Remove a word from a sentence (string) - c

Question

I am in the stage of preparing myself for exams, and the thing that I m least proud of are my skills with strings. What I need to do is remove a word from a sentence, without using <string.h> library at all.
This is what I've got so far. It keeps showing me that certain variables are not declared, such as start and end.
#include <stdio.h>
/* Side function to count the number of letters of the word we wish to remove */
int count(char *s) {
int counter = 0;
while (*s++) {
counter++;
s--;
return counter;
}
/* Function to remove a word from a sentence */
char *remove_word(const char *s1, const char *s2) {
int counter2 = 0;
/* We must remember where the string started */
const char *toReturn = s1;
/* Trigger for removing the word */
int found = 1;
/* First we need to find the word we wish to remove [Don't want to
use string.h library for anything associated with the task */
while (*s1 != '\0') {
const char *p = s1;
const char *q = s2;
if (*p == *q)
const char *start = p;
while (*p++ == *q++) {
counter2++;
if (*q != '\0' && counter2 < count(s2))
found = 0;
else {
const char *end = q;
}
}
/* Rewriting the end of a sentence to the beginning of the found word */
if (found) {
while (*start++ = *end++)
;
}
s1++;
}
return toReturn;
}
void insert(char niz[], int size) {
char character = getchar();
if (character == '\n')
character = getchar();
int i = 0;
while (i < size - 1 && character != '\n') {
array[i] = character;
i++;
character = getchar();
}
array[i] = '\0';
}
int main() {
char stringFirst[100];
char stringSecond[20];
printf("Type your text here: [NOT MORE THAN 100 CHARACTERS]\n");
insert(stringFirst, 100);
printf("\nInsert the word you wish to remove from your text.");
insert(stringSecond, 20);
printf("\nAfter removing the word, the text looks like this now: %s", stringFirst);
return 0;
}

Answer 1

your code is badly formed, i strongly suggest compiling with:
gcc -ansi -Wall -pedantic -Werror -D_DEBUG -g (or similar)
start with declaring your variables at the beginning of the function block, they are known only inside the block they are declared in.
your count function is buggy, missing a closing '}' (it doesn't compile)
should be something like
size_t Strlen(const char *s)
{
size_t size = 0;
for (; *s != '\n'; ++s, ++size)
{}
return size;
}
implementing memmove is much more efficient then copy char by char

Answer 2

I reformatted you code for small indentation problems and indeed indentation problems indicate real issues:
There is a missing } in count. It should read:
/* Side function to count the number of letters of the word we wish to remove */
int count(char *s) {
int counter = 0;
while (*s++) {
counter++;
}
return counter;
}
or better:
/* Side function to count the number of letters of the word we wish to remove */
int count(const char *s) {
const char *s0 = s;
while (*s++) {
continue;
}
return s - s0;
}
This function counts the number of bytes in the string, an almost exact clone of strlen except for the return type int instead of size_t. Note also that you do not actually use nor need this function.
Your function insert does not handle EOF gracefully and refuses an empty line. Why not read a line with fgets() and strip the newline manually:
char *input(char buf[], size_t size) {
size_t i;
if (!fgets(buf, size, stdin))
return NULL;
for (i = 0; buf[i]; i++) {
if (buf[i] == '\n') {
buf[i] = '\0';
break;
}
}
return buf;
}
In function remove_word, you should define start and end with a larger scope, typically the outer while loop's body. Furthermore s1 should have type char *, not const char *, as the phrase will be modified in place.
You should only increment p and q if the test succeeds and you should check that p and q are not both at the end of their strings.
last but not least: you do not call remove_word in the main function.
The complete code can be simplified into this:
#include <stdio.h>
/* Function to remove a word from a sentence */
char *remove_word(char *s1, const char *s2) {
if (*s2 != '\0') {
char *dst, *src, *p;
const char *q;
dst = src = s1;
while (*src != '\0') {
for (p = src, q = s2; *q != '\0' && *p == *q; p++, q++)
continue;
if (*q == '\0') {
src = p; /* the word was found, skip it */
} else {
*dst++ = *src++; /* otherwise, copy this character */
}
}
*dst = '\0'; /* put the null terminator if the string was shortened */
}
return s1;
}
char *input(char buf[], size_t size) {
size_t i;
if (!fgets(buf, size, stdin))
return NULL;
for (i = 0; buf[i]; i++) {
if (buf[i] == '\n') {
buf[i] = '\0';
break;
}
}
return buf;
}
int main() {
char stringFirst[102];
char stringSecond[22];
printf("Type your text here, up to 100 characters:\n");
if (!input(stringFirst, sizeof stringFirst))
return 1;
printf("\nInsert the word you wish to remove from your text: ");
if (!input(stringSecond, sizeof stringSecond))
return 1;
printf("\nAfter removing the word, the text looks like this now: %s\n",
remove_word(stringFirst, stringSecond));
return 0;
}

Answer 3

Your start and end pointers are defined within a block which makes their scope limited within that block. So, they are not visible to other parts of your code, and if you attempt to reference them outside their scope, the compiler will complain and throw an error. You should declare them at the beginning of the function block.
That said, consider the following approach to delete a word from a string:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int delete_word(char *buf,
const char *word);
int main(void)
{
const char word_to_delete[] = "boy";
fputs("Enter string: ", stdout);
char buf[256];
fgets(buf, sizeof(buf), stdin);
if (delete_word(buf, word_to_delete))
{
printf("Word %s deleted from buf: ", word_to_delete);
puts(buf);
}
else
{
printf("Word %s not found in buf: ", word_to_delete);
puts(buf);
}
system("PAUSE");
return 0;
}
int chDelimit(int ch)
{
return
(ch == '\n' || ch == '\t') ||
(ch >= ' ' && ch <= '/') ||
(ch >= ':' && ch <= '#') ||
(ch >= '[' && ch <= '`') ||
(ch >= '{' && ch <= '~') ||
(ch == '\0');
}
char *find_pattern(char *buf,
const char *pattern)
{
size_t n = 0;
while (*buf)
{
while (buf[n] && pattern[n])
{
if (buf[n] != pattern[n])
{
break;
}
n++;
}
if (!pattern[n])
{
return buf;
}
else if (!*buf)
{
return NULL;
}
n = 0;
buf++;
}
return NULL;
}
char *find_word(char *buf,
const char *word)
{
char *ptr;
size_t wlen;
wlen = strlen(word);
ptr = find_pattern(buf, word);
if (!ptr)
{
return NULL;
}
else if (ptr == buf)
{
if (chDelimit(buf[wlen]))
{
return ptr;
}
}
else
{
if (chDelimit(ptr[-1]) &&
chDelimit(ptr[wlen]))
{
return ptr;
}
}
ptr += wlen;
ptr = find_pattern(ptr, word);
while (ptr)
{
if (chDelimit(ptr[-1]) &&
chDelimit(ptr[wlen]))
{
return ptr;
}
ptr += wlen;
ptr = find_pattern(ptr, word);
}
return NULL;
}
int delete_word(char *buf,
const char *word)
{
size_t n;
size_t wlen;
char *tmp;
char *ptr;
wlen = strlen(word);
ptr = find_word(buf, word);
if (!ptr)
{
return 0;
}
else
{
n = ptr - buf;
tmp = ptr + wlen;
}
ptr = find_word(tmp, word);
while (ptr)
{
while (tmp < ptr)
{
buf[n++] = *tmp++;
}
tmp = ptr + wlen;
ptr = find_word(tmp, word);
}
strcpy(buf + n, tmp);
return 1;
}

Answer 4

If you have to do it manually, just loop over the indicies of your string to find the first one that matches and than you’ll have a second loop that loops for all the others that matches and resets all and jumps to the next index of the first loop if not matched something in order to continue the searching. If I recall accuretaly, all strings in C are accesible just like arrays, you’ll have to figure it out how. Don’t afraid, those principles are easy! C is an easy langugae, thiught very long to write.
In order to remove: store the first part in an array, store the second part in an array, alloc a new space for both of them and concatinate them there.
Thanks, hit the upvote button.
Vitali
EDIT: use \0 to terminate your newly created string.