Consider the following C code snippet.

/* Escapes all newlines in the input string, replacing them with " ". */

/* Requires: p != NULL; p is a valid \0-terminated string */

void escape(char *p)

{

while (*p != \0)

switch (*p)

{

case :

memcpy(p+2, p+1, strlen(p));

*p++ = \\; *p++ = n;

break;

default:

p++;

}

}

Question! You may assume that escape()'s argument is always non-null and points to a \0-terminated string. What's wrong with this code?

Hints:

The code '\0' terminates the string so the while statement reads each character until it reaches \0. The memcpy statement is "making room" for the to be inserted.

*p++ = '\\'; inserts a backslash \

*p++ = 'n'; inserts the n

p++; moves to the next character in the string. NOTE: The important fact here is that C++ does not detect when memory is overflowed (or underflowed).