Question

C++如何截取字符串中从a位置到b位置

Answer 1

方法1：
//...为了说明情况，假如有b>a>0，字符串str[]长度大于b
char *p;
p = &str[a];
*(p+(&str[b]-&str[a] + 1) = '\0'; //截取到指针p方法2：弊源
//...为了说明情况，假如有N>b>a>0，字符串str[]长度大于b
char dest[N];
char *p;
int i;
p=dest;
for(i=a;i<=b;i++,p++) *p = str[i];//截取到dest[N]
*p = '\0';
方法3：
//...为了说明情况，假如有N>b>a>0，字符串str[]长度大于谈大b
char dest[N];
strcpy(dest, &str[a]);//截取到dest[N]
dest[b-a] = '\0';
方法4：
//...为了说明情况，假如有N>b>a>含卜竖0，字符串str[]长度大于b
char dest[N];
strcat(dest, &str[a]);//截取到dest[N]
dest[b-a] = '\0';

Answer 2

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int a,b;
string s="灶枝012345678";
a=2;
b=5;
cout<洞辩郑<s.substr (a,b-a+1)<纳颂<endl;
return 0;
}

Answer 3

a、b都是下标那蔽慎么for(i = 0; a <= b; i++, a++)getstr[i]=str[a];getstr[b+1] = '\0';就可以了宏坦敬信返

Answer 4

include<iostream>
using namespace std;
void strncut(char str_target[],char str_source[],int start_position, int end_position)
{
int new_strlen=end_position-start_position+1;//get the len of new string
int i=start_position;
cout<<"\ntarget str is : ";
for(int j=0;j<new_strlen;j++)
{
str_target[j]=str_source[i-1];
i++;
cout<<str_target[j];
}/岁败/endfor
}/仔雀核/end function strncut
int main()
{
char a[10]="912345678"念掘;
char b[4];
strncut(b,a,2,5);
return 0;
}