我正在尝试捕获异常并在我的代码中的某个点引发更具体的错误:
try:
something_crazy()
except SomeReallyVagueError:
raise ABetterError('message')
这适用于Python 2,但在Python 3中,它显示了两个例外:
Traceback(most recent call last):
...
SomeReallyVagueError: ...
...
During handling of the above exception, another exception occurred:
Traceback(most recent call last):
...
ABetterError: message
...
有没有办法绕过这个,所以SomeReallyVagueError没有显示回溯？
说我有一个列表,如:
[Nothing, Just 1, Nothing, Just 2]
我想得到第一个Just(非错误)值; 在这种情况下,它是Just 1.我唯一能想到的是:
firstJust xs = case filter isJust xs of
[] -> Nothing
Just x -> Just x
有没有更好的/ monad-generic方法来做到这一点？
步骤:
1.sudo port boost
安装在/ opt/local/boost中的boost文件,库文件在/ opt/local/lib中
2.使用XCode创建c ++项目
#include <iostream>
#include <boost/asio.hpp>
int main () {
return 0;
}
3.设置XCode以
在"构建设置" - >"HEADER_SEARCH_PATHS"
中找到提升,在Debug和Release中添加路径/ opt/local/include
4."Build Settings" - >"LIBRARY_SEARCH_PATHS" - > add/opt/local/lib都用于调试和发布.
5.构建程序并失败.
错误消息,
Undefined symbols for architecture x86_64:
"boost::system::generic_category()", referenced from:
___cxx_global_var_init1 in main.o
___cxx_global_var_init2 in main.o
"boost::system::system_category()", referenced from:
___cxx_global_var_init3 in main.o
boost::asio::error::get_system_category() in main.o
"boost::asio::error::get_netdb_category()", referenced from:
___cxx_global_var_init5 in main.o <br>
"boost::asio::error::get_addrinfo_category()", referenced from:
___cxx_global_var_init6 in main.o <br>
"boost::asio::error::get_misc_category()", referenced from:
___cxx_global_var_init7 in main.o <br>
ld: …我正在编写一个输出 C++ 代码的语法翻译器，但遇到了一个有趣的问题。假设我有两个文件：ln.x和ln.cpp. 在ln.x：
abc
在ln.cpp：
#line 1 "ln.x"
(
当我尝试使用 GCC 编译它时，它会在ln.x以下位置打印相应的行：
ln.x:1:1: error: expected unqualified-id at end of input
abc
^
ln.x:1:1: error: expected ‘)’ at end of inpu
但是，Clang 只是打印同一文件的行：
ln.x:1:2: error: expected unqualified-id
(
^
ln.x:1:2: error: expected ')'
ln.x:1:1: note: to match this '('
(
^
2 errors generated.
有没有办法让 Clang 像 GCC 一样打印文件的行？
我正在学习Makefile！我包含了整个rm命令，以防某些参数相互依赖：
\n\nrm -f \\*~ core $(INCDIR)/\\*~\n我假设 C++ 生成一些以 \'~\' 结尾的文件，所以我们删除这些文件，但是core是什么？谷歌返回的唯一内容是似乎假设其功能已经已知的教程，但我找不到任何只是说“核心是......”的内容。
\n\nRider：假设“_OBJ”是目标文件列表，“ODIR”是一个目录。然后...
\n\n$(patsubst %, $(ODIR)/%, $(_OBJ))\n... 获取 \'_OBJ\' 中的任何文件名fname并将其替换为 \'$(ODIR)/fname\'\xe2\x80\x94，从而将其移动到目录 \'$(ODIR)\'名称fname，对吗？
\n我正在尝试制作这样的Python模块:
class square:
def _init_(self):
self._length = 0
self._perimeter = 0
self._area = 0
def setLength(self, length):
self._length = float(length)
self._perimeter = 0
self._area = 0
def getLength(self):
return self._length
def getPerimeter(self):
if self._perimeter == 0:
self._perimeter = float(self._length * 4)
return self._perimeter
def getArea(self):
if self._area == 0:
self._area = float(self._length * self._length)
return self._area
class rectangle:
def _init_(self):
self._length = 0
self._width = 0
self._perimeter = 0
self._area = 0
def setLength(self, length):
self._length = float(length)
self._perimeter = …我有以下Haskell程序:
catlines = unlines . zipWith (\(n,l) -> show n ++ l) [0..]
main = putStrLn $ catlines ["A", "B"]
当我尝试编译它时,GHC给出以下错误:
catlines.hs:1:41:
Couldn't match expected type `b0 -> String' with actual type `[a0]'
In the expression: show n ++ l
In the first argument of `zipWith', namely
`(\ (n, l) -> show n ++ l)'
In the second argument of `(.)', namely
`zipWith (\ (n, l) -> show n ++ l) [0 .. ]'
据我所知,这应该编译.我不知道出了什么问题.
我正在尝试运行这个Perl程序.
#!/usr/bin/perl
# --------------- exchange.pl -----------------
&read_exchange_rate; # read exchange rate into memory
# now let's cycle, asking the user for input...
print "Please enter the amount, appending the first letter of the name of\n";
print "the currency that you're using (franc, yen, deutschmark, pound) -\n";
print "the default value is US dollars.\n\n";
print "Amount: ";
while (<STDIN>) {
($amnt,$curr) = &breakdown(chop($_));
$baseval = $amnt * (1/$rateof{$curr});
printf("%2.2f USD, ", $baseval * $rateof{'U'});
printf("%2.2f Franc, ", $baseval * $rateof{'F'});
printf("%2.2f …