写一个电话号码中特殊字符的过滤的C++算法

[gccdy]

帮忙写一个电话号码中特殊字符的过滤的C++算法

[believefym]

需求不清

[gccdy]

比如: [phone]# 怎么样把+ - # (空格)过滤掉

[xalangying]

#include<string>
#include<algorithm>
using namespace std;

bool is(char c)
{
static const string str="+-# ";
return str.find(c)!=string::npos;
}
void main()
{
string snum="+86   20-1234567#";
erase(remove_if(snum.begin(), snum.end(), is), snum.end());
}

[believefym]

char c[20]="+86 20-1234567#";

int j=0;
for(int i=0; i<strlen(c); ++i)
{
if(c[i]=='+'||c[i]=='-'||c[i]=='#')
continue;
c[j++]=c[i];
}

c[j]=0;

cout<<c<<endl;

[believefym]

#include<string>
#include<algorithm>
using namespace std;

bool is(char c)
{
static const string str="+-# ";
return str.find(c)!=string::npos;
}
void main()
{
string snum="+86   20-1234567#";
erase(remove_if(snum.begin(), snum.end(), is), snum.end());
}

[believefym]

	char c[20]="+86   20-1234567#";

	int j=0;
	for(int i=0; i<strlen(c); ++i)
	{
		if(c[i]=='+'||c[i]=='-'||c[i]=='#')
			continue;
		c[j++]=c[i];
	}

	c[j]=0;

	cout<<c<<endl;

[qiuqiu173]

楼主胶泥一个最简单的方法:

用
int isdigit(int ch)//测试参数是否为数字 0 - 9

[xalangying]

我在三楼写错了，改正

#include <string>
#include <algorithm>
#include <cctype>
#include <functional>
#include <iostream>
using namespace std;

int main()
{
	string snum="+86   20-1234567#";
	snum.erase(remove_if(snum.begin(), snum.end(), not1( ptr_fun(isdigit))), snum.end());
	cout<<snum<<endl;
	return 0;
}

[gccdy]

呵呵，谢谢了！！！

[vrace]

#include <stdio.h>
#include <stdlib.h>

int GetNum(char *pszDest, const char *pszSource)
{
	int c;

	c = 0;
	while(*pszSource)
	{
		if(*pszSource >= '0' && *pszSource <= '9')
		{
			if(pszDest) pszDest[c] = *pszSource;
			c++;
		}
		pszSource++;
	}

	return c;
}

int main(void)
{
	char *str = "[phone]#";
	char *pDest;
	int c;

	c = GetNum(NULL, str);
	if(!c) return 1;

	pDest = (char*)malloc(sizeof(char) * (c + 1));
	GetNum(pDest, str);

	printf("%s\n", pDest);
	free(pDest);

	return 0;
}