append chars to get a string

I have the following code in C:

    char *str = "Hello World";
    char *s = malloc(strlen(str));
    int i =0;
    for(;i<strlen(str)-5;i++)
    {
        s += *(str+i);
    }
    printf(s);

It shows nothing. What I want is to get the substring of str stored in s.

In Java I would do the follow开发者_开发知识库ing:

    String str = "Hello World";
    String s="";

    for(int i=0;i<str.length()-5; i++)
        s+=str[i];

    System.out.println(s);

Or instead use the substring method. As String s = str.substring(1,2); for example.

How can I achieve it?

Use the strcpy function.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char *argv[] ) 
{
  char *str = "Hello World";
  size_t length = strlen(str);

  char *s = (char*)malloc(sizeof(char) * (length + 1));

  strcpy(s, str);

  s[length] = '\0'; // makes sure it's NUL terminated

  printf("%s", s);

  free(s);

  return 0;
}

When allocating the destination buffer, pay attention to the fact that strings are terminated by the NUL character.

To only copy a substring, use strncpy:

strncpy(s, str + 6, strlen(str) - 6);

will just copy "World" into s.

In any case, make sure your C strings are NUL terminated before using functions like printf.

See also strcat and strncat. And well, familiarize yourself with C arrays and pointers.

Others have said how to correctly solve this problem (strcat) but the trick is to think about types. C doesn't do ANYTHING magical for you. s is a char*. what happens when you add 3 to a char*? you get another char* that is pointing 3 characters farther down the line. Adding 'a' to a char* is the same as adding 97 (ascii of 'a') to the pointer, and thus pointing to another character far down the array...

I hope that explains what was happening...

You can't use the += operator on a char* object - it's just a pointer to char. Instead, you'll want to use strncpy, and pass it a pre-allocated buffer, plus a pointer to where in the string you want the copying to begin, and the number of characters you want to copy.

C strings are pointers to characters in memory, not objects like in other languages. I suggest you Google for an explanation of strings (this one, for example) and then look at the strcpy function defined in string.h.

For substrings in general where you want to omit characters from both the beginning and the end, you should use strncpy or strncat. (Be careful, though: strncpy doesn't necessarily NUL-terminate.)

Use strncpy():

char *str = "Hello World";         
size_t len = strlen(str) - 5;   
char *s = malloc(len + 1); // +1 for nul terminator
if (s)            
{
  strncpy(s, str, len); // copies strlen(str)-5 characters from str to s
  s[len] = 0; // add nul terminator
}

A somewhat more general function:

/**
 * Return a substring of _len_ characters starting at position _start_
 * substring("Hello, World", 2, 3) == "llo"
 */
char *substring(char *str, size_t start, size_t len)
{
  char *s = malloc(len + 1);
  if (s)
  {
    strncpy(s, str+start, len);
    s[len] = 0;
  }
  return s;
}

Add sanity checks as needed (e.g., start < strlen(str), len <= strlen(str) - start, etc.).

You need to allocate memory for the string that will hold the substring. Maybe something like this:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define LESS 4
#define MAX 10 /*Keep it atleast strlen(str) - LESS + 1*/

int main(int argc, char *argv[] )
{

        char *str = "Hello World";
        char s[MAX]; /* Declare the array to hold the substring.*/
        int i =0,j=0; /* i is index into original string and j in the result string. */
        for(;i<strlen(str)-LESS;i++)
        {
                s[j++] = *(str+i);
        }
        s[j] = '\0'; /* Terminate the result string.*/
        printf("%s\n",s); /* Output: Hello W */

        return 0;
}

You can do what you're asking to do with the following changes to your code:

int main (int argc, char *argv[])
{
    char *str = "Hello World";
    char *s = malloc(strlen(str));
    int i = 0;
    for(;i<strlen(str)-5;i++)
    {
        s[i] = *(str+i);
    }
    s[i] = 0;

    printf(s);

   return 0;
}

'+' is not a stribg concatenation operator in C