如何创建一个字符串,以便它将浮点数格式化为没有尾随的小数点或数字,当它是一个整数时,但不要切换到更大数字的科学记数法？
当我做:
float myFloat= 15.6f;
float myInt = 5.0f;
float myLarge = 7000000.0f;
sprintf(out, "my float is %g", myFloat);
sprintf(out, "my int is %g", myInt);
sprintf(out, "my large is %g", myLarge);
我有类似的东西:
my float is 15.6 my int is 5 my large is 7e+07f
我想要一个单独的格式字符串,它将给出15.6,5和700000.
编辑原因评论不做格式化:
那正是我所想.但是包装器非常不方便,因为格式字符串嵌入在更长格式的字符串中:
sprintf(buf, "%d %g", myInt, myFloat);
怎么包裹那个？
sprintf(buf, "%d %g", myInt, Wrapper(myFloat));??
Wrapper应该返回什么？或者更糟:
sprintf(buf, "%d %s", myInt, Wrapper(myFloat));??
我有以下XML结构:
<artists>
<artist>
<name></name>
<image size="small"></image>
<image size="big"></image>
</artist>
</artists>
我需要选择具有给定属性的名称和图像(size = big).
var q = from c in feed.Descendants("artist")
select new { name = c.Element("name").Value,
imgUrl = c.Element("image").Value };
如何在上面的查询中指定所需的图像属性(size = big)？
我还在学习金字塔,而我正在努力学习如何使用装饰器.下面是我的测试视图可调用的副本.
from pyramid.response import Response
from pyramid.view import view_config
from pyramid.renderers import render_to_response
def my_blog(request):
return {'project':'tricky'}
@view_config( renderer='templates/foo.pt' )
def foo_blog(request):
return {'name':'tricky'}
根据我对view_config装饰器的理解,它可用于设置应用程序配置,而无需在配置文件中实际设置它们.在本例中,我将渲染器设置为templates/foo.pt.这不起作用.
但是,如果我在配置文件(init .py)中设置渲染器,则:
config.add_route( 'foo_blog' , '/blog/{foo}' , view='tricky.views.Blog.blog.foo_blog' renderer='tricky:templates/mytemplate.pt' )
它会工作.
我做错了什么阻止我能够使用装饰器.谢谢!
码:
#include<iostream>
using namespace std;
template<class T, int N> class point {
T coordinate[N];
public:
point(const point<T,N>&);
const double& operator[](int i) const {
return coordinate[i];
}
};
template<class T, int N> point<T,N>::point(const point<T,N>&p)
{
for(int i=0;i<N;i++)
coordinate[i]=p.coordinate[i];
};
int main() {
point<int,2> P2;
point<double,3> P3;
cout<<P2[0]<<P3[1];
return 0;
}
输出:
prog.cpp: In function ‘int main()’:
prog.cpp:17: error: no matching function for call to ‘point<int, 2>::point()’
prog.cpp:11: note: candidates are: point<T, N>::point(const point<T, N>&) [with T =
int, int N = …在下面的示例中,我想删除该std::wstring(std::widen(...))部分,但'#'宏只返回一个char字符串文字 - 有没有办法容纳wchar？
#define FOO_MACRO(className)\
struct className##Factory : public OtherClass {\
// does some stuff here\
} className##Factory;\
someMap->add(std::wstring(std::widen(#className), className##Factory)))
我怎么用wchar做同样的事情？
正如我在最近的SO问题中所提到的那样,我正在通过项目Euler问题来学习F#.
我现在对问题3有一个功能正常的答案,如下所示:
let rec findLargestPrimeFactor p n =
if n = 1L then p
else
if n % p = 0L then findLargestPrimeFactor p (n/p)
else findLargestPrimeFactor (p + 2L) n
let result = findLargestPrimeFactor 3L 600851475143L
但是,由于有2个执行路径可能导致不同的调用findLargestPrimeFactor,我不确定它是否可以针对尾递归进行优化.所以我想出了这个:
let rec findLargestPrimeFactor p n =
if n = 1L then p
else
let (p', n') = if n % p = 0L then (p, (n/p)) else (p + 2L, n)
findLargestPrimeFactor p' n' …我正在尝试下载图像,并在完成时触发事件.我用这个:
BitmapImage btest = new BitmapImage(new Uri("http://www.google.com/images/srpr/logo4w.png"));
btest.ImageOpened += btest_ImageOpened;
void btest_ImageOpened(object sender, RoutedEventArgs e)
{
throw new NotImplementedException();
}
但是,该ImageOpened事件不会发生.如果我将Image Control的源设置为BitmapImageusing:
image.Source = btest;
它确实很火.ImageOpened除非将BitmapImage发件人设置为图像的来源,为什么不会触发事件？
我正在尝试编写一个带有菜单的程序,该菜单可以通过几种不同的方式从文本文件中读取.我只是在处理菜单选项#2(从文件末尾向后读),但我无法理解我做错了什么.我已经在这几天了,而且找不到任何好的资源来帮助解决这个问题.任何帮助,将不胜感激.
#include <iostream>
#include <string>
#include <iomanip>
#include <istream>
#include <math.h>
#include <fstream>
using namespace std;
const int SIZE = 20;
typedef char String[SIZE];
//prototypes
void Menu(int &);
void ReadFile(ifstream &);
void BackwardFile(ifstream &,long &);
long CountFile(ifstream &,long &);
int main()
{
char filename[]= "grades.txt";
int choice;
long numBytes;
ifstream InList;
InList.open(filename);
if(InList.good())
{
do
{
Menu(choice);
switch(choice)
{
case 1:
ReadFile(InList);
break;
case 2:
CountFile(InList,numBytes);
BackwardFile(InList,numBytes);
break;
case 3:
cout << "Pick a start position between 0 - " …我使用Python 2.7(iOS Pythonista App)和reportlab 2.7模块来创建带有表格的PDF.一切正常.RepotLab自动格式化列的宽度.但是在两种情况下,我无法理解为什么reportlab以它的方式格式化输出以及如何获得我想要的格式.
案例1:几乎我想要的一切......
+-----+----------+---------------+----------+---------------------------------------------------------------------------------------------------------------+
| Day | Date | Time | Duration | Notes |
+-----+----------+---------------+----------+---------------------------------------------------------------------------------------------------------------+
| Tue | 01.04.14 | 14:00 - 17:15 | 3.25 | Here are some notes. |
+-----+----------+---------------+----------+---------------------------------------------------------------------------------------------------------------+
| Wed | 02.04.14 | 18:00 - 20:15 | 2.25 | Sometime these notes are a little longer text so there must be a line break to let the whole note be visible! |
+-----+----------+---------------+----------+---------------------------------------------------------------------------------------------------------------+
| Thu | 02.04.14 | 14:00 - 17:15 | …