Check if a string is palindrome in C

问题

i've a question about this code i'm writing for an exercise. I've to check if a string is palindrome. I can't change the declaration of the function.The function only return 1 when all the letters are the same (like "aaaa") but if i charge the sentence with other palindrome (like "anna") the function return me 0 , i can't figure out why this appening.Thank you!

char* cargar (char*);
int pali (char*);

int main()
{ 
   char*texto=NULL;
   texto=cargar(texto);
   int res=pali(texto);
   if(res==1){printf("\nPalindrome");}
   else printf("\nNot palindrome");

   return 0;
}

char* cargar (char*texto)
{
   char letra;
   int i=0;
   texto=malloc(sizeof(char));
   letra=getche();
   *(texto+i)=letra;
   while(letra!='\r'){
      i++;
      texto=realloc(texto,(i+1)*sizeof(char));
      letra=getche();
      *(texto+i)=letra;}
   *(texto+i)='\0';      
   return texto;
}

int pali (char* texto)
{
   int i;
   for(i=0;*(texto+i)!='\0';i++){
   }i--;
   if(i==0||i==1){return 1;}

   if(*texto==*(texto+i)){
      return pali(++texto);
   }
   else return 0;
}

回答1:

Your function to determine whether a string is a palindrome is not well thought out.

Let's say you have a string s of length l. The characters in the string are laid out as:

Indices: 0    1    2    3            l-4  l-3  l-2  l-1
         +----+----+----+----+- ... -+----+----+----+----+
         |    |    |    |    |  ...  |    |    |    |    |   
         +----+----+----+----+- ... -+----+----+----+----+

If the string is a palindrome,

s[0] = s[l-1]
s[1] = s[l-2]

...

You can stop checking when the index of the LHS is greater or equal to the index of the RHS.

To translate that into code,

int is_palindrome(char const* s)
{
   size_t len = strlen(s);
   if ( len == 0 ) // An empty string a palindrome
   {
      return 1;
   }

   size_t i = 0;
   size_t j = len-1;
   for ( ; i < j; ++i, --j )
   {
      if ( s[i] != s[j] )
      {
         // the string is not a palindrome.
         return 0;
      }
   }

   // If we don't return from inside the for loop,
   // the string is a palindrome.
   return 1;
}

回答2:

MARCO try this.

#include <stdio.h>
#include <stdlib.h>
#include<string.h>
char* cargar (char*);
int pali (char*);

int main()
{

char*texto=NULL;

texto=cargar(texto);

int res=pali(texto);

if(res==strlen(texto)){printf("\nPalindrome");}
else printf("\nNot palindrome");

    return 0;
}


char* cargar (char*texto)
{
char letra;
int i=0;
texto=malloc(sizeof(char));
letra=getche();
*(texto+i)=letra;
while(letra!='\r')
{
    i++;
    texto=realloc(texto,(i+1)*sizeof(char));
    letra=getche();
    *(texto+i)=letra;
}
*(texto+i)='\0';      
return texto;
}

int pali (char* a)
{
int flag=0,i;
int len=strlen(a);
for (i=0;i<len;i++) 
    {
    if(a[i]==a[len-i-1])
         flag=flag+1;
    }
    return flag;
}

回答3:

You pali function tests if the first character of the string is equal to the last character and then invokes itself for a position of the second character of the string. Note however, it does not modify the end of a string, so the recursive invocation compares the second character again to the last one. Then you compare the third charcter to the last one... Finaly pali returns 1 if all characters are equal to the last one, that is if all are equal.

Try this:

int pali (char* texto)
{
   char* end;
   for(end = texto; *end != '\0'; end ++)
       ;

   for(--end; texto < end; ++texto, --end) {
      if(* texto != * end)
          return 0;
   }
   return 1;
}

来源：https://stackoverflow.com/questions/31397373/check-if-a-string-is-palindrome-in-c

标签

string

recursion

dynamic

palindrome