#include<stdio.h>
#include<unistd.h>
void ft_putchar(char x){
write(1, &x, 1);
}
void ft_print_comb()
{
char i, j, k;
i = '0';
while(i <= 7){
i++;
j = i+1;
while(j <= 8){
j++;
k = j+1;
while(k <= 9){
k++;
ft_putchar(i);
ft_putchar(j);
ft_putchar(k);
ft_putchar(',');
ft_putchar(' ');
}
}
}
}
int main(){
ft_print_comb();
return 0;
}
I have tried to do couple changes but it either broke the code or kept giving me no output. What I am trying to do is create a function that displays all different combinations of three different digits in
ascending order, listed by ascending order. for loop and printf functions are not allowed.
Since you are using chars, you should compare with character literals instead of integers. For instance, the while loop is never entered because the ASCII code for '0' is 48, which is greater than 7.
while (i <= '7') {
j = i + 1;
while (j <= '8') {
k = j + 1;
while (k <= '9') {
ft_putchar(i);
ft_putchar(j);
ft_putchar(k);
ft_putchar(',');
ft_putchar(' ');
k++;
}
j++;
}
i++;
}
i is a character with a value of 48 (the ASCII code of '0') so the while loop is never entered. Set i this way:
i = 0;
Related
I'm writing an entab program which is K&R exercise.
Exercise 1-21. Write a program entab that replaces strings of blanks by the minimum number of tabs and blanks to achieve the same spacing.
My solution is dividing each line by N character(where 1 tab == N whitespace) and checking from the rightmost character to the first character whether it is whitespace character or not.
#include <stdio.h>
void entab(int n);
int main(void)
{
entab(8);
}
void entab(int n) // n should be at least 2
{
int c;
int i, j;
int temp[n];
while (1) {
// Collect n characters
// If there is EOF, print all privious character and terminate the while loop.
// If there is new line character, print all privious and current character, then continue the while loop again.
for (i = 0; i < n; ++i) {
temp[i] = getchar();
if (temp[i] == EOF) {
if (i > 0)
for (j = 0; j < i; ++j)
putchar(temp[j]);
break;
}
else if (temp[i] == '\n') {
for (j = 0; j <= i; ++j)
putchar(temp[j]);
break;
}
}
if (temp[i] == EOF)
break;
if (temp[i] == '\n')
continue;
// the temp array has exactly 8 characters without newline character and EOF indicator.
// check a continuity of the whitespace character and put the tab character at the right space.
if (temp[n-1] != ' ') {
for (i = 0; i <= n-1; ++i)
putchar(temp[i]);
}
for (j = n-2; j >= 0; --j) {
else if (temp[j] != ' ') {
for (i = 0; i <= j; ++i)
putchar(temp[i]);
putchar('\t');
}
}
else
putchar('\t');
}
}
This part makes sense but is syntactically not allowed. But I have to use a loop syntax because I want the program to entab a arbitrary number of the whitespace character.
if (temp[n-1] != ' ') {
for (i = 0; i <= n-1; ++i)
putchar(temp[i]);
}
for (j = n-2; j >= 0; --j) {
else if (temp[j] != ' ') {
for (i = 0; i <= j; ++i)
putchar(temp[i]);
putchar('\t');
}
}
else
putchar('\t');
I tried a function-like macro but it didn't make significant difference. Is it just impossible? Should I consider another design or logic?
so this is my function. My main focus is to get the character frequencies and the highest character frequency.
The function below (get_letter_frequencies) is supposed to get a string example ("I am a big boy") and return the character frequencies and the highest character frequency.
The Function should return
i - 2
a - 2
m - 1
b - 2
g - 1
o - 1
y - 1
Highest character frequency would be " iab "
My problem is with the get_letter_frequencies function. What should I arrange from the function in order to return the above output?
void get_letter_frequencies(const char *text, size_t len, int freq[26], int *max_freq)
{
for(int i = 0; i<len; i++)
{
if(text[i] != ' ' || !(is_sentence_terminator(text[i]))) //this condition is set in order to ignore the spaces and the sentence terminators (! ? .)
{
if(text[i] >= 'a' && text[i] <= 'z')
{
freq[text[i] - 'a']++;
}
}
}
for(int j = 0; j < 26; j++)
{
if(freq[j] >= 1)
{
*max_freq = freq[j];
}
}
This function below(is_sentence_terminator). Here the function checks whether the sentence finishes with a " ! ? or . " if it does not finish with one of the terminators then it is not a sentence and ignores it.
int is_sentence_terminator(char ch)
{
if(ch == 33 || ch == 46 || ch == 63)
{
return 1;
}else
{
return 0;
}
}
There are some issues in your code:
there is no need to test for special characters, comparing text[i] to 'a' and 'z' is sufficient for ASCII systems.
in the second loop, you should update *max_freq only if freq[j] is greater than the current value, not 1. *max_freq should be initialized to 0 before the loop.
In the calling code, you would also
print the letters whose frequency is non 0.
print all letters with the maximum frequency using one final loop.
Here is a modified version:
void get_letter_frequencies(const char *text, size_t len, int freq[26], int *max_freq) {
for (int i = 0; i < 26; i++)
freq[i] = 0;
for (int i = 0; i < len; i++) {
if (text[i] >= 'a' && text[i] <= 'z') {
freq[text[i] - 'a']++; // assuming ASCII
}
}
*max_freq = 0;
for (int i = 0; i < 26; i++) {
if (*max_freq < freq[i]) {
*max_freq = freq[i];
}
}
}
Trying to write a c programm that writes a combination of 3 set of numbers and avoid reputation, not very good in C language. Something is wrong with one of my condition
int ft_putchar(char c){
write(1, &c, 1);
}
void ft_print_comb(void){
int numbers[3] = {48, 48, 48};
while(numbers[0] <= 55){
if((numbers[0] < numbers[1]) && (numbers[1] < numbers[2])){
ft_putchar(numbers[0]);
ft_putchar(numbers[1]);
ft_putchar(numbers[2]);
if(numbers[0] != 55){
ft_putchar(',');
}
if(numbers[0] != 55){
ft_putchar(' ');
}
if(numbers[2]++ >= 57){
numbers[2] = 48;
numbers[1]++;
}
if(numbers[1] >= 57){
numbers[1] = 48;
numbers[0]++;
}
}
}
}
int main(void){
ft_print_comb();
}
The issue is you create an infinite loop. Since all 3 numbers start out equal, the condition
if((numbers[0] < numbers[1]) && (numbers[1] < numbers[2]))
is false, so the body of the if never gets entered, numbers[0] never gets incremented, and while(numbers[0] <= 55){ is always true.
You can get (what I think is) the desired output with nested for loops:
int numbers[3] = {'0', '0', '0'};
for (int i = numbers[0]; i <= '9'; i++) {
for (int j = numbers[1]; j <= '9'; j++) {
for (int k = numbers[2]; k <= '9'; k++) {
if (i < j && j < k) {
printf("%c%c%c, ", i, j, k);
}
}
}
}
My program is designed to allow the user to input a string and my program will output the number of occurrences of each letters and words. My program also sorts the words alphabetically.
My issue is: I output the words seen (first unsorted) and their occurrences as a table, and in my table I don't want duplicates. SOLVED
For example, if the word "to" was seen twice I just want the word "to" to appear only once in my table outputting the number of occurrences.
How can I fix this? Also, why is it that i can't simply set string[i] == delim to apply to every delimiter rather than having to assign it manually for each delimiter?
Edit: Fixed my output error. But how can I set a condition for string[i] to equal any of the delimiters in my code rather than just work for the space bar? For example on my output, if i enter "you, you" it will out put "you, you" rather than just "you". How can I write it so it removes the comma and compares "you, you" to be as one word.
Any help is appreciated. My code is below:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
const char delim[] = ", . - !*()&^%$##<> ? []{}\\ / \"";
#define SIZE 1000
void occurrences(char s[], int count[]);
void lower(char s[]);
int main()
{
char string[SIZE], words[SIZE][SIZE], temp[SIZE];
int i = 0, j = 0, k = 0, n = 0, count;
int c = 0, cnt[26] = { 0 };
printf("Enter your input string:");
fgets(string, 256, stdin);
string[strlen(string) - 1] = '\0';
lower(string);
occurrences(string, cnt);
printf("Number of occurrences of each letter in the text: \n");
for (c = 0; c < 26; c++){
if (cnt[c] != 0){
printf("%c \t %d\n", c + 'a', cnt[c]);
}
}
/*extracting each and every string and copying to a different place */
while (string[i] != '\0')
{
if (string[i] == ' ')
{
words[j][k] = '\0';
k = 0;
j++;
}
else
{
words[j][k++] = string[i];
}
i++;
}
words[j][k] = '\0';
n = j;
printf("Unsorted Frequency:\n");
for (i = 0; i < n; i++)
{
strcpy(temp, words[i]);
for (j = i + 1; j <= n; j++)
{
if (strcmp(words[i], words[j]) == 0)
{
for (a = j; a <= n; a++)
strcpy(words[a], words[a + 1]);
n--;
}
} //inner for
}
i = 0;
/* find the frequency of each word */
while (i <= n) {
count = 1;
if (i != n) {
for (j = i + 1; j <= n; j++) {
if (strcmp(words[i], words[j]) == 0) {
count++;
}
}
}
/* count - indicates the frequecy of word[i] */
printf("%s\t%d\n", words[i], count);
/* skipping to the next word to process */
i = i + count;
}
printf("ALphabetical Order:\n");
for (i = 0; i < n; i++)
{
strcpy(temp, words[i]);
for (j = i + 1; j <= n; j++)
{
if (strcmp(words[i], words[j]) > 0)
{
strcpy(temp, words[j]);
strcpy(words[j], words[i]);
strcpy(words[i], temp);
}
}
}
i = 0;
while (i <= n) {
count = 1;
if (i != n) {
for (j = i + 1; j <= n; j++) {
if (strcmp(words[i], words[j]) == 0) {
count++;
}
}
}
printf("%s\n", words[i]);
i = i + count;
}
return 0;
}
void occurrences(char s[], int count[]){
int i = 0;
while (s[i] != '\0'){
if (s[i] >= 'a' && s[i] <= 'z')
count[s[i] - 'a']++;
i++;
}
}
void lower(char s[]){
int i = 0;
while (s[i] != '\0'){
if (s[i] >= 'A' && s[i] <= 'Z'){
s[i] = (s[i] - 'A') + 'a';
}
i++;
}
}
I have the solution to your problem and its name is called Wall. No, not the type to bang your head against when you encounter a problem that you can't seem to solve but for the Warnings that you want your compiler to emit: ALL OF THEM.
If you compile C code with out using -Wall then you can commit all the errors that people tell you is why C is so dangerous. But once you enable Warnings the compiler will tell you about them.
I have 4 for your program:
for (c; c< 26; c++) { That first c doesn't do anything, this could be written for (; c < 26; c++) { or perhaps beter as for (c = 0; c <26; c++) {
words[i] == NULL "Statement with no effect". Well that probably isn't what you wanted to do. The compiler tells you that that line doesn't do anything.
"Unused variable 'text'." That is pretty clear too: you have defined text as a variable but then never used it. Perhaps you meant to or perhaps it was a variable you thought you needed. Either way it can go now.
"Control reaches end of non-void function". In C main is usually defined as int main, i.e. main returns an int. Standard practice is to return 0 if the program successfully completed and some other value on error. Adding return 0; at the end of main will work.
You can simplify your delimiters. Anything that is not a-z (after lower casing it), is a delimiter. You don't [need to] care which one it is. It's the end of a word. Rather than specify delimiters, specify chars that are word chars (e.g. if words were C symbols, the word chars would be: A-Z, a-z, 0-9, and _). But, it looks like you only want a-z.
Here are some [untested] examples:
void
scanline(char *buf)
{
int chr;
char *lhs;
char *rhs;
char tmp[5000];
lhs = tmp;
for (rhs = buf; *rhs != 0; ++rhs) {
chr = *rhs;
if ((chr >= 'A') && (chr <= 'Z'))
chr = (chr - 'A') + 'a';
if ((chr >= 'a') && (chr <= 'z')) {
*lhs++ = chr;
char_histogram[chr] += 1;
continue;
}
*lhs = 0;
if (lhs > tmp)
count_string(tmp);
lhs = tmp;
}
if (lhs > tmp) {
*lhs = 0;
count_string(tmp);
}
}
void
count_string(char *str)
{
int idx;
int match;
match = -1;
for (idx = 0; idx < word_count; ++idx) {
if (strcmp(words[idx],str) == 0) {
match = idx;
break;
}
}
if (match < 0) {
match = word_count++;
strcpy(words[match],str);
}
word_histogram[match] += 1;
}
Using separate arrays is ugly. Using a struct might be better:
#define STRMAX 100 // max string length
#define WORDMAX 1000 // max number of strings
struct word {
int word_hist; // histogram value
char word_string[STRMAX]; // string value
};
int word_count; // number of elements in wordlist
struct word wordlist[WORDMAX]; // list of known words
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;
}