C++ 将前导零添加到字符串，没有 (s)printf
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/6143824/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Add leading zero's to string, without (s)printf
提问by Guus Geurkink
I want to add a variable of leading zero's to a string. I couldn't find anything on Google, without someone mentioning (s)printf, but I want to do this without (s)printf.
我想在字符串中添加一个前导零的变量。我在谷歌上找不到任何东西，没有人提到（s）printf，但我想在没有（s）printf的情况下做到这一点。
Does anybody of the readers know a way?
有读者知道方法吗？
采纳答案by Michael Burr
// assuming that `original_string` is of type `std:string`:
std::string dest = std::string( number_of_zeros, '0').append( original_string);
回答by Ruslan Guseinov
I can give this one-line solution if you want a field of n_zero zeros:
如果你想要一个 n_zero 零的字段，我可以给出这个单行解决方案：
std::string new_string = std::string(n_zero - old_string.length(), '0') + old_string;
For example: old_string = "45"; n_zero = 4; new_string = "0045";
例如： old_string = "45"; n_zero = 4; new_string = "0045";
回答by Rob?
You could use std::string::insert, std::stringstreamwith stream manipulators, or Boost.Format:
您可以使用std::string::insert,std::stringstream与流操纵器或Boost.Format：
#include <string>
#include <iostream>
#include <iomanip>
#include <boost/format.hpp>
#include <sstream>
int main() {
std::string s("12");
s.insert(0, 3, '0');
std::cout << s << "\n";
std::ostringstream ss;
ss << std::setw(5) << std::setfill('0') << 12 << "\n";
std::string s2(ss.str());
std::cout << s2;
boost::format fmt("%05d\n");
fmt % 12;
std::string s3 = fmt.str();
std::cout << s3;
}
回答by Jerry Coffin
You could do something like:
你可以这样做：
std::cout << std::setw(5) << std::setfill('0') << 1;
This should print 00001.
这应该打印00001.
Note, however, that the fill-character is "sticky", so when you're done using zero-filling, you'll have to use std::cout << std::setfill(' ');to get the usual behavior again.
但是请注意，填充字符是“粘性的”，因此当您使用零填充完成后，您将不得不再次使用它std::cout << std::setfill(' ');来获得通常的行为。
回答by Ivan Strelets
This works well for me. You don't need to switch setfill back to ' ', as this a temporary stream.
这对我很有效。您不需要将 setfill 切换回 ' '，因为这是一个临时流。
std::string to_zero_lead(const int value, const unsigned precision)
{
std::ostringstream oss;
oss << std::setw(precision) << std::setfill('0') << value;
return oss.str();
}
回答by Ozair Kafray
The C++ way of doing it is with setw, ios_base::widthand setfill
C++ 的方法是使用setw、ios_base::width和setfill
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
const int a = 12;
const int b = 6;
cout << setw(width) << row * col;
cout << endl;
return 0;
}
回答by Seth Robertson
memcpy(target,'0',sizeof(target));
target[sizeof(target)-1] = 0;
Then stick whatever string you want zero prefixed at the end of the buffer.
然后在缓冲区的末尾粘贴您想要零前缀的任何字符串。
If it is an integer number, remember log_base10(number)+1(aka ln(number)/ln(10)+1) gives you the length of the number.
如果它是一个整数，请记住log_base10(number)+1（又名ln(number)/ln(10)+1）为您提供数字的长度。
