Size definition of strcat() function

Question

The question is why should I define size of string (string[] should be string[some-number]) When the program is as following it gives me Abort trap: 6:

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

int main(void)
{
  char buffer1[] = "computer";
  char string[]="program";
  strcat( buffer1, string );
  printf( "buffer1 = %s\n", buffer1 );

}

This is the program from http://www.tutorialspoint.com/cprogramming/c_data_types.htm it works fine:

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

int main ()
{
   char str1[12] = "Hello";
   char str2[12] = "World";
   char str3[12];
   int  len ;

   /* copy str1 into str3 */
   strcpy(str3, str1);
   printf("strcpy( str3, str1) :  %s\n", str3 );

   /* concatenates str1 and str2 */
   strcat( str1, str2);
   printf("strcat( str1, str2):   %s\n", str1 );

   /* total lenghth of str1 after concatenation */
   len = strlen(str1);
   printf("strlen(str1) :  %d\n", len );

   return 0;
}

What is the mistake? Even if I define all of the sizes of strings in my program, my code still gives Abort trap:6?

Answer 1

Your strcat is buffer overflowing buffer1 which can hold only strlen("computer")+1 bytes. ommitting array size does not mean "dynamic" array! When you specify the size of the array, you are reserving as many bytes as you want: again you need to avoid bufferoverflow of course.

So,

 strcpy(str3, str1);

and

 strcat( str1, str2);

are ok since str3 size is enough for str1, and str1 is enough for strlen(str1) + strlen(str2) + 1, i.e. exactly 11: 5 (hello) + 5 (world) + 1 (terminator). The magic number 12 was choosen with a reason, big enough to hold both strings and a terminator.

About C strings

C-strings are array of chars where the last is "null", '\0', i.e. they are array of chars where the last one is 0. This terminator is needed so that string related functions can understand where the string ends.

If it happens that a null byte is found in the middle of a string, from the point of view of C string functions, the string will end at that point. E.g.

char buffer1[] = "computer\0program";
// array: { 'c', 'o', ... '\0', 'p', 'r', 'o', .., 'm', '\0' }

// ...
printf("%s\n", buffer1);

will print computer only. But at this point the buffer will be big enough to hold computer and program, a terminator (and another extra byte), since the compiler computed the size of the char array considering the literal sequence of characters which syntactically ends at the second ".

But for all C-string functions, the string contained in buffer1 is computer. Note also that sizeof buffer1 will give the correct size of the buffer, i.e. 17, opposed to the result of strlen(buffer1) which is just 8.