I am trying to write a program that checks to see if a word inputed to a program matches one of the predefined keywords. Input is going to be coming from a text file and the text file will have a single word in it. So far the text file I have just has the word 'frog' which means the program should clearly print 'No Match Found' but it in fact prints 'Match Found' It prints prints 'Match Found' even when the word in the text file is a string of random letters. Here is my code, does anything stand out to you guys? Thanks
#define NUM 4
#define SIZE 12
int isAlpha(char);
//Returns 1 if it is an Alphabetical character, 0 if it is not
int isAlpha(char c) {
return (c >= 'A' && c <= 'Z' || c >= 'a' && c <= 'z');
}
int main() {
char message[141];
int charCount = 0, c = 0, matchCheck = 0;
char keywords[NUM][SIZE] = {
"crackerjack",
"Hey",
"dog",
"feet"
};
//Removes non alphabetical characters
while((c = getchar()) != EOF && charCount <= 140) {
if(isAlpha(c)){
message[charCount] = c;
charCount++;
}
}
//checks if message matches keyword
for (int i = 0; i < NUM; i++) {
for (int j = 0; j < SIZE; j++) {
//Check if current two characters match
if (message[j] == keywords[i][j]) {
//Check if the two matched characters are the null terminator character
if (message[j] == '\0' && keywords[i][j] == '\0') {
matchCheck = 1;
break;
}
}
}
}
//prints "Match Found!" if there was a match
if (matchCheck == 1)
printf("Match Found!\n");
else
printf("No Match Found\n");
}
I havent programmed in C in a long time, and I dont have a C compiler with me right now. However, I think that there might be a problem with the logic. You need to check for every word individually. I would do somehting like the following for your last section ...
matchAnyWord = 0
for (int i = 0; i < NUM; i++) {
matchCheck = 1;
for (int j = 0; j < SIZE; j++) {
if (message[j] == '\0' && keywords[i][j] == '\0') break;
if (message[j] == '\0' && keywords[i][j] != '\0') { matchCheck = 0; break;}
if (message[j] != '\0' && keywords[i][j] == '\0') { matchCheck = 0; break;}
if (message[j] != keywords[i][j] ) { matchCheck = 0; break;}
}
if (matchCheck == 1) {matchAnyWord = 1; break;}
}
Note that here we check for the matching of individual words, and introduce a new variable matchAnyWord which is actually what you want. I might have errors in the syntax but I think the logic should be more-or-less there.
Cheers!
Also, not sure why the indentation isnt showing up correctly on my browser ...
Related
I am working on a question which requires me to print a string given a field-number at that position. The strings should be read from a file.
file.txt
C is a language.
lex lexical analyser
(blank line)
gcc is good
If the field-number is 2 (i.e the second word in the sentence). The program should output
is
lexical
(NULL)
is
I wrote a function but don't think its the correct way and that it would work for all cases. It should handle extra blanks or newlines.
while (fgets(buffer, MAX, file) != NULL) {
for (int i = 1; i < strlen(buffer); i++) {
if (count == field_number - 1) {
int j = i;
while (j < strlen(buffer) && buffer[j] != ' ') {
printf("%c", buffer[j++]);
}
printf("\n");
count = 0;
break;
}
if (buffer[i] == ' ' && buffer[i - 1] != ' ') {
count++;
}
}
}
I am a beginner. This code should be easy to understand.
This should work for all the cases,
int main() {
//FILE* file = fopen(__FILE__, "r");
//int field_number = 2;
int new_line = 0; // var to keep track of new line came or not
int word = 0;
int count = 0;
char c, prev_c;
while ((c = fgetc(file)) != EOF) {
// printf("[%c]", c);
// if a new line char comes it means you entered a new line
if(c == '\n') {
// you have to print the new line here on the output to handle
// empty line cases
printf("\n");
new_line = 1; // when line changes
word = 0; // no word has come in this new line so far
count = 0; // count becomes 0
} else if( c == ' ' && prev_c != ' ') {
if(word)
count++;
if(count == field_number) // if count exceeds field_number
new_line = 0; // wait till next line comes
} else if (new_line && count == field_number - 1) {
printf("%c", c);
} else {
word = 1; // fi a not new line or non space char comes, a word has come
}
prev_c = c;
}
return 0;
}
Write a program to "fold" long input lines into two or more shorter lines after the last non-blank character that occurs before the n-th column of input. Make sure your program does something intelligent with very long lines, and if there are no blanks or tabs before the specified column.
The algorithm I decided to follow for this was as follows:
If length of input line < maxcol (the column after which one would have to fold), then print the line as it is.
If not, from maxcol, I check towards it's left, and it's right to find the closest non-space character, and save them as 'first' and 'last'. I then print the character array from line[0] to line[first] and then the rest of the array, from line[last] to line[len] becomes the new line array.
Here's my code:
#include <stdio.h>
#define MAXCOL 5
int getline1(char line[]);
int main()
{
char line[1000];
int len, i, j, first, last;
len = getline1(line);
while (len > 0) {
if (len < MAXCOL) {
printf("%s\n", line);
break;
}
else {
for (i = MAXCOL - 1; i >= 0; i--) {
if (line[i] != ' ') {
first = i;
break;
}
}
for (j = MAXCOL - 1; j <= len; j++) {
if (line[j] != ' ') {
last = j;
break;
}
}
//printf("first %d last %d\n", first, last);
for (i = 0; i <= first; i++)
putchar(line[i]);
putchar('\n');
for (i = 0; i < len - last; i++) {
line[i] = line[last + i];
}
len -= last;
first = last = 0;
}
}
return 0;
}
int getline1(char line[])
{
int c, i = 0;
while ((c = getchar()) != EOF && c != '\n')
line[i++] = c;
if (c == '\n')
line[i++] = '\n';
line[i] = '\0';
return i;
}
Here are the problems:
It does not do something intelligent with very long lines (this is fine, as I can add it as an edge case).
It does not do anything for tabs.
I cannot understand a part of the output.
For example, with the input:
asd de def deffff
I get the output:
asd
de
def
defff //Expected until here
//Unexpected lines below
ff
fff
deffff
deffff
deffff
Question 1 - Why do the unexpected lines print? How do I make my program/algorithm better?
Eventually, after spending quite some time with this question, I gave up and decided to check the clc-wiki for solutions. Every program here did NOT work, save one (The others didn't work because they did not cover certain edge cases). The one that worked was the largest one, and it did not make any sense to me. It did not have any comments, and neither could I properly understand the variable names, and what they represented. But it was the ONLY program in the wiki that worked.
#include <stdio.h>
#define YES 1
#define NO 0
int main(void)
{
int TCOL = 8, ch, co[3], i, COL = 19, tabs[COL - 1];
char bls[COL - 1], bonly = YES;
co[0] = co[1] = co[2] = 0;
while ((ch = getchar()) != EOF)
{
if (ch != '\t') {
++co[0];
++co[2];
}
else {
co[0] = co[0] + (TCOL * (1 + (co[2] / TCOL)) - co[2]);
i = co[2];
co[2] = TCOL + (co[2] / TCOL) * TCOL;
}
if (ch != '\n' && ch != ' ' && ch != '\t')
{
if (co[0] >= COL) {
putchar('\n');
co[0] = 1;
co[1] = 0;
}
else
for (i = co[1]; co[1] > 0; --co[1])
{
if (bls[i - co[1]] == ' ')
putchar(bls[i - co[1]]);
else
for (; tabs[i - co[1]] != 0;)
if (tabs[i - co[1]] > 0) {
putchar(' ');
--tabs[i - co[1]];
}
else {
tabs[i - co[1]] = 0;
putchar(bls[i - co[1]]);
}
}
putchar(ch);
if (bonly == YES)
bonly = NO;
}
else if (ch != '\n')
{
if (co[0] >= COL)
{
if (bonly == NO) {
putchar('\n');
bonly = YES;
}
co[0] = co[1] = 0;
}
else if (bonly == NO) {
bls[co[1]] = ch;
if (ch == '\t') {
if (TCOL * (1 + ((co[0] - (co[2] - i)) / TCOL)) -
(co[0] - (co[2] - i)) == co[2] - i)
tabs[co[1]] = -1;
else
tabs[co[1]] = co[2] - i;
}
++co[1];
}
else
co[0] = co[1] = 0;
}
else {
putchar(ch);
if (bonly == NO)
bonly = YES;
co[0] = co[1] = co[2] = 0;
}
}
return 0;
}
Question 2 - Can you help me make sense of this code and how it works?
It fixes all the problems with my solution, and also works by reading character to character, and therefore seems more efficient.
Question 1 - Why do the unexpected lines print? How do I make my program/algorithm better?
You are getting the unexpected lines in the output because after printing the array, you are not terminating the new line array with null character \0 -
Here you are copying character from starting from last till len - last, creating a new line array:
for (i = 0; i < len - last; i++) {
line[i] = line[last + i];
}
You have copied the characters but the null terminating character is still at its original position. Assume the input string is:
asd de def deffff
So, initially the content of line array will be:
"asd de def deffff\n"
^
|
null character is here
Now after printing asd, you are copying characters from last index of line till len - last index to line array itself starting from 0 index. So, after copying the content of line array will be:
"de def deffff\n deffff\n"
|____ _____|
\/
This is causing the unexpected output
(null character is still at the previous location)
So, after for loop you should add the null character just after the last character copied, like this:
line [len - last] = '\0';
With this the content of line array that will be processed in the next iteration of while loop will be:
"de def deffff\n"
One more thing, in the line array you can see the \n (newline) character at the end. May you want to remove it before processing the input, you can do:
line[strcspn(line, "\n")] = 0;
Improvements that you can do in your program:
1. One very obvious improvement that you can do is to use pointer to the input string while processing it. With the help of pointer you don't need to copy the rest of the array, apart from processed part, again to the same array till the program process the whole input. Initialize the pointer to the start of the input string and in every iteration just move the pointer to appropriate location and start processing from that location where pointer is pointing to.
2. Since you are taking the whole input first in a buffer and then processing it. You may consider fgets() for taking input. It will give better control over the input from user.
3. Add a check for line array overflow, in case of very long input. With fgets() you can specify the maximum number of character to be copied to line array from input stream.
Question 2 - Can you help me make sense of this code and how it works?
The program is very simple, try to understand it at least once by yourself. Either use a debugger or take a pen and paper, dry run it once for small size input and check the output. Increase the input size and add some variations like multiple space characters and check the program code path and output. This way you can understand it very easily.
Here's another (and I think better) solution to this exercise :
#include <stdio.h>
#define MAXCOL 10
void my_flush(char buf[]);
int main()
{
int c, prev_char, i, j, ctr, spaceleft, first_non_space_buf;
char buf[MAXCOL+2];
prev_char = -1;
i = first_non_space_buf = ctr = 0;
spaceleft = MAXCOL;
printf("Just keep typing once the output has been printed");
while ((c = getchar()) != EOF) {
if (buf[0] == '\n') {
i = 0;
my_flush(buf);
}
//printf("Prev char = %c and Current char = %c and i = %d and fnsb = %d and spaceleft = %d and j = %d and buf = %s \n", prev_char, c, i, first_non_space_buf, spaceleft, j, buf);
if ((((prev_char != ' ') && (prev_char != '\t') && (prev_char != '\n')) &&
((c == ' ') || (c == '\t') || (c == '\n'))) ||
(i == MAXCOL)) {
if (i <= spaceleft) {
printf("%s", buf);
spaceleft -= i;
}
else {
putchar('\n');
spaceleft = MAXCOL;
for (j = first_non_space_buf; buf[j] != '\0'; ++j) {
putchar(buf[j]);
++ctr;
}
spaceleft -= ctr;
}
i = 0;
my_flush(buf);
buf[i++] = c;
first_non_space_buf = j = ctr = 0;
}
else {
if (((prev_char == ' ') || (prev_char == '\t') || (prev_char == '\n')) &&
((c != ' ') && (c != '\t') && (c != '\n'))) {
first_non_space_buf = i;
}
buf[i++] = c;
buf[i] = '\0';
}
prev_char = c;
}
printf("%s", buf);
return 0;
}
void my_flush(char buf[])
{
int i;
for (i = 0; i < MAXCOL; ++i)
buf[i] = '\0';
}
Below is my solution, I know the thread is no longer active but my code might help someone who's facing issues to grasp the already presented code snippets.
*EDIT
explaination
Keep reading input unless the input contains '\n', '\t' or there've been
atleast MAXCOl chars.
Incase of '\t', use expandTab to replace with required spaces and use printLine if it doesn't exceed MAXCOl.
Incase of '\n', directly use printLine and reset the index.
If index is 10:
find the last blank using findBlank ad get a new index.
use printLine to print the current line.
get new index as 0 or index of newly copied char array using the newIndex function.
code
/* fold long lines after last non-blank char */
#include <stdio.h>
#define MAXCOL 10 /* maximum column of input */
#define TABSIZE 8 /* tab size */
char line[MAXCOL]; /* input line */
int expandTab(int index);
int findBlank(int index);
int newIndex(int index);
void printLine(int index);
void main() {
int c, index;
index = 0;
while((c = getchar()) != EOF) {
line[index] = c; /* store current char */
if (c == '\t')
index = expandTab(index);
else if (c == '\n') {
printLine(index); /* print current input line */
index = 0;
} else if (++index == MAXCOL) {
index = findBlank(index);
printLine(index);
index = newIndex(index);
}
}
}
/* expand tab into blanks */
int expandTab(int index) {
line[index] = ' '; /* tab is atleast one blank */
for (++index; index < MAXCOL && index % TABSIZE != 0; ++index)
line[index] = ' ';
if (index > MAXCOL)
return index;
else {
printLine(index);
return 0;
}
}
/* find last blank position */
int findBlank(int index) {
while( index > 0 && line[index] != ' ')
--index;
if (index == 0)
return MAXCOL;
else
return index - 1;
}
/* re-arrange line with new position */
int newIndex(int index) {
int i, j;
if (index <= 0 || index >= MAXCOL)
return 0;
else {
i = 0;
for (j = index; j < MAXCOL; ++j) {
line[i] = line[j];
++i;
}
return i;
}
}
/* print line until passed index */
void printLine(int index) {
int i;
for(i = 0; i < index; ++i)
putchar(line[i]);
if (index > 0)
putchar('\n');
}
I'm trying to write a function to count occurrences of a particular word in a string. For example: Given string -
"Stop, time to go home. Todo fix me."
Letters "to" appeared three times (twice in different words); however, the word "to" appears only once. What should I do to count only word "to" (if will appear in string more times then count every single one). Any advice?
This is the code I was trying and playing around.
int word(char inputLine[]) {
int word = 0, i = 0, j = 0;
for (i = 0; inputLine[i] != '\0'; i++) {
if (inputLine[i] == 't' || inputLine[i] == 'o' || inputLine[i] != ' ') {
word++;
}
}
return word;
}
Try this:
int word(char inputLine[]) {
int word = 0, i = 0;
// stop before the last char
for (i = 0; inputLine[i] != '\0' && inputLine[i+1] != '\0'; i++) {
// is (T or t) and (O or o)
if ((inputLine[i] == 't' || inputLine[i] == 'T') && (inputLine[i+1] == 'o' || inputLine[i+1] == 'O')) {
// after the 'to' is not a letter
if ((inputLine[i+2] < 'a' || inputLine[i+2] > 'z') &&
(inputLine[i+2] < 'A' || inputLine[i+2] > 'Z')) {
// before is not a letter (or this is the start of the string)
if (i == 0 ||
((inputLine[i-1] < 'a' || inputLine[i-1] > 'z') &&
(inputLine[i-1] < 'A' || inputLine[i-1] > 'Z'))) {
word++;
}
}
}
}
return word;
}
The simplest way would be to use strtok. But, if you'd like to do it all by hand, the following will work. Although you only wanted the "to" this will work for any search string:
#include <stdio.h>
// word -- get number of string matches
int
word(char *input,char *str)
// input -- input buffer
// str -- string to search for within input
{
int chr;
int prev;
int off;
int stopflg;
int wordcnt;
off = -1;
stopflg = 0;
wordcnt = 0;
prev = 0;
for (chr = *input++; ! stopflg; prev = chr, chr = *input++) {
// we've hit the end of the buffer
stopflg = (chr == 0);
// convert whitespace characters to EOS [similar to what strtok might
// do]
switch (chr) {
case ' ':
case '\t':
case '\n':
case '\r':
chr = 0;
break;
}
++off;
// reset on mismatch
// NOTE: we _do_ compare EOS chars here
if (str[off] != chr) {
off = -1;
continue;
}
// we just matched
// if we're starting the word we must ensure we're not in the middle
// of one
if ((off == 0) && (prev != 0)) {
off = -1;
continue;
}
// at the end of a word -- got a match
if (chr == 0) {
++wordcnt;
off = -1;
continue;
}
}
return wordcnt;
}
void
tryout(int expcnt,char *buf)
{
int actcnt;
actcnt = word(buf,"to");
printf("%d/%d -- '%s'\n",expcnt,actcnt,buf);
}
// main -- main program
int
main(int argc,char **argv)
{
char *cp;
--argc;
++argv;
for (; argc > 0; --argc, ++argv) {
cp = *argv;
if (*cp != '-')
break;
switch (cp[1]) {
default:
break;
}
}
tryout(1,"to");
tryout(2,"to to");
tryout(1," to ");
tryout(1,"todo to");
tryout(2,"todo to to");
tryout(2,"doto to to");
tryout(1,"doto to doto");
tryout(0,"doto");
return 0;
}
If you must use only "basic" C functions the above solutions seems ok, but in the case you want to build a more scalable application (and you want to solve the problem in a smarter way) you can use a library that manipulate regular expressions. You can check this answer: Regular expressions in C: examples?
Regexes has the advantage that you can make the regex case unsensible (That is one of your issues).
I usually use pcre because it has the regex style of perl and java.
Here it is a very useful example that uses pcre: http://www.mitchr.me/SS/exampleCode/AUPG/pcre_example.c.html
Let's posit these rules:
"to" can be a word only when there is no char before and after it except the space char
If you accept those rules as valid and correct you need to check 4 conditions:
if (str[i]=='t'&& str[i+1]=='o'&& str[i-1]!='a-z'&& str[i+2]!='a-z'){
word++;
}
Two more conditions can be included to check for the upper case letters.
public class FindCountOfWordInString {
public static void main(String[] args) {
String str = "yhing ghingu jhhtring inghfg ajklingingd me";
String find = "ing";
int count = findCountOfWordInString(str, find);
System.out.println(count);
}
private static int findCountOfWordInString(String str, String find) {
String[] strArr = str.split(" ");
int count = 0, k = 0;
for (int i = 0; i < strArr.length; i++) {
if (strArr[i].contains(find)) {
String strCheck = strArr[i];
char[] findCharArr = find.toCharArray();
for (int j = 0; j < strCheck.length(); j++) {
if (strCheck.charAt(j) == findCharArr[k]) {
k++;
if (k == 3) {
count++;
k = 0;
}
} else {
k = 0;
}
}
}
}
return count;
}
}
I am creating a program that prints each line of a file one by one in reverse word order. i.e. "The big black moose" prints as "moose black big The".
However, in the code below, it does not print the last word of lines that do not have a delimiter before the line break. A delimiter in this case is defined as any whitespace character such as space or tab.
int main(int argc, char const *argv[]) {
if (argc != 2) return 0;
int i = 0, c;
int isD = 0, wasD = 1;
int count = 0;
FILE *file = fopen(argv[1], "r");
while ((c = getc(file)) != EOF) {
isD = c == ' ' || c == '\t' || c == '\n';
if (!isD) {
chars[i++] = c;
count++;
}
if (isD && !wasD) {
shiftInsert(i++, count);
count = 0;
}
wasD = isD;
}
fclose(file);
return 0;
}
int shiftInsert(int length, int shift) {
int word[shift+1], i;
printf("\n----------\nL:%d,S:%d\n", length, shift);
for (i = 0; i < shift; i++)
word[i] = chars[length-shift+i];
word[shift] = ' ';
for (i = 0; i < shift; i++)
printf("%c", word[i]);
for (i = length; i >= 0; i--)
chars[i+shift+1] = chars[i];
for (i = 0; i <= shift; i++)
chars[i] = word[i];
printf("|");
}
This happens, because you don't enter the loop when getc finds the end of the file. If wasD is false, you'll have one unprocessed word in the buffer.
You could treat EOF as whitespace and place the terminating condition at the end of the loop:
do {
c = getc(file);
isD = (c == ' ' || c == '\t' || c == '\n' || c == EOF);
// ...
} while (c != EOF);
This works, because you use the value of c only if it is not a delimiter. (The special value EOF is outside the valid range of (unsigned) chars and should not be inserted into strings or printed.)
stdout is not getting flushed because your last output didn't contain a newline...
Change this line
printf("|");
to
printf("|\n");
I got some help earlier fixing up one of the functions I am using in this program, but now I'm at a loss of logic.
I have three purposes and two functions in this program. The first purpose is to print a sentence that the user inputs backwards. The second purpose is to check if any of the words are anagrams with another in the sentence. The third purpose is to check if any one word is a palindrome.
I successfully completed the first purpose. I can print sentences backwards. But now I am unsure of how I should implement my functions to check whether or not any words are anagrams or palindromes.
Here's the code;
/*
* Ch8pp14.c
*
* Created on: Oct 12, 2013
* Author: RivalDog
* Purpose: Reverse a sentence, check for anagrams and palindromes
*/
#include <stdio.h>
#include <ctype.h> //Included ctype for tolower / toupper functions
#define bool int
#define true 1
#define false 0
//Write boolean function that will check if a word is an anagram
bool check_anagram(char a[], char b[])
{
int first[26] = {0}, second[26] = {0}, c = 0;
// Convert arrays into all lower case letters
while(a[c])
{
a[c] = (tolower(a[c]));
c++;
}
c = 0;
while(b[c])
{
b[c] = (tolower(b[c]));
c++;
}
c = 0;
while (a[c] != 0)
{
first[a[c]-'a']++;
c++;
}
c = 0;
while (b[c] != 0)
{
second[b[c]-'a']++;
c++;
}
for (c = 0; c < 26; c++)
{
if (first[c] != second[c])
return false;
}
return true;
}
//Write boolean function that will check if a word is a palindrome
bool palindrome(char a[])
{
int c=0, j, k;
//Convert array into all lower case letters
while (a[c])
{
a[c] = (tolower(a[c]));
c++;
}
c = 0;
j = 0;
k = strlen(a) - 1;
while (j < k)
{
if(a[j++] != a[k--])
return false;
}
return true;
}
int main(void)
{
int i = 0, j = 0, k = 0;
char a[80], terminator;
//Prompt user to enter sentence, store it into an array
printf("Enter a sentence: ");
j = getchar();
while (i < 80)
{
a[i] = j;
++i;
j = getchar();
if (j == '!' || j == '.' || j == '?')
{
terminator = j;
break;
}
else if(j == '\n')
{
break;
}
}
while(a[k])
{
a[k] = (tolower(a[k]));
k++;
}
k = 0;
while(k < i)
{
printf("%c", a[k]);
k++;
}
printf("%c\n", terminator);
//Search backwards through the loop for the start of the last word
//print the word, and then repeat that process for the rest of the words
for(j = i; j >= 0; j--)
{
while(j > -1)
{
if (j == 0)
{
for(k=j;k<i;k++)
{
printf("%c", a[k]);
}
printf("%c", terminator);
break;
}
else if (a[j] != ' ')
--j;
else if (a[j] == ' ')
{
for(k=j+1;k<i;k++)
{
printf("%c", a[k]);
}
printf(" ");
break;
}
}
i = j;
}
//Check if the words are anagrams using previously written function
for( i = 0; i < 80; i++)
{
if (a[i] == ' ')
{
}
}
//Check if the words are palindromes using previously written function
return 0;
}
I was thinking that perhaps I could again search through the array for the words by checking if the element is a space, and if it is, store from where the search started to the space's index-1 in a new array, repeat that process for the entire sentence, and then call my functions on all of the arrays. The issue I am seeing is that I can't really predict how many words a user will input in a sentence... So how can I set up my code to where I can check for anagrams/palindromes?
Thank you everyone!
~RivalDog
Would be better,if you first optimize your code and make it readable by adding comments.Then you can divide the problem in smaller parts like
1.How to count words in a string?
2.How to check whether two words are anagrams?
3.How to check whether a word is palindrome or not?
And these smaller programs you could easily get by Googling. Then your job will be just to integrate these answers. Hope this helps.
To check anagram, no need to calculate number of words and comparing them one by one or whatever you are thinking.
Look at this code. In this code function read_word() is reading word/phrase input using an int array of 26 elements to keep track of how many times each letter has been seen instead of storing the letters itself. Another function equal_array() is to check whether both array a and b (in main) are equal (anagram) or not and return a Boolean value as a result.
#include <stdio.h>
#include <ctype.h>
#include <stdbool.h>
void read_word(int counts[26]);
bool equal_array(int counts1[26],int counts2[26]);
int main()
{
int a[26] = {0}, b[26] = {0};
printf("Enter first word/phrase: ");
read_word(a);
printf("Enter second word/phrase: ");
read_word(b);
bool flag = equal_array(a,b);
printf("The words/phrase are ");
if(flag)
printf("anagrams");
else
printf("not anagrams");
return 0;
}
void read_word(int counts[26])
{
int ch;
while((ch = getchar()) != '\n')
if(ch >= 'A' && ch <= 'Z' || ch >= 'a' && ch <= 'z')
counts[toupper(ch) - 'A']++;
}
bool equal_array(int counts1[26],int counts2[26])
{
int i = 0;
while(i < 26)
{
if(counts1[i] == counts2[i])
i++;
else
break;
}
return i == 26 ? true : false;
}