(最简单)方法一：

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#include <stdio.h>
char s[]="123 ab 4";
char *p;
char str[10];
int num1, num2;
void main()
{
    p=s;
    sscanf(p,"%d %s %d",&num1, str, &num2);
    printf("num1 = %d\n",num1);
    printf("num2 = %d\n",num2);
    printf("string is %s\n",str);
}

output:

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num1 = 123
num2 = 4
string is ab

(逐个查找)方法二：

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#include <stdio.h>
char s[]="123 ab 4";
char *p;
int v,n,k;
void main()
{
    p=s;
    while (1) 
    {
        k=sscanf(p,"%d%n",&v,&n);
        printf("k,v,n=%d,%d,%d\n",k,v,n);
        if (1==k) //提取到数字
        {
            p+=n;
        } 
        else if (0==k) //提取到非数字
        {
            printf("skip char[%c]\n",p[0]);
            p++;
        } 
        else //EOF==k
            break;
    }
    printf("End.\n");
}

output:

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k,v,n=1,123,3
k,v,n=0,123,3
skip char[ ]
k,v,n=0,123,3
skip char[a]
k,v,n=0,123,3
skip char[b]
k,v,n=1,4,2
k,v,n=-1,4,2
End.

(正则表达式：匹配正负号和小数点)方法三：

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#include <iostream>  
#include <string>  
#include <regex>  

using namespace std;

int main()
{
	regex pattern("-?[0-9]\\d*\\.[0-9]\\d*");

	string str_1 = "v -1234.3/234.5/3.4567 3.4/5.6/7.88 67.8/8.901/7.65";
	string::const_iterator start = str_1.begin();
	string::const_iterator end = str_1.end();

	smatch mat;
	while (regex_search(start, end, mat, pattern))
	{
		string msg(mat[0].first, mat[0].second);
		cout << msg << endl;
		start = mat[0].second;
	}
	return 0;
}

output:

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-1234.3
234.5
3.4567
3.4
5.6
7.88
67.8
8.901
7.65