So I'm preparing for my exams by analysing past questions and I fell upon this question:
c. Sometimes some words like "structured programming language" or "computer science and engineering" are so long that written them many times in one text is quite tiresome. Let's consider a word too long if it's length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation. This abbreviation is made like this:
(i) The first and the last letter of a word
(ii) The number of letters (including spaces) between the first and the last letters.
Thus "structured programming language" will be spelled as "s29c" and "computer science and engineering" will be spelled as "c30g", otherwise the actual word will be printed. Now, construct a C program to implement the above scenario. The word length should not be more than 100. You can't use the built-in string functions.
It says to take a long string of less than 100 characters and convert it so that the final abbreviated form has its first character as the first character of the original string, the number of characters (including spaces) in between the first and last character and then end it off with the last character. We aren't allowed to use the built-in string functions.
My skill is still pretty beginner unfortunately, so my code might be far from efficient at the moment, but here's what I have so far:
#include <stdio.h>
int main()
{
int len, n, i;
char str[100], abb[5];
char ws, we;
gets(str);
for (i = 0; str[i] != '\0'; i++)
{
len = i + 1;
}
if (len > 10)
{
ws = str[0];
abb[0] = ws;
for (i = 1; str[i] != '\0'; i++)
{
n = i - 1;
we = str[i];
}
}
else
puts(str);
}
I've managed to find the first and last characters as well as the number of characters between the two, but I'm not sure how to place that into another string. I just did a slapstick abb[0] = ws for the first character, but I don't know how to move forward from there.
There are problems in your code:
the assignment says The word length should not be more than 100, which means a length of 100 is accepted. You should make the array one byte longer for the null terminator.
you must never use gets(): this function is inherently unsafe as any sufficiently long input will cause a buffer overflow and many times can be exploited by attackers. It has been removed from the C Standard. You should use fgets() instead and make the array one byte longer to accommodate for the trailing newline and remove that after input.
you are not allowed to use the string functions from the C library, but you can write your own function to compute the length of the string and use that in your program.
The way you compute the length of the string is not foolproof: you only
modify len if the string is not empty. If the string is empty, len is left uninitialized, so the rest of the code will have undefined behavior.
to convert a number into its representation as a string in base 10, you should divide by 10 to get the digits and add '0' to convert a digit into the corresponding character in the character set. It is tricky for the general case, but for a positive number below 10, a single digit is easy to produce and for a number len between 10 and 99, the first character is obtained by dividing len by 10 and the second by taking the modulo len % 10. Given that the number to convert is in the range 9 to 98, you can use a test to determine which method to use.
the abbreviation should be constructed in the abb array, and a null terminator set at the end of the string.
Here is a modified version:
#include <stdio.h>
size_t my_strlen(const char *s) {
size_t len = 0;
while (s[len] != '\0')
len++;
return len;
}
int main(void)
{
char str[102], abb[5];
size_t len;
if (!fgets(str, sizeof str, stdin)) {
printf("missing input\n");
return 1;
}
len = my_strlen(str);
/* strip the trailing newline if present */
if (len > 0 && str[len - 1] == '\0')
str[--len] = '\0';
if (len > 10) {
size_t num = len - 2;
size_t i = 0;
abb[i++] = str[0];
if (num < 10) {
abb[i++] = '0' + num;
} else {
abb[i++] = '0' + num / 10;
abb[i++] = '0' + num % 10;
}
abb[i++] = str[len - 1];
abb[i] = '\0';
puts(abb);
} else {
puts(str);
}
return 0;
}