Strings with malloc in C

Question

I am writing a very simple program to copy a string using malloc.

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

char * copyStr(char s[])
    {
      int len = strlen(s); //find length of s
      char * copy; 
      copy = (char *)malloc(len); //dynamically allocate memory 
      int i; 
      for(i=0;i<len;i++) 
       {
         copy[i]=s[i];  //copy characters               
       }
       return copy; //return address
    }

int main(int argc, char ** argv)
    {
     char * str;
     str = "music is my aeroplane";
     char * res;
     res = copyStr(str);
     printf("The copied string is : %s",res);
     getch();
    }

The desired output is:

The copied string is : music is my aeroplane

The current output is:

The copied string is : music is my aeroplaneOMEeJ8«≤╝

Any advice is appreciated.

Your copy is not a C string, because it is not null terminated. — M Oehm
– M Oehm, Commented Apr 18, 2016 at 12:42
Please see this discussion on why not to cast the return value of malloc() and family in C.. — Sourav Ghosh
– Sourav Ghosh, Commented Apr 18, 2016 at 12:47
I love casting malloc. The misguided, lenient C type system is no reason to be sloppy. C++ tightened it wherever it could for a reason. — Peter - Reinstate Monica
– Peter - Reinstate Monica, Commented Apr 18, 2016 at 13:06
To clarify what @PeterA.Schneider wrote: Every unnecessary cast inhibits type-checking by the compiler. Which in turn has potential to not detect a type error automatically. Thus the general rule only to use a cast iff 1) it is absolutely necessary, 2) you fully understand all implications, and 3) completely accept them. Said than, casting void * definitively is a bad idea. — too honest for this site
– too honest for this site, Commented Apr 18, 2016 at 13:13

Boiethios · Accepted Answer · 2016-04-18 12:47:08Z

8

A C string is null terminated. Add a null character (ie the char which ASCII code is 0) at end of the string :

char * copyStr(char s[])
{
  size_t len = strlen(s); //find length of s
  char * copy; 
  copy = (char *)malloc(len + 1); //dynamically allocate memory
                           /* One more char must be allocated for the null char */

  size_t i; 
  for(i=0;i<len;i++) 
   {
     copy[i]=s[i];  //copy characters               
   }
   copy[i] = '\0'; // HERE
   return copy; //return address
}

It is better to use size_t for the lengths because it is unsigned.

edited Apr 18, 2016 at 12:47

answered Apr 18, 2016 at 12:44

Boiethios

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11 Comments

Boiethios Over a year ago

You could insofar it will compile, but don't. NULL is not a char (1 byte) but a pointer not set : (void *)0

alk Over a year ago

@Arjun: A simple 0 would do as well. And please note that the ASCII character named NUL is something completely different then the null-pointer constant NULL.

Art Over a year ago

@Arjun No, it isn't. It would work for complex reasons that we don't need to discuss. But a NULL pointer and a null character are two different beasts that are confusingly named in the C standard.

Arjun Over a year ago

@Art Okay, thanks. I think size-wise we prefer '\0' right? 4 bytes vs 1 byte?

Art Over a year ago

@hroptatyr Is it? I can't find the string "NUL " anywhere in the C11, C99 or C89 standards. "null character" is used all over the place though.

|

Lacobus · Accepted Answer · 2016-04-18 14:09:56Z

6

Strings in C are null-terminated.

The C programming language has a set of functions implementing operations on strings (character strings and byte strings) in its standard library. Various operations, such as copying, concatenation, tokenization and searching are supported. For character strings, the standard library uses the convention that strings are null-terminated: a string of n characters is represented as an array of n + 1 elements, the last of which is a "NUL" character.

In this case, you should have enough memory to store the original string contents and the "NUL" character (length+1). And don't forget to ensure the presence of the "NUL" character after the end of the string.

There are many possible ways to implement char * copyStr(char[]) function:

1) Your way (corrected):

char * copyStr(char s[])
{
    int i;
    size_t len = strlen( s );

    char * p = (char*) malloc( len + 1 );

    for( i = 0; i < len; i++ )
        p[i] = s[i];

    p[i] = '\0';

    return p;
}

2) Using memcpy():

char * copyStr(char s[])
{
    size_t len = strlen( s );

    char * p = (char*) malloc( len + 1 );

    memcpy( p, s, len );

    p[ len ] = '\0';

    return p;
}

3) Using strcpy():

char * copyStr(char s[])
{
    size_t len = strlen( s );
    char * p = (char*) malloc( len + 1 );
    strcpy( p, s );
    return p;
}

4) Using strdup():

char * copyStr(char s[])
{
    return strdup(s);
}

Note: For every malloc() function call you need a free() function call, your main() needs a little modification for correctness:

int main( int argc, char ** argv )
{
    char * str;
    char * res;

    str = "music is my aeroplane";

    res = copyStr( str );

    printf( "The copied string is : %s", res );

    free(res); /* freedom */

    getch();

    return 0;
}

Hope it Helps!

edited Apr 18, 2016 at 14:09

answered Apr 18, 2016 at 13:55

Lacobus

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2 Comments

Arjun Over a year ago

Brilliant, I'm kind of lost for words. Yes, I understand that I should free the dynamically allocated memory. I forgot previously..

Shark Over a year ago

I see a bright future for you on StackOverflow and the yet unrevealed Documentation ;) keep up the good work soldier ;)

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