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- Write a function htoi(s), which converts a string of hexa-decimal digits (including an optional 0x or 0X) into its equivalent integer value.
(The allowed digits are 0 through 9, a through f, and A through F)
Input: – oxAA
Output: – 170
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define EXIT_SUCCESS 0
long htoi(char *);
void strrev(char *);
int main()
{
long number=0;
char inputstr[30];
printf(“Enter a hexadecimal string “);
scanf(“%s”,inputstr);
printf(“string is %s\n”,inputstr);
number=htoi(inputstr);
printf(“Integer is %ld\n”,number);
return EXIT_SUCCESS;
}
long htoi(char * inputstr)//converts the hexrepresentation to integer
{
int length=0, base=1,i=0,k;
long hnumber=0;
char str[30];
strrev(inputstr);
length=strlen(inputstr);
//to include optional ox
if((inputstr[length-1]==’O’ || inputstr[length-1]==’o’)&&(inputstr[length-2]==’X’ || inputstr[length-2]==’x’))
{
length=length-2;
}
while(i<=length-1)
{if(inputstr[i]==’a’||inputstr[i]==’A’)
hnumber=hnumber+base*10;
else if(inputstr[i]==’b’||inputstr[i]==’B’)
hnumber=hnumber+base*11;
else if(inputstr[i]==’c’||inputstr[i]==’C’)
hnumber=hnumber+base*12;
else if(inputstr[i]==’d’||inputstr[i]==’D’)
hnumber=hnumber+base*13;
else if(inputstr[i]==’e’||inputstr[i]==’E’)
hnumber=hnumber+base*14;
else if(inputstr[i]==’f’||inputstr[i]==’F’)
hnumber=hnumber+base*15;
else
{
str[0]=inputstr[i];
str[1]=’\0′;
hnumber=hnumber+(base*atoi(str));
}
base=base*16;
i++;
}
printf(“%ld\n”,hnumber);
return hnumber;
}
void strrev(char *string)
{
int length, c;
char *begin, *end, temp;
length = strlen(string);
begin = string;
end = string;
for ( c = 0 ; c < ( length – 1 ) ; c++ )
end++;
for ( c = 0 ; c < length/2 ; c++ )
{temp = *end;
*end = *begin;
*begin = temp;
begin++;
end–;
}
}
Output:
[[email protected] ~]$ gcc hextoint.c -o htoi
[[email protected] ~]$ ./htoi
Enter a hexadecimal string oxAA
Integer is 170
- Write a function squeeze(s1,s2) that deletes each character in s1 that matches any character in the string s2.
Input: –
S1 = “character”
S2 = “at”
Output: –
S1=”chrcer”
S2
:q=”at”
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void squeeze(char*,char*);
int main()
{
char s1[15],s2[15]=””;
printf(“Enter first string:”);
scanf(“%s”,s1);
printf(“Enter the second string:”);
scanf(“%s”,s2);
squeeze(s1,s2);
printf(“Final result after squeezing is %s\n”,s1);
return 0;
}
void squeeze(char *s1,char *s2)
{
int i,j,k=0;
for(i=0;i<strlen(s1);i++)
for(j=0;j<strlen(s2);j++)
{
if(s1[i]==s2[j])
{
for(k=i;k<(strlen(s1))-1;k++)
{
s1[k]=s1[k+1];
}
s1[k]=’\0′;
}
}
}
Output:
[[email protected] ~]$ gcc squeeze.c
[[email protected] ~]$ ./a.out
Enter first string:character
Enter the second string:at
Final result after squeezing is chrcer
- Write a function any(s1,s2) which returns
the first location in the string s1 where any character from the string s2 occurs, or
-1 if s1 contains no characters from s2.
i) Input: – S1 = “character” S2 = “abc”
Output: – Function should return “3”
ii)Input: – S1 = “character” S2 = “xyz”
Output: – Function should return “-1”
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int any(char *,char *);
int main()
{
char s1[15],s2[15];
int pos;
printf(“Enter first string:”);
scanf(“%s”,s1);
printf(“Enter the second string:”);
scanf(“%s”,s2);
pos=any(s1,s2);
printf(“position at which it occurs is %d\n”,pos);
return 0;
}
int any(char *s1,char *s2)
{
int i,j=0;
int pos=-1;
for(i=0;i<(strlen(s2));i++)
for(j=0;j<(strlen(s1));j++)
if(s1[i]==s2[j])
{
pos=i+1;
break; }
return pos;
}
Output:
[[email protected] ~]$ gcc any.c -o any
[[email protected] ~]$ ./any
Enter first string:character
Enter the second string:abc
pos at which it occurs is 3
[[email protected] ~]$ ./any
Enter first string:character
Enter the second string:xyz
pos at which it occurs is -1
- Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted(i.e., 1 changed to 0 and vice versa), leaving the others unchanged.
Input: – Invert(45,3,2)
Output:- 53
- Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n bit positions.
Input: – rightrot(48,3)
Output: – x=9
- In a two’s complement number system, x &= (x-1) deletes the rightmost 1-bit in x. Explain.
Input: – X= 45
Output: – X=44
- .
(Use conditional expression instead of if-else)
Input: – S1=”APPLE”
Output:= S1=”apple”
#include<stdio.h>
#include<string.h>
int lowerfunc(int);
int main()
{
char bound[30];
char *p;
printf(“enter the string “);
scanf(“%s”,bound);
p=bound;
int result=0;
while(‘\0’!=*p)/*while the string is not null*/
{
result=lowerfunc(*p);
printf(“%c”,result);
++p;
}
return 0;
}
int lowerfunc(int val)
{
{
if(val>=65 && val<=90)/*ASCII value of A-65,Z-90,if the input is in between these ascii values*/
return val+97-65;/*difference will be the same value to get lower case letter*/
else
return val;
}
Output:
[[email protected] ~]$ gcc tolow_func.c -o tolow
[[email protected] ~]$ ./tolow
enter the string APPLE
apple
- Write a function escape(s,t) that converts characters like newline and tab into visible escape sequences like \n and \t as it copies the string t to s. Use a switch.
Input : – S1=” Encapsulation
Inheritance polymorphism”
- Output: – S1=”Encapsulation\nInheritance\tpolymorphism”;
#include<stdio.h>
#define exit 0
void escape(char *s,char *t);
int main()
{
char s1[100]=”Encapsulation \n Inheritance \t \t Polymorphism”;
char s2[100];
printf(“entered string is:%s :\n”,s1);
escape(s2,s1);
printf(“Escaped string is:\n %s \n”,s2);
return exit;
}
void escape(char *s,char *t)
{
int i=0,j=0;
while(t[i])
{
switch(t[i])
{
case ‘\n’:
s[j++]=’\\’;
s[j]=’n’;
break;
case ‘\t’:
s[j++]=’\\’;
s[j]=’t’;
break;
default:
s[j]=t[i];
break;
}
++i;
++j;
}
s[j]=t[i];
}
Output:
[[email protected] ~]$ gcc escape.c
[[email protected] ~]$ ./a.out
entered string is:Encapsulation
Inheritance Polymorphism :
Escaped string is:
Encapsulation \n Inheritance \t \t Polymorphism
- Write a function which converts escape sequences into the real characters.
- Input: – S1=”Encapsulation\nInheritance\tpolymorphism”;
- output : – S1=” Encapsulation
- Inheritance polymorphism”
#include<stdio.h>
#define exit 0
void escapetoreal(char *s,char *t);
int main()
{
char s1[100]=”\nEncapsulation\nInheritance\t\tPolymorphism”;
char s2[100]=””;
printf(“Entered string is %s\n:”,s1);
escapetoreal(s1,s2);
printf(“after converting the escape sequences to real:%s”,s1);
return exit;
}
void escapetoreal(char *s,char *t)
{
int i=0,j=0;
while(t[i])
{
switch(t[i])
{
case ‘\\’:
switch(t[++i])
{
case ‘n’:
s[j]=’\n’;
break;
case ‘t’:
s[j]=’\t’;
break;
/*
case ‘\”‘:
s[j]=’\”‘;
break; */
default:
s[j++]=’\\’;
s[j]=t[i];
}
break;
default:
s[j]=t[i];
}
i++;
j++;
}
}
Output:
[[email protected] ~]$ gcc Escape_to_real.c -o escape
[[email protected] ~]$ ./escape
Entered string is
Encapsulation
Inheritance Polymorphism
:after converting the escape sequences to real:
Encapsulation
Inheritance Polymorphism
