我想知道是否无论如何我都可以使std::cout << std::endl;
重载endl
来不仅换行,而且还要打印一个'-'
,换行应该在该位置然后打印另一个换行符。
就像我做了std::cout << std::endl << '-' << std::endl;
因此,我假设我必须重载<<
,但是我不确定从哪里可以使用endl
。
答案 0 :(得分:2)
受epic question about indenting std::ostream
instances的启发,这是一个编解码器类,它将添加其他字符。
该类由@MartinYork的popular answer改编而成:我复制粘贴了该类,使它改编为使用不同的字符,然后将for循环重写为我发现更自然的形式。
这里是working example。
#include <iostream>
#include <locale>
class augmented_newline_facet : public std::codecvt<char, char, std::mbstate_t>
{
const char addition = '-';
public:
explicit augmented_newline_facet(const char addition, size_t refs = 0) : std::codecvt<char,char,std::mbstate_t>(refs), addition{addition} {}
using result = std::codecvt_base::result;
using base = std::codecvt<char,char,std::mbstate_t>;
using intern_type = base::intern_type;
using extern_type = base::extern_type;
using state_type = base::state_type;
int& state(state_type& s) const {return *reinterpret_cast<int*>(&s);}
protected:
virtual result do_out(state_type& addition_needed,
const intern_type* rStart, const intern_type* rEnd, const intern_type*& rNewStart,
extern_type* wStart, extern_type* wEnd, extern_type*& wNewStart) const override
{
result res = std::codecvt_base::noconv;
while ((rStart < rEnd) && (wStart < wEnd))
{
// The last character seen was a newline.
// Thus we need to add the additional character and an extra newline.
if (state(addition_needed) == 1)
{
*wStart++ = addition;
*wStart++ = '\n';
state(addition_needed) = 0;
res = std::codecvt_base::ok;
continue;
}
else
{
// Copy the next character.
*wStart = *rStart;
}
// If the character copied was a '\n' mark that state
if (*rStart == '\n')
{
state(addition_needed) = 1;
}
++rStart;
++wStart;
}
if (rStart != rEnd)
{
res = std::codecvt_base::partial;
}
rNewStart = rStart;
wNewStart = wStart;
return res;
}
virtual bool do_always_noconv() const throw() override
{
return false;
}
};
int main(int argc, char* argv[]) {
std::ios::sync_with_stdio(false);
std::cout.imbue(std::locale(std::locale::classic(),
new augmented_newline_facet{'-'}));
for (int i = 0; i < 5; ++i)
{
std::cout << "Line " << i << std::endl;
}
}
答案 1 :(得分:0)
如何定义类似于std::endl
的自己的函数(或函数模板):
std::ostream& enddash(std::ostream& os)
{
// Can use std::endl in place of '\n' if flushing desired.
os << "\n-\n";
return os;
}
用法示例:
int main(int argc, char* argv[]) {
std::cout << "Hello, world." << enddash;
std::cout << "Hello, world, once again" << enddash;
}
输出:
Hello, world.
-
Hello, world, once again
-
答案 2 :(得分:0)
这不像@NicholasM的答案那么酷,但是如果主体实际上仅包含std::cout << std::endl;
,则会更容易:
#include <stdio.h> // We don't want any iostream stuff, so we use the C header.
namespace std
{
struct dummy{};
dummy cout;
dummy endl;
dummy& operator<<(dummy &d, const dummy& other){
printf("\n-\n");
return d;
}
}
int main()
{
std::cout << std::endl;
}
我只是写这段代码感到有点脏...
编辑:非常清楚:我不鼓励任何人以任何有意义的方式使用此代码(如@uneven_mark所述,它包含UB)。在我看来,该问题是由OP引起的,是一种插科打or或谜语,因此,我认为类似这样的问题有可能作为答案。
答案 3 :(得分:0)
您可以使用endl
的宏定义来做到这一点。
我不建议您使用它,但是如果您必须...
#include <iostream>
#include <string>
#define endl string("\n-\n") << std::flush
int main()
{
std::cout << std::endl;
return 0;
}