for (std::string_view const key : { "uint32_t", "long", "lONg" }) {
if (auto match = vw.find(key); match != vw.end()) {
std::cout << std::quoted(key) << " -> " << match->second << "\n";
} else {
std::cout << std::quoted(key) << " not found\n";
}
}

打印
"uint32_t" not found
"long" -> 5
"lONg" -> 5

关于c++ - 使用 map 将多行 if 语句转换为单行,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/62299008/

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