C++ c++;从流中提取双精度
我对学校的运动有一个有趣的问题。我得到了纬度和经度,我必须确保它的格式正确:C++ c++;从流中提取双精度,c++,iostream,C++,Iostream,我对学校的运动有一个有趣的问题。我得到了纬度和经度,我必须确保它的格式正确:\(\d+\.\d+[NS],\d\+.\d+[EW]\)。如您所见,我必须检查 有 char lBracket, rBracket, comma, NS, EW; int nLat, nLon; double lat, lon; istringstream iss ("(51.5N, 0.0E)"); iss >> fixed >> lBracket >> nLa
\(\d+\.\d+[NS],\d\+.\d+[EW]\)
。如您所见,我必须检查
有
char lBracket, rBracket, comma, NS, EW;
int nLat, nLon;
double lat, lon;
istringstream iss ("(51.5N, 0.0E)");
iss >> fixed >> lBracket >> nLat >> lat >> NS >> comma >>
nLon >> lon >> EW >> rBracket;
lat += nLat;
lon += nLon;
>lon
提取“.0E”
部分(这是一个有效的双精度),但我需要将其提取到EW
我提出的唯一解决方案是用其他字母替换E
,这可能会起作用,但不是很好的代码
这个问题还有其他更优雅的解决方案吗
PS:您猜对了,不允许使用正则表达式:D您实际上看到的是一个解析问题<代码>运算符>>(istream&,string&)是一个非常简单的解析器,只对空白进行标记 如果您的格式规范足够严格,您应该拒绝
(51.5N,0.0E)
(逗号前的额外空格),那么就不要提取逗号。相反,在提取之后,您应该检查nLat
是否包含尾随逗号并将其删除。您不再需要逗号
如果必须在任何位置支持可选空格,则可以通过在逗号前后插入额外空格(无条件)来帮助预处理字符串。如果已经有一个空间,那么现在将有两个。这没有问题,因为跳过空格将跳过任何数量。您的问题不是您所想的
.0E
不是有效的浮点数字
甚至在科学记数法中。它没有任何指数
发生的事情是流解析器被提交到解释中
当.0E
到达E
时,作为一个科学的浮点数字,然后
找不到指数;认为其解释是伪造的;将0分配给
目标加倍并在iss
中设置failbit
,因此不再从
流是可能的。您可以通过将0.0E
更改为1.1E
和
在尝试提取lon之后立即测试iss.fail()
。
仍然发现lon
设置为0,而不是0.1,并且iss.fail()==true
我认为在上下文中没有办法避免这一点w.fE
(w
=整个部分,f
=小数部分)
如果试图使用>
将w.f
或仅将.f
提取到浮点变量中。
在这种情况下,您已经超出了格式化>
提取的舒适区
而且还需要变得更加灵活。事实上,这种需要并不仅仅来自于
w.fE
角盒,给定模式\(\d+\.\d+[NS],\d\+.\d+[EW]\)
你告诉我们参考资料必须满足我们的要求。\d+\.\d+[NS]
部分也很重要
对>double>>char
或>>[unsigned | int]>>double>>char
进行精确计算:
整数或浮点提取器将使用前导的+
或-
而浮点提取器也不会坚持使用
小数点或前后的非零数字计数
解析映射引用的繁琐(没有正则表达式的帮助)会
提示我为它们创建一个map\u ref
类,以便您可以尝试提取
map\u ref
来自输入流(可能也来自字符串);可以询问map\u ref
它是好的还是坏的(例如,在尝试提取之后),并且可以插入
格式化的映射到输出流中
以下是此类课程的示意图:
#include <ostream>
#include <istream>
#include <iomanip>
#include <utility>
#include <limits>
struct map_ref
{
map_ref() = default;
map_ref(char NS, double lat, char EW, double lon)
: _NS(NS == 'N' || NS == 'S' ? NS : '?'),
_EW(EW == 'E' || EW == 'W' ? EW : '?'),
_lat(lat >= 0.0 && lat <= 90.0 ? lat : -1),
_lon(lon >= 0.0 && lon <= 180.0 ? lon : -1),
_good(_NS != '?' && _EW != '?' && _lat != -1 && _lon != -1){}
// Todo: string ctor
std::pair<char,double> latitude() const {
return std::make_pair(_NS,_lat);
}
std::pair<char,double> longitude() const {
return std::make_pair(_EW,_lon);
}
// Todo: setters, getters, equality etc.
bool good() const {
return _good;
}
bool read(std::istream & in);
std::ostream & write(std::ostream & out, std::size_t precision = 4) const {
return out << std::fixed << std::setprecision(precision) <<
'(' << _lat << _NS << ", " << _lon << _EW << ')';
}
void clear() {
*this = map_ref();
}
private:
double
read_fixed_point(std::istream & in, char & dest, std::string const & delims);
char _NS = '?';
char _EW = '?';
double _lat = -1;
double _lon = -1;
bool _good = false;
};
double
map_ref::read_fixed_point(
std::istream & in, char & dest, std::string const & delims)
{
std::string s;
unsigned whole_digs = 0, frac_digs = 0;
while(in >> dest &&
dest != '.' && delims.find(dest) == std::string::npos)
{
whole_digs += std::isdigit(dest) != 0;
s += dest;
}
if (dest != '.') {
return -1;
}
s += dest;
while(in >> dest && delims.find(dest) == std::string::npos)
{
frac_digs += std::isdigit(dest) != 0;
s += dest;
}
if (whole_digs == 0 || frac_digs == 0 ||
whole_digs + frac_digs > std::numeric_limits<double>::digits10 ||
s.length() != 1 + whole_digs + frac_digs) {
return -1;
}
return std::stod(s);
}
bool map_ref::read(std::istream & in)
{
char lparen = 0;
clear();
in >> std::noskipws >> lparen;
if (lparen != '(') {
return _good = false;
}
_lat = read_fixed_point(in,_NS,"NS");
if (_lat < 0.0 || _lat > 90.0) {
return _good = false;
}
char comma = 0;
in >> comma;
if (comma != ',') {
return _good = false;
}
while (std::isspace(in.peek())) {
_EW = in.get();
}
_lon = read_fixed_point(in,_EW,"EW");
if (_lon < 0.0 || _lon > 180.0) {
return _good = false;
}
char rparen = 0;
in >> rparen;
return _good = rparen == ')';
}
std::istream & operator>>(std::istream & in, map_ref & mr)
{
mr.read(in);
return in;
}
std::ostream & operator<<(std::ostream & out, map_ref const & mr)
{
return mr.write(out);
}
用于指定纬度和经度的精度
按照或在输出流中表示
接受默认的4
。如果将其设置为0,则将丢失所有小数位
在输出上,所以不要这样做
该类带有全局运算符的重载>
和,您可以在输入字符串中将E
替换为X
(例如),运行代码后,可以在EW
变量中将X
替换为E
char lBracket, rBracket, comma, NS, EW;
int nLat, nLon;
double lat, lon;
std::string inputString = "(51.5N, 0.0E)";
size_t e_pos = inputString.find( 'E' );
if ( e_pos != std::string::npos ) {
inputString.replace( e_pos, e_pos + 1, 'X' );
}
istringstream iss ( inputString );
iss >> fixed >> lBracket >> nLat >> lat >> NS >> comma >>
nLon >> lon >> EW >> rBracket;
if ( EW == 'X' ) {
EW = 'E';
}
lat += nLat;
lon += nLon;
更新:对不起,没有看到您的评论
我提出的唯一解决方案是用其他字母替换E,这可能会起作用,但不是很好的代码
如果你愿意,我将删除一个答案,这是典型的数学方法:如果你已经有了X的解,就足以把你的实际问题Y变成X。不要删除它。这是一个有效的答案,虽然不是我想要的答案,但它仍然对某些人有用:)你如何处理7位数的分数,并区分50.0000123和50.123?这段代码需要认真的工作来管理这些细节。顺便提一下,当用户键入50.123时,代码返回62.3。另请参见@JonathanLeffler I stand correct。答案出奇地错误。我希望更新不会发生。
#include <iostream>
#include <sstream>
static unsigned tests = 0, pass = 0, fail = 0;
static void expect_good(char NS, double lat, char EW, double lon )
{
std::cout << "Testing (" << ++tests << ") "
<< NS << ',' << lat << ',' << EW << ',' << lon << '\n';
map_ref mr(NS,lat,EW,lon);
if (!mr.good()) {
std::cerr << "Failed (" << tests << "): Is good, got bad\n";
++fail;
} else {
++pass;
std::cout << "Passed (" << tests << "): Is good. Got \"" << mr << "\"\n";
}
}
static void expect_bad(char NS, double lat, char EW, double lon )
{
std::cout << "Testing (" << ++tests << ") "
<< NS << ',' << lat << ',' << EW << ',' << lon << '\n';
map_ref mr(NS,lat,EW,lon);
if (mr.good()) {
std::cerr << "Failed (" << tests << "): Is bad, got good\n";
++fail;
} else {
++pass;
std::cout << "Passed (" << tests << "): Is bad, got bad\n";
}
}
static void expect_good(std::string const & s)
{
std::cout << "Testing (" << ++tests << ") \"" << s << "\"\n";
std::istringstream iss(s);
map_ref mr;
iss >> mr;
if (!mr.good()) {
std::cerr << "Failed (" << tests << "): Is good, got bad\n";
++fail;
} else {
++pass;
std::cout << "Passed (" << tests << "): Is good. Got \"" << mr << "\"\n";
}
}
static void expect_bad(std::string const & s)
{
std::cout << "Testing (" << ++tests << ") \"" << s << "\"\n";
std::istringstream iss(s);
map_ref mr;
iss >> mr;
if (mr.good()) {
++fail;
std::cerr << "Failed (" << tests << "): Is bad, got good\n";
} else {
++pass;
std::cout << "Passed (" << tests << "): Is bad, got bad\n";
}
}
int main()
{
expect_bad('E',1.00,'S',1.00);
expect_bad('N',-1.00,'W',1.00);
expect_bad('N',90.00,'W',180.01);
expect_bad('S',90.01,'W',180.00);
expect_good('S',90.00,'E',180.00);
expect_good('S',0.0,'E',0.0);
expect_bad("");
expect_bad("1.1N, 2.2W");
expect_bad("(1.1N, 2.2W");
expect_bad("1.1N, 2.2W)");
expect_bad("[1.1N, 2.2W)");
expect_bad("(1.1N, 2.2W(");
expect_bad("(N)");
expect_bad("(N, W)");
expect_bad("(0N, 1W)");
expect_bad("(1.0N, 2W)");
expect_bad("(1.0N, .2W)");
expect_bad("(.01N, 1.2E)");
expect_bad("(1.N, 1.2W)");
expect_bad("(N1.1, E1.2)");
expect_bad("(1.0N, 1.2 W)");
expect_bad("(1.0X, 1.2W)");
expect_bad("(1.0N, 1.2Z)");
expect_bad("(+1.0N, 1.2E)");
expect_bad("(1.+0N, 1.2E)");
expect_bad("(1.0N, -1.2E)");
expect_bad("(1.0N, 1.-2E)");
expect_bad("(1.1N, 2.3.4E)");
expect_bad("(0.0NN, 0.0E)");
expect_bad("(0.0N, 0.0EW)");
expect_bad("(0.0 1N, 0.0E)");
expect_bad("(0.01N, 0 2.0E)");
expect_bad("(0 .01N, 2.0E)");
expect_bad("(0.01N, 2. 0E)");
expect_bad("(12.34567890123456N, 2.0E)");
expect_good("(0.0N, 0.0E)");
expect_good("(1.0N,1.2W)");
expect_good("(01.0N,01.2W)");
expect_good("(1.0N, 1.2W)");
expect_good("(0.123456789N, 0.000000001E)");
expect_good("(0.000000001S, 0.123456789W)");
expect_good("(0.123456789N, 0.000000001W)");
expect_good("(0.000000001S, 0.123456789E)");
expect_good("(1.1N, 12.3456789012345E)");
expect_bad("(0.1E, 0.1N)");
expect_bad("(0.1W, 0.1S)");
expect_bad("(0.1W, 0.1N)");
expect_bad("(0.1E, 0.1S)");
expect_good("(90.0N, 180.0E)");
expect_good("(90.0S, 180.0W)");
expect_good("(90.0N, 180.0W)");
expect_good("(90.0S, 180.0E)");
expect_bad("(90.000001N, 180.0E)");
expect_bad("(90.000001S, 180.0W)");
expect_bad("(90.0N, 180.000001W)");
expect_bad("(90.0S, 180.000001E)");
std::cout << "Tests: " << tests << std::endl;
std::cout << "Passed: " << pass << std::endl;
std::cout << "Failed: " << fail << std::endl;
return 0;
}
char lBracket, rBracket, comma, NS, EW;
int nLat, nLon;
double lat, lon;
std::string inputString = "(51.5N, 0.0E)";
size_t e_pos = inputString.find( 'E' );
if ( e_pos != std::string::npos ) {
inputString.replace( e_pos, e_pos + 1, 'X' );
}
istringstream iss ( inputString );
iss >> fixed >> lBracket >> nLat >> lat >> NS >> comma >>
nLon >> lon >> EW >> rBracket;
if ( EW == 'X' ) {
EW = 'E';
}
lat += nLat;
lon += nLon;