C ++可以setw和setfill填充字符串的结尾? [英] C++ can setw and setfill pad the end of a string?
问题描述
有没有办法让 setw
和 setfill
填充字符串的结尾而不是前面? / p>
我有这样一个打印的情况。
CONSTANT TEXT variablesizeName1 .....:number1
CONSTANT TEXT varsizeName2 ..........:number2
我想添加一个'。'
到$ /
CONSTANT TEXT variablesizeName#
所以我可以使:number#
排队屏幕。
$ b
注意:我有一个数组variablesizeName#
,所以我知道最广泛的情况。 / p>
或
手动设置 setw
like this
$ b
for(int x = 0; x< ARRAYSIZE; x ++)
{
string temp = string(CONSTANT TEXT)+ variabletext [x];
cout<<温度;
cout<< (MAXWIDTH - temp.length)<< setfill('。')<<:;
cout<<数<< \\\
;
$ b $ p
$ b 我想这样做可以做,但是感觉有点笨拙。 p>
想法?
解决方案您可以使用操纵器 std :: left
, std :: right code>和 std :: internal
来选择填充字符的位置。
对于您的具体情况,类似这样的事情可以做到这一点:
#include< iostream>
#include< iomanip>
#include< string>
const char * C_TEXT =常量文本;
const size_t MAXWIDTH = 10;
void print(const std :: string& var_text,int num)
{
std :: cout<< C_TEXT
//将输出对齐到左边,填充到右边
<< std :: left<< std :: setw(MAXWIDTH)<< std :: setfill('。')
<< var_text<< :<< num<< \\\
;
}
int main()
{
print(1234567890,42);
print(12345,101);
$ b $ p $输出:
常量文本1234567890:42
常量文本12345 .....:101
编辑:
如链接所述, std :: internal
仅适用于整数,浮动点和货币产量。例如对于负整数,它会在负号和最左边的数字之间插入填充字符。
这个:
int32_t i = -1;
std :: cout<< std :: internal
<< std :: setfill('0')
<< std :: setw(11)//最多10位数字+负号
<<我 \\\
;
i = -123;
std :: cout<< std :: internal
<< std :: setfill('0')
<< std :: setw(11)
<<一世;
会输出
-0000000001
-0000000123
Is there a way to make setw
and setfill
pad the end of a string instead of the front?
I have a situation where I'm printing something like this.
CONSTANT TEXT variablesizeName1 .....:number1
CONSTANT TEXT varsizeName2 ..........:number2
I want to add a variable amount of '.'
to the end of
"CONSTANT TEXT variablesizeName#"
so I can make ":number#"
line up on the screen.
Note: I have an array of "variablesizeName#"
so I know the widest case.
Or
Should I do it manually by setting setw
like this
for( int x= 0; x < ARRAYSIZE; x++)
{
string temp = string("CONSTANT TEXT ")+variabletext[x];
cout << temp;
cout << setw(MAXWIDTH - temp.length) << setfill('.') <<":";
cout << Number<<"\n";
}
I guess this would do the job but it feels kind of clunky.
Ideas?
解决方案 You can use manipulators std::left
, std::right
, and std::internal
to choose where the fill characters go.
For your specific case, something like this could do:
#include <iostream>
#include <iomanip>
#include <string>
const char* C_TEXT = "Constant text ";
const size_t MAXWIDTH = 10;
void print(const std::string& var_text, int num)
{
std::cout << C_TEXT
// align output to left, fill goes to right
<< std::left << std::setw(MAXWIDTH) << std::setfill('.')
<< var_text << ": " << num << '\n';
}
int main()
{
print("1234567890", 42);
print("12345", 101);
}
Output:
Constant text 1234567890: 42
Constant text 12345.....: 101
EDIT:
As mentioned in the link, std::internal
works only with integer, floating point and monetary output. For example with negative integers, it'll insert fill characters between negative sign and left-most digit.
This:
int32_t i = -1;
std::cout << std::internal
<< std::setfill('0')
<< std::setw(11) // max 10 digits + negative sign
<< i << '\n';
i = -123;
std::cout << std::internal
<< std::setfill('0')
<< std::setw(11)
<< i;
will output
-0000000001
-0000000123
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