[英]How do I use iomanip's setw, setfill, and left/right? Setfill isn't stopping its output
I'm trying to get my output to look like this:我试图让我的输出看起来像这样:
size time1 time2
-------------------------------
10 4 8
100 48 16
1000 2937 922
10000 123011 3902
100000 22407380 830722
And I know I need to use setw()
, setfill()
, and left
.而且我知道我需要使用
setw()
、 setfill()
和left
。 However, my attempts keep giving me incorrect output.但是,我的尝试不断给我错误的输出。 Here is one example of my code:
这是我的代码的一个示例:
std::cout << "size" << std::setw(20) << "time" << std::setw(20) << "time2\n";
std::cout << std::setfill('-') << std::setw(60) << "-" << std::endl;
run = 10;
for(int i = 0; i < 5; i++) {
std::cout << run;
run *= 10;
std::cout << std::setw(20) << std::left << time1[i];
std::cout << std::setw(20) << std::left << time2[i] << "\n";
}
And here's the output:这是输出:
size time1 time2
------------------------------------------------------------
103-------------------13------------------
100171-----------------199-----------------
100013183---------------667-----------------
10000670130--------------8205----------------
10000014030798-------------1403079888---------
I've tried changing the order that I'm using setw()
, setfill()
, and left
, but I'm just flying blind right now.我已经尝试更改我使用
setw()
、 setfill()
和left
的顺序,但我现在只是盲目飞行。 I've also searched iomanip tutorials.我还搜索了 iomanip 教程。 I'm following the directions--as far as I can tell--but I'm still not getting it.
我正在遵循指示——据我所知——但我仍然没有明白。
How do I stop the setfill()
from running over?如何阻止
setfill()
溢出? How do I justify left but use setw()
to stop the numbers from running into each other?我如何证明左对齐但使用
setw()
来阻止数字相互碰撞?
How about:怎么样:
std::cout << "size" << std::setw(20) << "time" << std::setw(20) << "time2\n";
std::cout << std::setfill('-') << std::setw(60) << "-" << std::endl;
run = 10;
std::cout << std::setfill(' '); //fill with spaces
for(int i = 0; i < 5; i++) {
std::cout << std::setw(20) << std::left << run; // fill the run column
run *= 10;
std::cout << std::setw(20) << std::left << time1[i];
std::cout << std::setw(20) << std::left << time2[i] << "\n";
}
sj0h's answer is great except the titles don't quite line up. sj0h 的回答很好,只是标题不太一致。 To fix it I started the title line with "left" and "setw", I also had to end with "endl" instead of "\\n".
为了修复它,我以“left”和“setw”开始标题行,我还必须以“endl”而不是“\\n”结尾。
std::cout << std::left << std::setw(20) << "size" << std::setw(20) << "time" << std::setw(20) << "time2" << std::endl;
std::cout << std::setfill('-') << std::setw(60) << "-" << std::endl;
run = 10;
std::cout << std::setfill(' '); //fill with spaces
for(int i = 0; i < 10; i++) {
std::cout << std::setw(20) << std::left << run; // fill the run column
run *= 10;
std::cout << std::setw(20) << std::left << time1[i];
std::cout << std::setw(20) << std::left << time2[i] << std::endl;
}
In C++20 you'll be able to use std::format
to do this:在 C++20 中,您将能够使用
std::format
来执行此操作:
std::cout << std::format("{:20}{:20}{:20}\n", "size", "time", "time2");
std::cout << std::format("{:-<60}\n", "");
int run = 10;
for(int i = 0; i < 5; i++) {
std::cout << std::format("{:<20}{:<20}{:<20}\n", run, time[i], time2[i]);
run *= 10;
}
Output:输出:
size time time2
------------------------------------------------------------
10 4 8
100 48 16
1000 2937 922
10000 123011 3902
100000 22407380 830722
In the meantime you can use the {fmt} library , std::format
is based on.同时,您可以使用{fmt} 库,
std::format
基于。 {fmt} also provides the print
function that makes this even easier and more efficient ( godbolt ): {fmt} 还提供了
print
功能,使这更容易和更高效( Godbolt ):
fmt::print("{:20}{:20}{:20}\n", "size", "time", "time2");
fmt::print("{:-<60}\n", "");
int run = 10;
for(int i = 0; i < 5; i++) {
fmt::print("{:<20}{:<20}{:<20}\n", run, time[i], time2[i]);
run *= 10;
}
Disclaimer : I'm the author of {fmt} and C++20 std::format
.免责声明:我是 {fmt} 和 C++20
std::format
。
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