C ++ code to find if a string is a palindrome

A string is called a palindrome if it reads the same if they read both forward and backward. For eg:

noon civic radar redivider

All the above words spell the same when read forward and backward and are called as palindromes. The following code takes a input string and finds you whether the string is a palindrome or not.

To find if string is a palindrome or not we will start comparing the characters using two pointers, one titrating from the beginning and other from the end of the string. At each step we will increment the pointer that starts from the beginning by one character and decrementing the pointer that starts from the end by one character. At any step if the comparison returns a false that is characters don't match then we can conclude that the sting is not a palindrome.

The following figure shows the same steps for finding if the string CIVIC is a palindrome or not.

The pointer passed to the function , str, will be pointing to the beginning of the string. To get a new pointer pointing to the end of string, we will assign the value of new pointer to

rev=(st+(len-1));

where len is the length of the string. We do len-1 to skip comparing the null character.

We can start comparing the rev and st, and only len/2 comparisons are enough cause once we reach the middle of the string we would have covered the whole string as shown the example above.

Thus the full code in c++ for finding out if a string is palindrome or not is

 
#include&ltiostream&gt 
#include&ltcstring&gt 
using namespace std;

void palindrome(char *st){
int len=strlen(st);
cout << len << "\n";
bool palin=1;
char *rev;
rev=(st+(len-1)); 
for(int i=0;i&ltlen/2;i++){
	if(*(st++) != *(rev--)) {
		palin=0;
		break;
		}
}
if(palin){
	cout <<"The string is a palindrome\n";
}
else {
	cout <<"The string is not a palindrome\n";
}
	
}


int main() {
char str[20];
cout << "Enter the string to be checked \n";
cin >> str; 
if(strlen(str)>10) {
	cout <<"String length excedded\n";
	return -1;
}
palindrome(str);
return 0;
}

Need for copy constructor...

The following examples try to bring out the need for having a copy constructor. With out a copy constructor if we allocate on object to other, i.e. object1=object2, the compiler internally does a bitwise copy,which works fine as long as there are no pointers present in the member variables of the object. A bit wise copy ends up copying the pointer too as it is, making pointers of both the objects point ot the same memory location.

This is depicted in the example below.

#include<iostream> using namespace std; class foo{ public: int *bar1; foo(int val1) { bar1= new int; *bar1=val1; } }; int main() { foo obj1(10); foo obj2=obj1; cout <<"Address of pointer in object 1 " << obj1.bar1<<"\n"; cout <<"Address of pointer in object 2 " << obj2.bar1<<"\n"; *obj1.bar1=0; cout <<"Value of bar 1 in object 1 " << *obj1.bar1<<"\n"; cout <<"Value of bar 1 in object 2 " << *obj2.bar1<<"\n"; }

Output :

Address of pointer in object 1 0x8466008 Address of pointer in object 2 0x8466008 Value of bar 1 in object 1 0 Value of bar 1 in object 2 0

We can clearly see in the output above that the bar1 variable in both the objects is pointing to the same memory location so even when we modify only the variable in one object, the change is reflected in other object too which should no be the case as the two subjects are not related to each other.

The following example illustrates the same with the use of copy constructor.

#include<iostream> using namespace std; class foo{ public: int *bar1; foo(int val1) { bar1= new int; *bar1=val1; } foo(const foo &obj){ bar1= new int; *bar1=*obj.bar1; } }; int main() { foo obj1(10); foo obj2=obj1; cout <<"Address of pointer in object 1 " << obj1.bar1<<"\n"; cout <<"Address of pointer in object 2 " << obj2.bar1<<"\n"; *obj1.bar1=0; cout <<"Value of bar 1 in object 1 " << *obj1.bar1<<"\n"; cout <<"Value of bar 1 in object 2 " << *obj2.bar1<<"\n"; }

Output:

Address of pointer in object 1 0x9286008 Address of pointer in object 2 0x9286018 Value of bar 1 in object 1 0 Value of bar 1 in object 2 10

In the above output it is clearly visible that the two objects have bar1 pointing to different memory location and changing the value in one object does not affect the other.