The Rules of Buffer In C

December 14, 2015

This rules does not cover of all buffer operation yet.

Buffer or String in C is represented as an array of char terminated by null character \0.

char buff[6] = {'h', 'e', 'l', 'l', 'o', '\0'};

/* or */
char *buff = "hello"; /* \0 implicitly appended */

/* or */
char buff[] = "hello";

Null character is not the part of buff length but it does requires a memory space.

The problem is in the function for handling a string operation like allocation malloc, copy strcpy(dest, src) and concatination strcat(dest, src) is performed in unsafe way.

malloc only allocate memory for us without initialize it which potentially have a garbage values (a values from previous operation that are not cleared). Where strcpy and strcat is assume that dest is big enough which lead us buffer overflow problem if the src is larger than dest.

This is a list of rules that I always follow when dealing with buffer

Allocation

Use calloc instead

char *buff = (char *)calloc(LEN, sizeof(char));

calloc will allocate LEN contiguos block of memory with size sizeof(char) bytes for each block and initialize it to zero. This will make sure that buff is not contains any garbage values or sensitive information from the previous operation.

Copy

Use strncpy and terminate.

char buff[6];
strncpy(buff, "helllo is extra hello", sizeof(buff) - 1);
buff[sizeof(buff) - 1] = '\0';

strncpy is copy all the first sizeof(buff) - 1 from src. This help us prevent buffer overflow and corrupting the heap.

Concat

Use strncat

char path[11];
char fname = "hello";
char *ext = ".txt";
strncat(path, fname, sizeof(path) - strlen(fname) - 1);
strncat(path, ext, sizeof(path) - strlen(ext) - 1);

This will make sure that the concat operation is not write to more than allocated memory for path buffer.

That’s it.