Copying argv and checking it for a palindrome - c

Question

I'm writing a program to check for palindromes. I recently picked up C and was wondering is there a reason why my take on it won't work? Does it have something to do with my use of directly copying argv into a char array
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(int argc,char *argv[]){
int i;
int a;
int size;
for(a = 1; a < argc; a++){
char *reverseThis = argv[a];
char *normal = argv[a];
size = strlen(reverseThis);
for(i = 0; i < size; i++){
reverseThis[i] = normal[size - i - 1];
}
for(i = 0; i < size; i++){
reverseThis[i] = tolower(reverseThis[i]);
normal[i] = tolower(normal[i]);
}
if(strcmp(reverseThis,normal)==0){
printf("\"%s\": on palindromi\n",argv[i]);
}
else
printf("\"%s\": ei ole palindromi\n",argv[i]);
}
return 0;
}

Answer 1

In your code you are not copying the strings, you assigned both normal and reverseThis to same string argv[a].In reverseThis you need to copy argv[a] after allocating memory.
Just modify your code in forloop:
for(a = 1; a < argc; a++){
char *normal = argv[a];
size = strlen(normal);
char *reverseThis = (char*)malloc((size+1)*sizeof(char));
int j=0;
for(i = size-1; i >= 0; i++){
reverseThis[j++] = normal[i];
}
reverseThis[j]='\0';
.
.

Answer 2

You are using an incorrect approach.
For starters this loop
for(i = 0; i < size; i++){
reverseThis[i] = normal[size - i - 1];
}
copies the right half of the string in reverse order in the left half of the string totally overwriting its left part.
For example if you have a string like this "123456" then after the loop it will look like "654456"
This comparison also does not make sense
if(strcmp(reverseThis,normal)==0){
because the both pointers point to the same string. So the condition always yields true.
Take into account that these declarations
char *reverseThis = argv[a];
char *normal = argv[a];
do not copy the original string pointed to by argv[a]. The declared pointers just point to the first character of the same string.
And here is a typo
printf("\"%s\": on palindromi\n",argv[i]);
^^^
The task can be done simpler without changing the strings.
For example
size_t n = strlen( argv[a] );
size_t i = 0;
while ( i < n / 2 && tolower( ( unsigned char )argv[i] ) == tolower( ( unsigned char )argv[n -i - 1] ) ) ++i;
if ( i == n / 2 )
{
printf("\"%s\": on palindromi\n", argv[a]);
}
else
{
printf("\"%s\": ei ole palindromi\n",argv[a]);
}
If you indeed need to copy the strings then either declare variable length arrays (if the compiler supports them) or allocate arrays dynamically. For example (declaring variable length arrays):
size = strlen( argv[a] );
char reverseThis[size + 1];
char normal[size + 1];
strcpy( reverseThis, argv[a] );
strcpy( normal, argv[a] );

Answer 3

You don't need to reverse the string and compare to find out whether the input string is palindrome or not.
You can simply compare the characters of the string starting from both ends of the string and move one character forward from the start of the string and one character backward from the end of the string. If all character matches till you reach to mid of string then the string is palindrome otherwise not a palindrome.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(int argc,char *argv[]){
int i, a, size;
for(a = 1; a < argc; a++){
char *ptr= argv[a];
int notpalindrom = 0;
size = strlen(ptr) - 1;
for(i = 0; i < size; ){
if (tolower(ptr[i++]) != tolower(ptr[size--])){
notpalindrom = 1;
break;
}
}
if (notpalindrom)
printf ("%s is not palindrom\n", ptr);
else
printf ("%s is palindrom\n", ptr);
}
return 0;
}

Answer 4

The first "if" take from argv in index "i", i == to the last change that has changed in the last "for", in your case i == 4, and the program crashed cause their isn't string in that member, to fix that you should change the "i" to 0 before the "if".