I want to write a function that finds the biggest index of the character in a string. But when the index is already found once, it may not return it.
For example:
word: "hello"
character: l
indexset={};
expected return value = 3
when I do this function again:
word: "hello"
character: l
indexset={3};
expected return value = 2
This is how I allocated spaces for my indexset:
*indexset = (int*) malloc(strlen(am->transposition)*sizeof(int));
In my code, that's enough space for each index of each character.
My function to find the biggest index:
int findBiggestIndex(char karakter,char* woord,int* indexset, int *size){
int i=0,index;
for(i=0;i<strlen(woord);i++){
if(karakter==woord[i] && !inArray(indexset,i,*size)){
index=i;
printf("%d",*size);
indexset[*size]=index;
(*size)++;
}
}
return index;
}
And the method inArray():
#define TRUE 1
#define FALSE 0
int inArray(int* arr, int a, int size){
int i=0;
for(i=0;i<size;i++){
if(arr[i]==a){
return TRUE;
}
return FALSE;
}
}
The method inArray is also declared before findBiggestIndex so that's not the problem.
Please help
I want to write a function that finds the biggest index of the character in a string.
If you want to find the last occurrence of a character in a string...
loop through string and compare char elements:
Something like:
int findLastIndex(char *str, char c)
{
int keep = 0, i, len;
if(str)
{
len = strlen(str);
for(i=0;i<len ;i++)
{
if(str[i] == c) keep = i;
}
return keep;//0 means nothing found
}
return -1;//means problem with input string
}
int main(void)
{
int cnt= findLastIndex("only 1 g in line", 'g');
//cnt should return as 7 for these inputs
cnt= findLastIndex("2 g's in line(g)", 'g');
//cnt should return as 14 for these inputs
return 0;
}
literal string arguments:
|0|1|2|3|4|5|6|7|8|9|0|1|2|3|4|5|
|o|n|l|y| |1| |g| |i|n| |l|i|n|e|
^
|2| |g|'|s| |i|n| |l|i|n|e|(|g|)|
^
Therefore, should see index as 7 for first arg, 14 for second arg
Related
I know how to print a single string vertically.
char test[100] = "test";
int i;
for(i=0;i<strlen(test);i++){
printf("%c\n",test[i]);
}
Which will give me:
t
e
s
t
But how can I print an array of strings vertically? For example:
char listOfTest[2][10] = {"testing1","quizzing"};
So it can return:
tq
eu
si
tz
iz
gi
1g
Simply loop through the first string and print each character at index i in the first string and the second string till you reach the null terminator of first string
NOTE: this only work when string 1 and string2 are equal in length and will need modification for other test cases
#include <stdio.h>
int main(void)
{
char listOfTest[2][10] = {"testing1","quizzing"};
int i = 0;
//loop through string 1 till NULL is reach
while (listOfTest[0][i])
{
//prints char at index i in string 1 and 2
printf("%c%c\n", listOfTest[0][i], listOfTest[1][i]);
//increment the index value
i++;
}
return (0);
}
Simply print a character from both strings.
(Better to test for the null character rather than repeatedly call strlen().)
for(i = 0; listOfTest[0][i] && listOfTest[1][i]; i++) {
printf("%c%c\n", listOfTest[0][i], listOfTest[1][i]);
}
To extend to n strings ...
size_t num_of_strings = sizeof listOfTest/sizeof listOfTest[0];
bool done = false;
for (size_t i = 0; listOfTest[0][i] && !done; i++) {
for (size_t n = 0; n < num_of_strings; n++) {
if (listOfTest[n][i] == '\0') {
done = true;
break;
}
printf("%c", listOfTest[n][i]);
}
printf("\n");
}
This program, tokenizes a user input string, removes extra spaces and saves each word into a 2D array and then print the tokens
EXAMPLE:
input: " Hello world string house and car"
output and EXPECTED output:
token[0]: Hello
token[1]: world
token[2]: string
token[3]: house
token[4]: and
token[5]: car
THE PROBLEM:
the problem is that I achieved this by using strlen() function when printing the tokens(code located at the very bottom), I am not supposed to use any other library than stdio.h and stdlib.h, since strlen() function is defined in string.h i tried to use sizeof(arr) / sizeof(arr[0]); but it does not work as I want, the result using sizeof is :
token[0]: Hello
token[1]: world
token[2]: string
token[3]: house
token[4]: and
token[5]: car
�oken[6]: ��
token[7]: �
token[8]: ����
token[9]: �
token[10]:
I WOULD LIKE TO HAVE THE EXPECTED OUTPUT WITHOUT USING STRLEN()
#include<stdio.h>
#include <stdlib.h>
#define TRUE 1
char tokenize(char *str, char array[10][20])
{
int n = 0, i, j = 0;
for(i = 0; TRUE; i++)//infinite loop until is the end of the string '\0'
{
if(str[i] != ' '){
//position 1, char 1
array[n][j++] = str[i];// if, it is not space, we save the character
}
else{
array[n][j++] = '\0';//end of the first word
n++;// position for next new word
j=0;// start writting char at position 0
}
if(str[i] == '\0')
break;
}
return 0;
}
//removes extra spaces
char* find_word_start(char* str){
/*also removes all extra spaces*/
char *result = (char*) malloc(sizeof(char) *1000);
int c = 0, d = 0;
// no space at beginning
while(str[c] ==' ') {
c++;
}
while(str[c] != '\0'){ // till end of sentence
result[d++] = str[c++]; //take non-space characters
if(str[c]==' ') { // take one space between words
result[d++] = str[c++];
}
while(str[c]==' ') { //
c++;
}
}
result[d-1] = '\0';
//print or return char?
return result;
free(result);
}
int main()
{
char str[]=" Hello world string dudes and dudas ";
//words, and chars in each word
char arr[10][20];
//call the method to tokenize the string
tokenize(find_word_start(str),arr);
int row = sizeof(arr) / sizeof(arr[0]);
/*----------------------------------------------------------------------*/
/*----------------------------------------------------------------------*/
for(int i = 0;i <= strlen(arr);i++)
/*----------------------------------------------------------------------*/
/*----------------------------------------------------------------------*/
printf("token[%d]: %s\n", i, arr[i]);
return 0;
}
Your code using strlen() may appear the work in this instance but it is not correct.
strlen(arr) makes no semantic sense because arr is not a string. It happens in this case to return 5 because arr has the same address as arr[0], then you kludged it to work for the 6 word output by using the test i <= strlen(arr) in the for loop. The two values strlen(arr) and the number of strings stored in arr are not related.
The expression sizeof(arr) / sizeof(arr[0]) determines the run-time constant number arrays within the array of arrays arr (i.e. 10), not the number of valid strings assigned. It is your code's responsibility to keep track of that either with a sentinel value such as an empty string, or by maintaining a count of strings assigned.
I suggest you change tokenize to return the number of strings (currently it is inexplicably defined to return a char, but in fact only ever rather uselessly returns zero):
int tokenize( char* str, char array[][20] )
{
...
return n ;
}
Then:
int rows = tokenize( find_word_start(str), arr ) ;
for( int i = 0; i < rows; i++ )
{
printf( "token[%d]: %s\n", i, arr[i] ) ;
}
How can I complete the function canArrangeWords() ?
Question : Given a set of words check if we can arrange them in a list such that the last letter of any word and first letter of another word are same. The input function canArrangeWords shall contain an integer num and array of words arr. num denotes the number of word in the list (1<=num<=100). arr shall contain words consisting of lower case letters between 'a' - 'z' only . return 1 if words can be arranged in that fashion and -1 if cannot.
Input : 4 pot ten nice eye
output : 1
input : 3 fox owl pond
output: -1
Please help me complete this program .
**
#include<stdio.h>
#include<string.h>
int canArrangewords(int,char [100][100]);
void main(){
int n ,count=0 , i ;
char arrayS[100][100];
scanf("%d",&n);
for (i = 0; i < n; ++i)
{
scanf("%s",arrayS[i]);
}
for(i=0;i<n;i++)
{
printf("%s",arrayS[i]);
printf("\n");
}
printf("%c\n",arrayS[2][4]);
canArrangewords(n , arrayS);
}
int canArrangewords(int n,char arrayS[100][100]){
int i , j ;
for ( i = 0; i < n; i++)
{
for ( j = i+1 ; j < strlen(arrayS[j+1]); i++)
{
int flag = strlen(arrayS[j+1]) - 1;
int temp = strcmp(arrayS[i][0],arrayS[j][flag]);
}
}
}
}
Well, first of all think of the way you can reach that answer.
If you only need to know if they can or can not be arranged and you do not have to do so your self you can use an empty array of int array[26] for each letter a-z.
The rule is that from all the first and last letters for all the words only two MAY appear an odd amount of times - the first letter of first word in list and the last letter in the last word in the list, the rest MUST appear an even amount of times. I would add a check to make sure the letters are lowercase as well. good luck!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MINASCII 97
#define LETTERS 26
void UpdateArray(char letter, int* arr)
{
if(arr[letter - MINASCII] == 0)
{
++arr[letter - MINASCII];
}
else
{
--arr[letter - MINASCII];/*for each second time same letter is seen reduce back to zero */
}
}
int canArrangewords(int wordNum, char* wordArr[])
{
int arr[LETTERS] = {0};
int i = 0;
int count = 0 ;
char first;
char last;
char* string;
for (i= 0; i< wordNum; ++i)
{
string = wordArr[i];
first = string[0];
last = string[strlen(string)-1];
UpdateArray(first, &arr[0]);
UpdateArray(last, &arr[0]);
}
for(i = 0; i< LETTERS; ++i)
{
count+=arr[i];
}
if(count == 2 || count == 0)/*either once each or twice -see word1 example in main*/
{
return 1;
}
return -1;
}
int main()
{
int i = 0;
char* words[] = {"pot", "ten", "nice", "eye"};
char* words1[] = {"pot", "ten", "nip"};
char* words2[] = {"fox", "owl", "pond"};
i = canArrangewords(4,words);
printf("%d\n", i);
i = canArrangewords(3,words1);
printf("%d\n", i);
i = canArrangewords(3,words2);
printf("%d\n", i);
return 0;
}
Change your array of words into an array of pointers to words. Then you can simply exchange the pointers.
To speed things up, instead of a pointer to a word, have it point to a structure:
struct WORD {
char *firstchar; // begin of word
char *lastchar; // last char of word
} *words[100]; // array of 100 pointers to words
To read the words:
char buf[100];
for (i = 0; i < n; ++i)
{
scanf("%s",buf);
int len= strlen(buf);
words[i]= malloc(sizeof(struct WORDS));
words[i]->firstchar= malloc(len+1);
strcpy(words[i]->firstchar, buf);
words[i]->lastchar= words[i]->firstchar + len-1;
}
Now compare and sort:
if (*words[i]->lastchar == *words[j]->firstchar) {
struct WORDS *tmp= words[i+1];
words[i+1]= words[j];
words[j]= tmp;
}
Do this in a loop, a kind of bubble sort. I leave that to you.
For an assignment in class, we have been instructed to write a program which takes a string and a delimiter and then takes "words" and stores them in a new array of strings. i.e., the input ("my name is", " ") would return an array with elements "my" "name" "is".
Roughly, what I've attempted is to:
Use a separate helper called number_of_delimeters() to determine the size of the array of strings
Iterate through the initial array to find the number of elements in a given string which would be placed in the array
Allocate storage within my array for each string
Store the elements within the allocated memory
Include directives:
#include <stdlib.h>
#include <stdio.h>
This is the separate helper:
int number_of_delimiters (char* s, int d)
{
int numdelim = 0;
for (int i = 0; s[i] != '\0'; i++)
{
if (s[i] == d)
{
numdelim++;
}
}
return numdelim;
}
`This is the function itself:
char** split_at (char* s, char d)
{
int numdelim = number_of_delimiters(s, d);
int a = 0;
int b = 0;
char** final = (char**)malloc((numdelim+1) * sizeof(char*));
for (int i = 0; i <= numdelim; i++)
{
int sizeofj = 0;
while (s[a] != d)
{
sizeofj++;
a++;
}
final[i] = (char*)malloc(sizeofj);
a++;
int j = 0;
while (j < sizeofj)
{
final[i][j] = s[b];
j++;
b++;
}
b++;
final[i][j+1] = '\0';
}
return final;
}
To print:
void print_string_array(char* a[], unsigned int alen)
{
printf("{");
for (int i = 0; i < alen; i++)
{
if (i == alen - 1)
{
printf("%s", a[i]);
}
else
{
printf("%s ", a[i]);
}
}
printf("}");
}
int main(int argc, char *argv[])
{
print_string_array(split_at("Hi, my name is none.", ' '), 5);
return 0;
}
This currently returns {Hi, my name is none.}
After doing some research, I realized that the purpose of this function is either similar or identical to strtok. However, looking at the source code for this proved to be little help because it included concepts we have not yet used in class.
I know the question is vague, and the code rough to read, but what can you point to as immediately problematic with this approach to the problem?
The program has several problems.
while (s[a] != d) is wrong, there is no delimiter after the last word in the string.
final[i][j+1] = '\0'; is wrong, j+1 is one position too much.
The returned array is unusable, unless you know beforehand how many elements are there.
Just for explanation:
strtok will modify the array you pass in! After
char test[] = "a b c ";
for(char* t = test; strtok(t, " "); t = NULL);
test content will be:
{ 'a', 0, 'b', 0, 'c', 0, 0 }
You get subsequently these pointers to your test array: test + 0, test + 2, test + 4, NULL.
strtok remembers the pointer you pass to it internally (most likely, you saw a static variable in your source code...) so you can (and must) pass NULL the next time you call it (as long as you want to operate on the same source string).
You, in contrast, apparently want to copy the data. Fine, one can do so. But here we get a problem:
char** final = //...
return final;
void print_string_array(char* a[], unsigned int alen)
You just return the array, but you are losing length information!
How do you want to pass the length to your print function then?
char** tokens = split_at(...);
print_string_array(tokens, sizeof(tokens));
will fail, because sizeof(tokens) will always return the size of a pointer on your local system (most likely 8, possibly 4 on older hardware)!
My personal recommendation: create a null terminated array of c strings:
char** final = (char**)malloc((numdelim + 2) * sizeof(char*));
// ^ (!)
// ...
final[numdelim + 1] = NULL;
Then your print function could look like this:
void print_string_array(char* a[]) // no len parameter any more!
{
printf("{");
if(*a)
{
printf("%s", *a); // printing first element without space
for (++a; *a; ++a) // *a: checking, if current pointer is not NULL
{
printf(" %s", *a); // next elements with spaces
}
}
printf("}");
}
No problems with length any more. Actually, this is exactly the same principle C strings use themselves (the terminating null character, remember?).
Additionally, here is a problem in your own code:
while (j < sizeofj)
{
final[i][j] = s[b];
j++; // j will always point behind your string!
b++;
}
b++;
// thus, you need:
final[i][j] = '\0'; // no +1 !
For completeness (this was discovered by n.m. already, see the other answer): If there is no trailing delimiter in your source string,
while (s[a] != d)
will read beyond your input string (which is undefined behaviour and could result in your program crashing). You need to check for the terminating null character, too:
while(s[a] && s[a] != d)
Finally: how do you want to handle subsequent delimiters? Currently, you will insert empty strings into your array? Print out your strings as follows (with two delimiting symbols - I used * and + like birth and death...):
printf("*%s+", *a);
and you will see. Is this intended?
Edit 2: The variant with pointer arithmetic (only):
char** split_at (char* s, char d)
{
int numdelim = 0;
char* t = s; // need a copy
while(*t)
{
numdelim += *t == d;
++t;
}
char** final = (char**)malloc((numdelim + 2) * sizeof(char*));
char** f = final; // pointer to current position within final
t = s; // re-assign t, using s as start pointer for new strings
while(*t) // see above
{
if(*t == d) // delimiter found!
{
// can subtract pointers --
// as long as they point to the same array!!!
char* n = (char*)malloc(t - s + 1); // +1: terminating null
*f++ = n; // store in position pointer and increment it
while(s != t) // copy the string from start to current t
*n++ = *s++;
*n = 0; // terminate the new string
}
++t; // next character...
}
*f = NULL; // and finally terminate the string array
return final;
}
While I've now been shown a more elegant solution, I've found and rectified the issues in my code:
char** split_at (char* s, char d)
{
int numdelim = 0;
int x;
for (x = 0; s[x] != '\0'; x++)
{
if (s[x] == d)
{
numdelim++;
}
}
int a = 0;
int b = 0;
char** final = (char**)malloc((numdelim+1) * sizeof(char*));
for (int i = 0; i <= numdelim; i++)
{
int sizeofj = 0;
while ((s[a] != d) && (a < x))
{
sizeofj++;
a++;
}
final[i] = (char*)malloc(sizeofj);
a++;
int j = 0;
while (j < sizeofj)
{
final[i][j] = s[b];
j++;
b++;
}
final[i][j] = '\0';
b++;
}
return final;
}
I consolidated what I previously had as a helper function, and modified some points where I incorrectly incremented .
Im trying to find character frequency in a string,i wrote the following code,but it does'nt show any output.All im trying is to fill the character array with respective counts.
When i tried to debug,it some how gives output,but prints some garbage value.
#include<stdio.h>
/* Program for printing character frequency in string */
charcount(char *,int *);
int main()
{
int n,i=0;
printf("Enter n :");
scanf("%d",&n);
char var[n];
int count[100]; // array for storing character frequency
printf("Enter string :");
fflush(stdin);
scanf("%s",var);
charcount(var,count); // calling frequeny function
for(i=0;i<strlen(count);i++)
{
printf("%d\n",count[i]);
}
getch();
return 0;
}
charcount(char *p,int *q)
{
for(;*p;p++)
{
q[*p]++;
}
}
You have few problems in your code:
count array is not initialized.
You are applying strlen() on an integer array.
count array should be 256 (count[256]) to cover all possible ascii chars. For example, if your input is abcd you'll go out of bound of array as d is 100.
You are printing the wrong count:
printf("%d\n",count[i]); should be printf("%d\n",count[var[i]]);
Declare proper prototype for charcount().
After fixing these:
#include<stdio.h>
/* Program for printing character frequency in string */
void charcount(char *,int *);
int main()
{
int n,i=0;
printf("Enter n :");
scanf("%d",&n);
char var[n];
int count[256]={0}; // array for storing character frequency
printf("Enter string :");
scanf("%s",var);
charcount(var,count); // calling frequeny function
for(i=0;i<strlen(var);i++)
{
printf("%d\n",count[var[i]]);
}
return 0;
}
void charcount(char *p,int *q)
{
for(;*p;p++)
{
q[*p]++;
}
}
Make sure to compile in C99 or C11 (e.g. gcc -std=c99 file.c) mode as VLAs are not supported in earlier standards of C.
You need to initialize your count array. Otherwise it will have garbage values in it by default. You can initialize the whole array to 0 like so:
int count[100] = {0};
Your count array may not be large enough to hold all printable values (even assuming ASCII), and it should be 0 initialized. Your for loop should be checking against the length of var, not count, since you cannot sensibly treat the count integer array as a string.
int count[1<<CHAR_BIT] = {};
/*...*/
for(i=0;i<strlen(var);i++)
{
printf("%d\n",count[var[i]]);
}
Well, it really depends on what you want to output. If you intend to output all of count, then:
for(i=0;i<sizeof(count)/sizeof(count[0]);i++)
{
printf("%d\n",count[i]);
}
int count is nothing but a hashmap
Your code will not work for this string "abcd"
count['a'] = val // Works fine ASCII value of a is 97
count['b'] = val // Works fine ASCII value of a is 98
count['c'] = val // Works fine ASCII value of a is 99
count['d'] = val ; // Undefined Behaviour ASCII value of d is 100
The size should be equal to ASCII set length
int count[128] = {};
JAVA program to print "*" as much times as the occurrence of a character in String.
Or char_in_String : frequency of char_in_String
Instead of * you can print frequency count
public class CharFreq
{
public static void main(String[] args)
{
String s = "be yourself "; System.out.println(s);
int r=0;
char[] str = s.toCharArray();
for (int i = 0; i < str.length; i++)
{
int cnt = 0;
if (str[i] != ' ')
{
for (int j = 0; j < str.length; j++)
{
if (str[i] == str[j])
{
cnt++; r=j;
}
}
if(i==r)
{
System.out.print(str[i] + ":");
for (int k = 1; k <=cnt; k++)
System.out.print("*");
System.out.println();
}
}
}
}
}
Output:
be yourself
b:*
y:*
o:*
u:*
r:*
s:*
e:**
l:*
f:*
Your count[100] is not large enough. Assume you only enter "A - Z" or "a - z" it's still not large enough because 'z' is 122, then your charcount will increase count[122].
You should consider change int count[100] to int count[128] = { 0 }