[英]String pattern matching and insertion C++
I am trying to match and insert a patern in string. 我正在尝试匹配并在字符串中插入模式。
Here in the Good peo Good peo
and I am searching for peo
and inserting ple
. 在这里,在
Good peo Good peo
,我正在寻找peo
和插入ple
。
But the Output is coming this: 但是输出结果如下:
Good people Good peo /n
Good peo Good people
I need to have output to be like 我需要有输出才能像
Good people Good people
My code: 我的代码:
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<string.h>
using namespace std;
int length(char s[])
{
int len=0;
int i=0;
while(s[i]!='\0')
{
i++;
len++;
}
return len;
}
void concatenate(char s1[], char s2[])
{
int i=length(s1);
int j=length(s2);
int count=0;
while(count<=j)
{
s1[i]=s2[count];
i++;
count++;
}
}
void substring(char s[], char dest[], int ip, int len)
{
int i=ip;
int count=0;
while(count<len)
{
dest[count]=s[i];
count++;
i++;
}
dest[count]='\0';
}
void ins(char T[], int ip, char P[])
{
char temp1[100];
char temp2[100];
substring(T, temp1, 0, ip);
substring(T, temp2, ip, length(T)-ip);
concatenate(temp1, P);
concatenate(temp1, temp2);
T=temp1;
cout<<T<<endl;
}
void del(char T[], int ip, int L)
{
char temp1[100];
char temp2[100];
substring(T, temp1, 0, ip);
substring(T, temp2, ip+L, length(T)-ip-L);
concatenate(temp1, temp2);
T=temp1;
cout<<T<<endl;
}
//where T is the original string and P is the pattern to be deleted whereever it appears in the original string.
void delpat(char T[], char P[])
{
char temp[100];
for(int i=0; i<=length(T); i++)
{
substring(T, temp, i, length(P));
if(strcmp(temp, P)==0)
del(T, i, length(P));
}
}
//where T is the original string, Q is the pattern to be inserted and P is the pattern after which it is inserted.
void inspat(char T[], char P[], char S[])
{
char temp[100];
for(int i=0; i<=length(T); i++)
{
substring(T, temp, i, length(P));
if(strcmp(temp, P)==0)
ins(T, i+length(P), S);
}
}
int main()
{ char a[100];
char T[]="Good peo Good peo";
char P[]="peo";
char S[]="ple";
inspat(T, P, S);
gets(a);
}
1) The assignment in the function ins()
doesn't change the value at the caller: 1)函数
ins()
中的赋值不会更改调用方的值:
T=temp1;
cout<<T<<endl;
You would need to use strcpy()
to the copy the temp1
char array: 您将需要使用
strcpy()
复制temp1
char数组:
strcpy(T, temp1);
cout<<T<<endl;
2) Since you want to print the after inserting all occurrences, the above cout
needs to go and you can print T
either in main()
or at the of inspat()
(outside the for
loop): 2)既然你想将所有出现后打印的,上面
cout
需要去,你可以打印T
无论是在main()
或在的inspat()
外面for
环):
cout<<T<<endl;
3) Since the insertion happens in the original array, you need to ensure the array is big enough. 3)由于插入发生在原始数组中,因此需要确保数组足够大。 Do something like in
main()
: 在
main()
做类似的事情:
char T[256]="Good peo Good peo"; // 256 is some arbitrary size
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