#include “stdio.h”
#include “conio.h”
#include “string.h”
void main()
{
char input[100],output[100],c;
int i=0,j=0;
clrscr();
printf(“Enter the string\n”);
gets(input);
for(i=0;i<strlen(input);i++)
{
if(input[i]==' '&&input[i+1]!=' ')
output[j++]=input[i];
else if(input[i]!=' ')
output[j++]=input[i];
}
output[j]='';
printf("\noutput string is\n");
puts(output);
getch();
}

Output:

Enter the string
How          are                 you         ?
Output string is:
How are you ?

1
2
3
4
5
6
7
8
9
10
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12
13
14
15
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18
19
20
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26
27
28
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31
32
33
34

#include "stdio.h"
#include "conio.h"
#include "string.h"
void main()
{
 int matchany(char str1[],char str2[]);//function declaration
 char str1[50],str2[50];
 int s;
 clrscr();
 printf("Enter string 1 and string 2\n");
 scanf("%s%s",&str1,&str2);
 s=matchany(str1,str2);
 if(s==-1)
 printf("\nNo match found");
 else
 printf("\nThe location where the first match occured is %d",s);
 getch();
}
 
int matchany(char str1[],char str2[])
{
 int i,j;
 for(i=0;i<strlen(str2);i++)
 {
  for(j=0;j<strlen(str1);j++)
  {
   if(str2[i]==str1[j])
   {
    return j+1;
   }
  }
 }
 return -1;
}

output 1:

Enter string 1 and string 2
hello
world
The location where the first match occured is 5

Output 2:

Enter string 1 and string 2
ice
smooth
No match found

Logic: The approach here is to reverse the string using the pointers. Reversing the string includes the reversing the each and every words in it. After accepting a string from user, it calls a function “strev” with two string pointer arguments, the source and destination. It has an iterative loop, which traces from the EOL through the beginning. Each time it copies the current letter to the destination. Finally it displays the resultant string.

The earlier program implements the same by direct method, i.e. without pointers.

#include<stdio.h>
#include<conio.h>
void strev(char *str1, char *str2);
void main()
{
        char *str1, *str2;
        clrscr();
        printf("\n\n\t ENTER A STRING...: ");
        gets(str1);
        strev(str1,str2);
        printf("\n\t THE REVERSED STRING IS...: ");
        puts(str2);
        getch();
}

void strev(char *str1, char *str2)
{
        int i = 0, len = 0, r = 0;
        while(*(str1+len)!='\0')
                len++;
        for(i=len-1; i>=0; i--)
        {
                *(str2+r) = *(str1+i);
                r++;
        }
        *(str2+r) = '\0';
}

Logic: Reversing the string includes the reversing the each and every words in it. After accepting a string from user, it calls a function “strev” with two string arguments, the source and destination. It has an iterative loop, which traces from the EOL through the begining. Each time it copies the current letter to the destination. Finally it displays the resultant string.

The same program is implemented using the string pointers.

#include<stdio.h>
#include<conio.h>
void strev(char str1[50], char str2[50]);
void main()
{
char str1[50], str2[50];
clrscr();
printf(“\n\n\t ENTER A STRING…: “);
gets(str1);
strev(str1,str2);
printf(“\n\t THE REVERSED STRING IS…: “);
puts(str2);
getch();
}
void strev(char str1[50], char str2[50])
{
int i = 0, len = 0, r = 0;
while(str1[len]!=’\0′)
len++;
for(i=len-1; i>=0; i–)
{
str2[r] = str1[i];
r++;
}
str2[r] = ‘\0′;
}
Download exe and source code here.

Logic : The program asks the user to input the string to copy and stores using the pointer str1. The inner function “stcpy” takes 2 string pointers as arguments. By keeping the length as the reference, it traces till EOF , for each i of this iteration it copies the i-th  letter to the destination string, i.e str2. After the EOL, it puts a NULL to the second line. That gives the duplicated string of the entered.

This function copies the string using the pointer, where earlier program copies the string directly.

#include<stdio.h>
#include<conio.h>
void stcpy(char *str1, char *str2);
void main()
{
char *str1, *str2;
clrscr();
printf(“\n\n\t ENTER A STRING…: “);
gets(str1);
stcpy(str1,str2);
printf(“\n\t THE COPIED STRING IS…: “);
puts(str2);
getch();
}
void stcpy(char *str1, char *str2)
{
int i, len = 0;
while(*(str1+len)!=’\0′)
len++;
for(i=0;i<len;i++)
*(str2+i) = *(str1+i);
*(str2+i) = ‘\0′;
}
Download exe and source code here.

Logic : The program asks the user to input the string to copy and stores this as str1. The inner function “stcpy” takes 2 strings as arguments. By keeping the length as the reference, it traces till EOL , for each i of this iteration, it copies the i-th  letter to the destination string, i.e str2.

After the EOL, it puts a NULL to the second line. That gives the duplicated string of the entered.

#include<stdio.h>
#include<conio.h>
void stcpy(char str1[50], char str2[50]);
void main()
{
char str1[50], str2[50];
clrscr();
printf(“\n\n\t ENTER A STRING…: “);
gets(str1);
stcpy(str1,str2);
printf(“\n\t THE COPIED STRING IS…: “);
puts(str2);
getch();
}
void stcpy(char str1[50], char str2[50])
{
int i, len = 0;
while(str1[len]!=’\0′)
len++;
for(i=0;i<len;i++)
str2[i] = str1[i];
str2[i] = ‘\0′;
}
Download exe and source code here.

Logic :  The given string is referred by two names, where one is traced by the SOL (Start Of Line) and the other is traced from EOL (End Of Line). In every iteration, the condition is checked, if both the characters are the same. If so, sets a flag, “pal” as true (i.e. 1), and hence prints out the result.

The problem can also solved in another method, where we check  by reversing the string and then comparing.

#include<stdio.h>
#include<conio.h>
int stpal(char str[50]);
void main()
{
char str[50];
int pal;
clrscr();
printf(“\n\n\t ENTER A STRING…: “);
gets(str);
pal = stpal(str);
if(pal)
printf(“\n\t THE ENTERED STRING IS A PALINDROME”);
else
printf(“\n\t THE ENTERED STRING IS NOT A PALINDROME”);
getch();
}
int stpal(char str[50])
{
int i = 0, len = 0, pal = 1;
while(str[len]!=’\0′)
len++;
len–;
for(i=0; i<len/2; i++)
{
if(str[i] == str[len-i])
pal = 1;
else
{
pal = 0;
break;
}
}
return pal;
}
Download exe and source code here.

Logic :  The given string is reversed using a method, which puts the reversed string in the resultant string from the EOL (End Of Line). After reversing, it compares if strings are one and the same. If so, sets a flag, “pal” as true (i.e. 1), and hence prints out the result.

The problem can also solved in a bit sophisticated method, where we check  without reversing the string, but directly.

#include<stdio.h>
#include<conio.h>
int stpal(char str1[50], char str2[50]);
void main()
{
char str[50], rev[50];
int pal;
clrscr();
printf(“\n\n\t ENTER A STRING…: “);
gets(str);
pal = stpal(str, rev);
printf(“\n\t THE REVERSED STRING IS…: “);
puts(rev);
if(pal)
printf(“\n\t THE ENTERED STRING IS A PALINDROME”);
else
printf(“\n\t THE ENTERED STRING IS NOT A PALINDROME”);
getch();
}
int stpal(char str1[50], char str2[50])
{
int i = 0, len = 0, r = 0, pal = 1;
while(str1[len]!=’\0′)
len++;
for(i=len-1; i>=0; i–)
{
str2[r] = str1[i];
r++;
}
str2[r] = ‘\0′;
for(i=0; i<=len-1; i++)
{
if(str1[i] == str2[i])
pal = 1;
else
{
pal = 0;
break;
}
}
return pal;
}

Download exe and source code here.