void Rstring( char *str,int len)
{
for(int i = 0; i < (len/2); i++)
{
str[i] ^= str[len-i-1];
str[len-i-1] ^= str[i];
str[i] ^= str[len-i-1];
}
}
int main( void )
{
char str[] = "my string";
printf("Actual string is [%s]\n", str);
Rstring(str,strlen(str));
printf("Reversed string is [%s]\n", str);
}
I dont call this swaping, coz it's not, recursive creates
new incarnations of the reverse func, EXTRA MEMORY BIG
TIME!!!
#include <iostream>
#include <conio>
void reverse(char a[], int s, int sc );
void reverse(char a[], int s, int sc ){
if ((sc-s)<(s-1))
{
a[sc-s]^=a[s-1];
a[s-1]^=a[sc-s];
a[sc-s]^=a[s-1];
reverse (a, s-1, sc) ;
}
}
void main (){
char a[]="ABCDEFG";
reverse(a, 7, 7);
cout<<a;
getch(); //i just use it to freeze the screen
}
#include <iostream>
#include <string>
using namespace std;
char * reverse (char *); //function prototype
int length(char *); //function prptotype
int main()
{
int i;
char *str = new char[6], *rev = new char[6];
cin >> str;
strcpy(rev,reverse(str));
cout <<"Original "<< str << " reverse " << rev << endl;
free(str);
free(rev);
return 0;
}
int length(char *s)
{
int i;
for (i=0; *(s+i)!='\0' ; ++i);
return i;
}
char *reverse(char *s)
{
char *t;
int i, n;
n=length(s);
t = new char[n];
for (i=0; i<n; ++i)
{
*(t+i)=*(s+n-1-i);
}
*(t+i)='\0';
return t;
}
/*based off of answer 8 i took this an intialized the
pointers so that it would run, and switched it over to the
C++ standard output commands. His algorithm was correct, he
just forgot to setup the memory*/
#include <iostream>
#include <string>
using namespace std;
char * reverse (char *); //function prototype
int length(char *); //function prptotype
int main()
{
int i;
char *str = new char[6], *rev = new char[6];
cin >> str;
strcpy(rev,reverse(str));
cout <<"Original "<< str << " reverse " << rev << endl;
free(str);
free(rev);
return 0;
}
int length(char *s)
{
int i;
for (i=0; *(s+i)!='\0' ; ++i);
return i;
}
char *reverse(char *s)
{
char *t;
int i, n;
n=length(s);
t = new char[n]; //opps have to add 1 here or there
wont be room for a null!
for (i=0; i<n; ++i)
{
*(t+i)=*(s+n-1-i);
}
*(t+i)='\0';
return t;
}
//can only handle words 5 letters or less.
/*based off of answer 8 i took this an intialized the
pointers so that it would run, and switched it over to the
C++ standard output commands. His algorithm was correct, he
main()
{
char str[10];
cin>>str;
int len=strlen(str);
reverse(len);
cout<<"Reversed string is: "<<str;
}
void reverse(int len)
{
static int i=0;
str[i]=str[len-1]; //put the char in last pos to first pos
for(j=len-1;j>i;j--)
str[j]=str[j-1]; //shift to right
i++;
if(i==len)
return;
reverse(len);
}
what will be the result of the following program ?
char *gxxx()
{
static char xxx[1024];
return xxx;
}
main()
{
char *g="string";
strcpy(gxxx(),g);
g = gxxx();
strcpy(g,"oldstring");
printf("The string is :
%s",gxxx());
}
a) The string is : string
b) The string is :Oldstring
c) Run time error/Core dump
d) Syntax error during compilation
e) None of these
let's take a code
struct FAQ
{
int a;
char b;
float c;
double d;
int a[10];
}*temp;
now explain me how the memory will be allocated for the
structure FAQ and what address will be in the structure
pointer (temp)....................
#include<stdio.h>
int fun();
int i;
int main()
{
while(i)
{
fun();
main();
}
printf("hello \n");
return 0;
}
int fun()
{
printf("hi");
}
answer is hello.how??wat is tat while(i) mean?