Checking a string if it is palindrome or not using pointer - c

What does the bold part stand for in this code? I've managed to fulfill most of the template (given by my teacher as assignment), can't find the reason behind that bold part.
I could use the strlen part before it appeared, I know one can still write the code without strlen at all.
#include <stdio.h>
int strlen(char *str)
{
int c = 0;
while (str[c] != '\0')
{
c++;
}
return c;
}
int isPalindrome(char *str)
{
char *ptr1 = str;
char *ptr2 = str + strlen(str) - 1;
while (ptr2 > ptr1)
{
if (tolower(*ptr1) != tolower(*ptr2))
{
return (0);
}
ptr1++;
ptr2--;
}
return (1);
}
/**
This function will
-return 1 if the given string is Palindrome,
-return 0 if the given string is not Palindrome.
*/
**int n = strlen(str);
// Write your code here**
}
int main()
{
char *str = "ABCCBA ABCCBA"; //your input
if (isPalindrome(str))
{
printf("%s is a palindrome", str);
}
else
{
printf("%s is not a palindrome", str);
}
}

The teacher hinted that using the length of the zero-terminated input string will be needed. They stored that useful value inside a variable.
If you are required to use the code provided by the teacher, then move all your code AFTER that line and within your code replace the strlen(str) with n.
int isPalindrome(char *str)
{
/**
This function will
-return 1 if the given string is Palindrome,
-return 0 if the given string is not Palindrome.
*/
int n = strlen(str);
// Write your code here
char *ptr1 = str;
char *ptr2 = str + n - 1;
while (ptr2 > ptr1)
{
if (tolower(*ptr1) != tolower(*ptr2))
{
return (0);
}
ptr1++;
ptr2--;
}
return (1);
}

You can implement isPalindrome without any knowledge about pointers.
I prefer to use indices instead of pointers, look at the next example
#include <stdio.h>
#include <assert.h>
int strLen(const char *str)
{
int n = 0;
while(str[n]) { ++n; }
return n;
}
int isPalindrome(const char *str) {
int i = 0;
int j = strLen(str) - 1;
while (i < j) {
if (str[i++] != str[j--]) { return 0; }
}
return 1;
}
int main() {
assert(isPalindrome(""));
assert(isPalindrome("a"));
assert(isPalindrome("aa"));
assert(isPalindrome("aba"));
assert(!isPalindrome("ab"));
assert(!isPalindrome("aab"));
}

I think that the function template starts from the declaration of the variable n that gets the size of the passed string.
int isPalindrome(char *str)
{
/**
This function will
-return 1 if the given string is Palindrome,
-return 0 if the given string is not Palindrome.
*/
int n = strlen(str);
// Write your code here*
}
But you ignored this declaration and used the expression strlen( str ) in this declaration
char *ptr2 = str + strlen(str) - 1;
instead of writing for example
int n = strlen(str);
//...
char *ptr2 = str + n - 1;
In any case the assignment looks not very well. For example this declaration
char *ptr2 = str + strlen(str) - 1;
can invoke undefined behavior when the passed string is empty.
The functions' parameters should be declared with the qualifier const because passed strings as arguments are not changed in the functions.
The argument of the standard function tolower must be converted to the type unsigned char.
The functions could look the following way
size_t strlen( const char *str )
{
size_t n = 0;
while ( str[n] ) ++n;
return n;
}
int isPalindrome( const char *str )
{
const char *ptr1 = str;
const char *ptr2 = str + strlen( str );
if ( ptr1 != ptr2 )
{
while ( ptr1 < --ptr2 &&
tolower( ( unsigned char )*ptr1 ) == tolower( ( unsigned char )*ptr2 ) )
{
++ptr1;
}
}
return !( ptr1 < ptr2 );
}
Also if the function strlen also must use pointers then it can look the following way
size_t strlen( const char *str )
{
const char *p = str;
while ( *p ) ++p;
return p - str;
}
Here is a demonstration how your functions should look.
#include <stdio.h>
#include <ctype.h>
size_t strlen( const char *str )
{
const char *p = str;
while ( *p ) ++p;
return p - str;
}
int isPalindrome( const char *str )
{
const char *ptr1 = str;
const char *ptr2 = str + strlen( str );
if ( ptr1 != ptr2 )
{
while ( ptr1 < --ptr2 &&
tolower( ( unsigned char )*ptr1 ) == tolower( ( unsigned char )*ptr2 ) )
{
++ptr1;
}
}
return !( ptr1 < ptr2 );
}
int main(void)
{
char *str = "ABCCBA ABCCBA"; //your input
if ( isPalindrome( str ) )
{
printf("\"%s\" is a palindrome\n", str);
}
else
{
printf("\"%s\" is not a palindrome\n", str);
}
return 0;
}
The program output is
"ABCCBA ABCCBA" is a palindrome

Related

applying cipher code given as argv to userinput

I am trying to write a function that takes in variables cipher, userinput, alphabet and outputs the replace letters. for example at execution it takes in an argument argv ZYXWVUTSRQPONMLKJIHGFEDCBA. requests a user to input anything then switches only the letters for example input aBc should output zYx
// function that takes in cipher(c), userinput(u) and alphabet(a) and outputs cipher text
void cipher_text(string c, string u, string a)
{
string result = u;
for (int i = 0; i < strlen(u); i++)
{
for (int k = 0; k < strlen(a); k++)
{
if (tolower(u[i]) == a[k])
{
if(islower(u[i]))
{
result[i] = tolower(c[k]);
printf("%s %s\n",result,c);
}
else
{
result[i] = toupper(c[k]);
}
}
}
}
// printf("ciphertext: %s\n", result);
}
with cipher as YUKFRNLBAVMWZTEOGXHCIPJSQD
and userinput as abcd
I was expecting yukf but got qidc
The problem of the function is this inner for loop
for (int k = 0; k < strlen(a); k++)
{
if (tolower(u[i]) == a[k])
{
if(islower(u[i]))
{
result[i] = tolower(c[k]);
printf("%s %s\n",result,c);
}
else
{
result[i] = toupper(c[k]);
}
}
}
You need to break it as soon as a letter is found in the string pointed to by the pointer a. Otherwise there can be a chain of changes.
Also using the function strlen is inefficient and redundant.
Instead of the for loop you could use standard function strchr.
The function can be declared and defined the following way
string cipher_text( string c, string u, string a )
{
for ( string p = u; *p != '\0'; p++ )
{
string pos = strchr( a, tolower( ( unsigned char )*p ) );
if ( pos != NULL )
{
size_t i = pos - a;
*p = islower( ( unsigned char )*p )
? tolower( c[i] )
: c[i];
}
}
return u;
}
Pay attention to that instead of the typedef name string it is better to use its underlaying type char *.
In this case the function can look the following way as shown in the demonstration program below.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
char * cipher_text( const char *c, char *u, const char *a )
{
for (char *p = u; *p != '\0'; p++)
{
const char *pos = strchr( a, tolower( ( unsigned char )*p ) );
if (pos != NULL)
{
size_t i = pos - a;
*p = islower( ( unsigned char )*p )
? tolower( c[i] )
: c[i];
}
}
return u;
}
int main( void )
{
const char *c = "XYZ";
const char *a = "abc";
char u[] = "c_B_a";
puts( cipher_text( c, u, a ) );
}
The program output is
z_Y_x

Doubling all occurrences of a specific character in a string

I want to write a C program, that returns a new string allocated on the heap. This string is obtained by doubling all occurrences of “c” in a string (“abcdc” becomes “abccdcc” after doubling “c”).
This is my code and I don't really see where the problem is to fix it!
size_t taille = stringLength(str);
size_t k=0;
size_t q=0;
while (*str!='\0')
{
if (*str == c)
{
k=k+1;
}
++str;
}
char *nouvelle=malloc(taille+1+k);
int i,j= 0;
while(*str !='\0')
{
if (str[i] != c)
{
j=i;
nouvelle[j]=str[i];
}
else
{
j=i;
++q;
nouvelle[j]=str[i];
j=i+q;
nouvelle[j++]=str[i];
}
++i;
}
nouvelle[taille+1+k]='\0';
return nouvelle;
}
There are two problems with your code.
The first one is that after this while loop
while (*str!='\0')
{
if (*str == c)
{
k=k+1;
}
++str;
}
the pointer str points to the end of the string that is to the terminating zero character '\0'.
The second one is that you are using the uninitialized variable i
int i,j= 0;
while(*str !='\0')
{
if (str[i] != c)
{
j=i;
//..
This declaration
int i,j= 0;
is not the same as
int i = 0,j= 0;
That is only the variable j is initialized by 0.
And the statement
j = i;
does not make sense.
Also it is unclear whether c denotes a variable or the character 'c'. If you mean the character 'c' then you need to write at least like
if (*str == 'c')
You could define the function for example the following way as shown in the demonstration program below.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char * duplicate_char( const char *s, char c )
{
size_t n = 0;
for (const char *p = s; *p; ++p)
{
if (*p == c) ++n;
}
char *nouvelle = malloc( strlen( s ) + n + 1 );
if (nouvelle)
{
if (n == 0)
{
strcpy( nouvelle, s );
}
else
{
char *p = nouvelle;
while (*s)
{
*p++ = *s++;
if (p[-1] == c) *p++ = c;
}
*p = '\0';
}
}
return nouvelle;
}
int main( void )
{
char *nouvelle = duplicate_char( "abcdc", 'c' );
if (nouvelle != NULL) puts( nouvelle );
free( nouvelle );
}
The program output is
abccdcc
If you want to use your own function stringLength instead of the standard C function strlen then it can look like
size_t stringLength( const char *s )
{
const char *p = s;
while ( *p ) ++p;
return p - s;
}

pointer string not returning expected output

#include <stdio.h>
void revstr(char str[])
{
char temp;
int size = 0;
while(*str != '\0'){
size++;
str++;
}
for (int i = 0; i < size/2; i++)
{
temp = str[i];
str[i] = str[size-1-i];
str[size-1-i] = temp;
}
for(int i = 0; i < size; i++)
{
printf("%c\n", *(str+i));
}
}
int main()
{
char str[20];
printf("enter a string: \n");
scanf("%s", &str);
revstr(str);
return 0;
}
why is my rev string not printing the reverse of the string it is printing out some garbage value.
can you point out why?
After this while loop
while(*str != '\0'){
size++;
str++;
}
the pointer str does not point to the beginning of the string.
Instead you could write for example
while( str[size] != '\0'){
size++;
}
Nevertheless such a function should do only one thing: to reverse a string. It is the caller of the function that decides whether to output the reversed string.
So the function can look like
char * revstr( char s[] )
{
size_t n = 0;
while ( s[n] ) ++n;
for ( size_t i = 0; i < n / 2; i++ )
{
char c = s[i];
s[i] = s[n - i - 1];
s[n - i - 1] = c;
}
return s;
}
and in main you could write
puts( revstr( str ) );
Here is a demonstrative program.
#include <stdio.h>
char * revstr( char s[] )
{
size_t n = 0;
while ( s[n] ) ++n;
for ( size_t i = 0; i < n / 2; i++ )
{
char c = s[i];
s[i] = s[n - i - 1];
s[n - i - 1] = c;
}
return s;
}
int main(void)
{
char s[] = "Hello";
puts( s );
puts( revstr( s ) );
return 0;
}
The program output is
Hello
olleH
Of course instead of the while loop you could use the standard string function strlen.
size_t n = strlen( s );
Pay attention to that in the call of scanf
scanf("%s", &str);
the second argument shall be the expression str
scanf("%s", str);
I would implement it another way:
char *reverse(char *str)
{
char *wrk = str, *end = str;
if(str && *str)
{
while(*(end + 1)) end++;
while(end > wrk)
{
char tmp = *end;
*end-- = *wrk;
*wrk++ = tmp;
}
}
return str;
}
It is good to check if the parameter is correct.
I would return the value. It allows you to use the reversed string in expressions and as a function parameter. Example below:
int main(void)
{
char s[] = "Hello world";
printf("%s\n", reverse(s));
}
https://godbolt.org/z/fbo4nsn38
In your code you advance the pointer str to calculate its length but you forgot to reset it to the start after that. Also, your call to scanf should pass str, not its address. Moreover scanf will write beyond the end of str if the user enters a string longer than 19 characters. Try this instead:
#include <stdio.h>
#include <string.h>
void revstr(char str[])
{
char temp;
int size = strlen(str);
for (int i = 0; i < size / 2; i++) {
temp = str[i];
str[i] = str[size - 1 - i];
str[size - 1 - i] = temp;
}
}
int main()
{
char str[20];
printf("enter a string: \n");
fgets(str, sizeof(str), stdin);
str[strcspn(str, "\n")] = '\0';
revstr(str);
printf("%s\n", str);
return 0;
}

How to write function that return dynamic allocated string?

I have assignment that to use function char *slova(const char *s) which returns dynamically allocated string that is only made of small and big letters. It's forbidden to use string.h library.
char *slova(const char *s)
{
char *new;
int br = 0;
new = (char *)malloc(sizeof(s));
for (int i = 0; i != '\0'; i++)
if(s[i] >= 'A' && s[i] <= 'z')
{
new[br] = s[i];
br++;
}
return new;
}
I know there are some other characters other than small and big letters between A and z in ASCII code, so don't worry about that. For some reason this code doesn't work and I don't know why.
sizeof(s) will return not the buffer size but the size of the pointer s.
i != '\0' is wrong. This means i != 0 and prevent it from entering the loop because initial value of i is 0.
You forgot to terminate the resulting string by adding a terminating null-character.
Casting the result of malloc() in C is discouraged.
Fixed code:
char *slova(const char *s){
char *new;
int br = 0;
// calculate the length
for (int i = 0; s[i] != '\0'; i++)
if(s[i] >= 'A' && s[i] <= 'z'){
br++;
}
// allocate buffer
new = malloc(br + 1);
if (new == NULL) return NULL;
// actually create the string
br = 0;
for (int i = 0; s[i] != '\0'; i++)
if(s[i] >= 'A' && s[i] <= 'z'){
new[br] = s[i];
br++;
}
new[br] = '\0';
return new;
}
Look carefully at your function declaration
char *slova(const char *s){
^^^^^^^^^^^^^
Its parameter has the pointer type const char *. Thus in this statement
new = (char *)malloc(sizeof(s));
the expression sizeof(s) yields the size of a pointer that usually is equal to 8 or 4 bytes depending on the used system. That is this expression does not provide the length of the passed string.
Also the body of this loop
for (int i = 0; i != '\0'; i++)
never gets the control because the condition i != '\0' at once evaluates to false because the variable i was initialized by zero.
The function can look the following way as it is shown in the demonstrative program below. It does not use functions from the header <string.h>.
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
char * slova( const char *s )
{
size_t n = 0;
for ( const char *t = s; *t != '\0'; ++t )
{
if ( isalpha( ( unsigned char )*t ) ) ++n;
}
char * result = malloc( ( n + 1 ) *sizeof( char ) );
if ( result != NULL )
{
char *p = result;
for ( ; *s; ++s)
{
if ( isalpha( ( unsigned char )*s ) )
{
*p++ = *s;
}
}
*p = '\0';
}
return result;
}
int main(void)
{
const char *s = "H#e#l#l#o W#o#r#l#d";
char *p = slova( s );
if ( p ) puts( p );
free( p );
return 0;
}
The program output is
HelloWorld
If you are not allowed also to use functions from the header <ctype.h> then the function can look the following way as it is shown in the demonstrative program below.
#include <stdio.h>
#include <stdlib.h>
char * slova( const char *s )
{
const char *upper_case = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const char *lower_case = "abcdefghijklmnopqrstuvwxyz";
size_t n = 0;
for ( const char *t = s; *t != '\0'; ++t )
{
const char *letter = lower_case;
if ( *t < lower_case[0] )
{
letter = upper_case;
}
while ( *letter && *letter < *t ) ++letter;
if ( *letter == *t ) ++n;
}
char * result = malloc( ( n + 1 ) *sizeof( char ) );
if ( result != NULL )
{
char *p = result;
for ( ; *s; ++s)
{
const char *letter = lower_case;
if ( *s < lower_case[0] )
{
letter = upper_case;
}
while ( *letter && *letter < *s ) ++letter;
if ( *letter == *s )
{
*p++ = *s;
}
}
*p = '\0';
}
return result;
}
int main(void)
{
const char *s = "H#e#l#l#o W#o#r#l#d";
char *p = slova( s );
if ( p ) puts( p );
free( p );
return 0;
}
Again the program output is
HelloWorld

Is there any way to return a string starting at a certain index without using the library functions

char* string = "Hello, what's up?";
and I want to just return
"at's up?"
You can write a function. If the function shall not use standard C string functions then it can look for example the following way
char * substr( char *s, size_t pos )
{
size_t i = 0;
while ( i < pos && s[i] ) ++i;
return s + i;
}
As C does not support function overloading then the function above can be also written like
char * substr( const char *s, size_t pos )
{
size_t i = 0;
while ( i < pos && s[i] ) ++i;
return ( char * )( s + i );
}
Here is a demonstrative program.
#include <stdio.h>
char * substr( const char *s, size_t pos )
{
size_t i = 0;
while ( i < pos && s[i] ) ++i;
return ( char * )( s + i );
}
int main(void)
{
char * s = "Hello, what's up?";
puts( substr( s, 9 ) );
return 0;
}
The program output is
at's up?
If you already know the length of the string (and therefore know that going n characters ahead you will not go past the end), you can use string + n or &string[n], which are equivalent ways of "skipping" the first n characters of the string.
#include <stdio.h>
int main(void) {
char *str = "Hello!";
printf("%s\n", str + 1); // ello!
printf("%s\n", str + 3); // lo!
char *str2 = str + 3;
printf("%s\n", str2); // lo!
}
On the other hand, if you do not know the length of the string, you will have to make sure you don't go past the end by using a loop first. You can write a function for this:
#include <stdio.h>
char *skip(char *s, size_t n) {
while (*s && n--)
s++;
return s;
}
int main(void) {
char *str = "Hello!";
printf("%s\n", skip(str, 1)); // ello!
printf("%s\n", skip(str, 3)); // lo!
printf("%s\n", skip(str, 100)); // empty line
char *str2 = skip(str, 3);
printf("%s\n", str2); // lo!
}

Resources