ANSI C - append chars to get a string

Question

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 following:

    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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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...

Answer 5

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.)

Answer 6

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;
}

Answer 7

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;
}

Answer 8

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.).

Answer 9

'+' is not a stribg concatenation operator in C