Answer / ivr

char *a="india";
for(i=0;a[i]!='\0';i++);
printf("the length is :%d",i);

Answer / vignesh1988i

#include<stdio.h>
#include<conio.h>
int str_len(char *)
void main()
{
char s[30];
int count;
printf("enter the string :");
gets(s);
count=str_len(s);
printf("the length is :%d",count);
getch();
}
int str_len(char *a)
{
int i=0;
while(*a!='\0')
a++;
i++;
}
return i;
}

thank u

Answer / j.j.anand kamlesh

void main()
{
int i;
char ch[100];
printf("ENTER THE STRING :");
gets(ch);
for(i=1;ch[i];i++);
printf("THIS IS THE LENGTH : %d",i);
}

Answer / ankitecian

int StrLen(const char *_input)
{
int _len = 0;
while( *(_input + _len) != NULL)
{
_len++;
}
return _len;
}

Answer / ruchi

#include<conio.h>
#include<stdio.h>
int main()
{
char a[10];
int i=0,c,length=0;
printf("\nEnter the string ");
while((a[i++]=getchar())!='\n');
printf("\nThe lengh of the string is ");
i=i-1;
printf("%d",i);
getch();
}

Answer / ramesh

# include <stdio.h>
# include <conio.h>

void main()
{
int i;
char sring[100];
printf("ENTER THE STRING :");
gets(string);
for(i=1;string[i];i++);
printf("LENGTH IS : %d",i);
}

In cryptography, you could often break the algorithm if you know what was the original (plain) text that was encoded into the current ciphertext. This is called the plain text attack. In this simple problem, we illustrate the plain text attack on a simple substitution cipher encryption, where you know each letter has been substituted with a different letter from the alphabet but you don’t know what that letter is. You are given the cipherText as the input string to the function getwordSets(). You know that a plain text "AMMUNITION" occurs somewhere in this cipher text. Now, you have to find out which sets of characters corresponds to the encrypted form of the "AMMUNITION". You can assume that the encryption follows simple substitution only. [Hint: You could use the pattern in the "AMMUNITION" like MM occurring twice together to identify this]

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