1. int to string
一、使用atoi
说明:
itoa( int value, char *string, int radix );
第一个参数:你要转化的int;
第二个参数:转化后的char*;
第三个参数:你要转化的进制;
举例:
//-------------------------------------
//功能:C++ int 转 string (使用atoi)
//环境:VS2005
//-------------------------------------
#include "stdafx.h"
#include <iostream>
using namespace std;
int main(int argc, char* argv[])
{
int n = 30;
char c[10];
itoa(n, c, 2);
cout << "2-> " << c << endl;
itoa(n, c, 10);
cout << "16-> " << c << endl;
itoa(n, c, 16);
cout << "10-> " << c << endl;
system("pause");
return 0;
}
输出:
2-> 11110
16-> 30
10-> 1e
请按任意键继续. . .
二、使用sprintf
头文件 #include<stdio.h>
语法: int sprintf(string format, mixed [args]...);
返回值:字符串长度(strlen)
转换字符
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
% 印出百分比符号,不转换。
b 整数转成二进位。
c 整数转成对应的 ASCII 字元。
d 整数转成十进位。
f 倍精确度数字转成浮点数。
o 整数转成八进位。
s 整数转成字串。
x 整数转成小写十六进位。
X 整数转成大写十六进位。
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
举例:
//-------------------------------------
//功能:C++ int 转 string (使用sprintf)
//环境:VS2005
//-------------------------------------
#include "stdafx.h"
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n = 30;
char c[20];
sprintf(c, "%d", n);
cout << c << endl;
sprintf(c, "%o", n);
cout << c << endl;
sprintf(c, "%X", n);
cout << c << endl;
sprintf(c, "%c", n);
cout << c << endl;
float f = 24.678;
sprintf(c, "%f", f);
cout << c << endl;
sprintf(c, "%.2f", f);
cout << c << endl;
sprintf(c, "%d-%.2f", n, f);
cout << c << endl;
system("pause");
return 0;
}
输出:
30
36
1E
//注:这里是个特殊符号
24.677999
24.68
30-24.68
请按任意键继续. . .
三、使用stringstream
举例:
//-------------------------------------
//功能:C++ int 转 string (使用stringstream)
//环境:VS2005
//-------------------------------------
#include "stdafx.h"
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
stringstream strStream;
int a = 100;
float f = 23.5566;
strStream << a << "----"<< f ;
string s = strStream.str();
cout << s << endl;
system("pause");
return 0;
}
输出:
100----23.5566
请按任意键继续. . .
四、其它
1.sprintf可能引起缓冲区溢出,可以考虑使用 snprintf 或者非标准的 asprintf
2.如果是mfc程序,可以使用 CString::Format
3.如果使用boost,则可以直接使用: string s = boost::lexical_cast <string>(a);
4.atoi 也是不可移植的。
五、其它NB方法
//-----------------------------------------------------------------------------------
// 参考引用 :
// http://baike.baidu.com/view/982195.htm?fr=ala0_1_1
// http://baike.baidu.com/view/1295144.htm?fr=ala0_1
// http://pppboy.blog.163.com/blog/static/3020379620085511954382/
//-----------------------------------------------------------------------------------
2. string to int
//第一种方法
const char* str = "123";
int i;
if(sscanf(str, "%d", &i) == EOF )
{ /* error */}
//第二种方法----C++ with standard library stringstream:
int str2int (const string &str)
{
stringstream ss(str);
int num;
if((ss >> num).fail())
{ //ERROR
}
return num;
}
//第三种方法----With boost library:
#include <boost/lexical_cast.hpp>
#include <string>
try{
std::string str = "123";
int number = boost::lexical_cast< int >( str );
}catch( const boost::bad_lexical_cast & )
{ // Error
}
=============My Test========================
#include "stdafx.h"
#include <cstdio>
#include <iostream>
#include <strstream>
#include <string>
#include <sstream>
using namespace std;
template<class NumType>
bool ConvetStringToNumber(const string& str, NumType& num)
{
stringstream strStream(str);
if((strStream>>num).fail())
{
cout<<"Sorry, convert failed!"<<endl;
return false;
}
return true;
}
template<class NumType>
bool ConvetNumberToString(const NumType& num, string& str)
{
stringstream strStream;
strStream<<num;
strStream>>str; //str = strStream.str(); is ok too.
return true;
}
int _tmain(int argc, _TCHAR* argv[])
{
string strNum = "22.56";
float fNum;
if(ConvetStringToNumber(strNum, fNum))
{
cout<<"fNum = "<<fNum<<endl;
}
double dNum;
if(ConvetStringToNumber(strNum, dNum))
{
cout<<"dNum = "<<dNum<<endl;
}
int iNum;
if(ConvetStringToNumber(strNum, iNum))
{
cout<<"iNum = "<<iNum<<endl;
}
string aa="abc";
if(ConvetStringToNumber(aa, iNum))
{
cout<<"iNum = "<<iNum<<endl;
}
string str;
ConvetNumberToString(fNum, str);
cout<<"str from fNum = "<<str<<endl;
ConvetNumberToString(dNum, str);
cout<<"str from dNum = "<<str<<endl;
ConvetNumberToString(iNum, str);
cout<<"str from iNum = "<<str<<endl;
char c;
cin>>c;
return 0;
}
Output is :
fNum = 22.56
dNum = 22.56
iNum = 22
Sorry, convert failed!
str from fNum = 22.56
str from dNum = 22.56
str from iNum = 22