Searching for block of characters (word) in a text - c

Question

I want to search for a block of characters (word) in a text.
For example, I have the next text "Hello xyz world", and I want to search for "xyz ", note the space after the word.
// The Text
const char * text = "Hello xyz world";
// The target word
const char * patt = "xyz ";
size_t textLen = strlen(text),
pattLen = strlen(patt), i, j;
for (i = 0; i < textLen; i++) {
printf("%c", text[i]);
for (j = 0; j < pattLen; j++) {
if (text[i] == patt[j]) {
printf(" <--");
break;
}
}
printf("\n");
}
The result must be like following:
But unfortunately, the result as the following:
It collects all the similar characters in the whole text, not just the target characters (the word).
How to fix that problem?

Answer 1

You have to do a full substring match before you print; mark the applicable characters on a first pass, and then have a second pass to print the results. In your case, you'd create a second array, with boolean values corresponding to the first, something like
text = "Hello xyz world";
match 000000111100000
I assume that you can find a basic substring match program online. Printing on the second pass will be easy: you already have the logic. Instead of if (text[i] == patt[j]), just use if match[i].
Is that enough of a hint?

Answer 2

You need to make sure that there is a full match before starting to print any <--. And to avoid to do accesses passed end of array on patt, you will have to stop searching when less than pattLen characters remain in array.
Then when you have found a full match, you can print the content of patt followed with <-- and increment position of pointer of pattLen-1. And at the end you will have to copy remaining characters from text.
Code could become:
// The Text
const char * text = "Hello xyz world";
// The target word
const char * patt = "xyz ";
size_t textLen = strlen(text),
pattLen = strlen(patt), i, j;
for (i = 0; i <= textLen - pattLen; i++) { // don't search if less that pattLen remains
printf("%c", text[i]);
if (text[i] == patt[0]) { // first char matches
int found = 1; // be optimistic...
for (j = 1; j < pattLen; j++) {
if (patt[j] != text[i + j]) {
found = 0;
break; // does not fully match, go on
}
}
if (found) { // yeah, a full match!
printf(" <--"); // already printed first char
for (j = 1; j < pattLen; j++) {
printf("\n%c <--", patt[j]);// print all others chars from patt
}
i += pattLen - 1; // increase index...
}
}
printf("\n");
}
while (i < textLen) {
printf("%c\n", text[i++]); // process the end of text
}
Above code gives expected output for "xyz " and also "llo"...

Answer 3

You should check every letter of your pattern from the beginning (and not check the whole pattern). Try this (not tested):
int currIndex = 0;
for (i = 0; i < textLen; i++) {
printf("%c", text[i]);
if (text[i] == patt[currIndex]) {
for (j = 0; j < pattLen; j++) {
if(text[i+j] != patt[j]){
continue;
}
}
printf(" <--");
currIndex++;
if(currIndex==pattLen)
currIndex = 0;
}
else{
currIndex = 0;
}
printf("\n");
}
Note: It is not the best way to achieve this but the easiest with your example
Note 2: This question should be closed as it is:
Questions seeking debugging help ("why isn't this code working?") must
include the desired behavior, a specific problem or error and the
shortest code necessary to reproduce it in the question itself.
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