I'm trying to make a palindrome finder in C and I don't know where it is going wrong, no matter what I get the output false on the 2 different ways that I have tried to code this. I have only just started C (in the past week) so if you could explain things simply that'd be great, thanks!
//way1
#include <stdio.h>
int read_char() { return getchar(); }
void read_string(char* s, int size) { fgets(s, size, stdin); }
void print_char(int c) { putchar(c); }
void print_string(char* s) { printf("%s", s); }
int is_palin(char word[]) {
int m = 0;
int arr_len = sizeof(word) / sizeof(char); //change to char_index
int n = arr_len;
int t = 1;
if(n % 2 != 0) {
for (m=0; m < ((n-1)/2); m++) {
if(word[m] != word[n-m-2]) {
t = 0;
}
else {
t = 1;
}
}
}
else {
for (m=0; m < (n/2)-1; m++) {
if(word[m] != word[n-m-2]) {
t = 0;
}
else {
t = 1;
}
}
}
if(t == 1) {
return 1;
}
else {
return 0;
}
}
int main(void) {
char word[6] = "civic";
int arr_len = sizeof(word)/sizeof(char);
if (is_palin(word) == 1) {
printf("is palin\n");
}
else {
printf("is not palin\n");
}
printf(word);
printf("\n");
printf("%d\n", arr_len);
return 0;
}
////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////
//way2
#include <stdio.h>
int read_char() { return getchar(); }
void read_string(char* s, int size) { fgets(s, size, stdin); }
void print_char(int c) { putchar(c); }
void print_string(char* s) { printf("%s", s); }
int is_palin(char word[]) {
int m = 1;
int input_length = sizeof(word);
int j = input_length-1;
int i = 0;
for(i=0; i <= j; i++) {
if(word[i] != word[j]) {
m = 0;
j--;
}
}
if(m == 1) {
return 1;
}
else {
return 0;
}
}
int main(void) {
char word[6] = "civic";
int input_length = sizeof(word);
if (is_palin(word) == 1) {
printf("is palin\n");
}
else {
printf("is not palin\n");
}
printf(word);
printf("\n");
printf("%d\n", input_length);
return 0;
}
Please try this, it works fine.
#include <stdio.h>
int main( )
{
int flag = 0;
int length = 0;
int len2 = 0;
int i = 0;
char name[130];
char p[130];
char q[130];
printf( "please enter a name or sentence\n" );
scanf( "%[^\n]", name );
length = strlen( name );
len2 = length;
strcpy( p, name );
memset( q, '.', length ); // handy to debug comparaison
q[length] = '\0';
for ( i = 0; i < length; i++ )
{
q[--len2] = p[i];
}
printf( "\n p==%s", p );
printf( "\n q==%s", q );
getchar( );
if ( !strcmp( p, q ) )
flag = 1;
if ( flag == 1 )
printf( "\npalindrome\n" );
else
printf( "\nnot a palindrome\n" );
return 0;
}
Take a look at this code, that's how I have implemented it (remember to #include <stdbool.h> or it will not work):
for(i = 0; i < string_length; i++)
{
if(sentence[i] == sentence[string_lenght-1-i])
palindrome = true;
else
{
palindrome = false;
break;
}
}
Doing that it will check if your sentence is palindrome and, at the first occurence this is not true it will break the for loop. You can use something like
if(palindrome)
printf(..);
else
printf(..);
for a simple prompt for the user.
Example :
radar is palindrome
abba is palindrome
abcabc is not palindrome
Please , pay attention to the fact that
Abba
is not recognized as a palindrome due to the fact that ' A ' and 'a' have different ASCII codes :
'A' has the value of 65
'a' has the value of 97
according to the ASCII table. You can find out more here.
You can avoid this issue trasforming all the characters of the string to lower case characters.
You can do this including the <ctype.h> library and calling the function int tolower(int c); like that :
for ( ; *p; ++p) *p = tolower(*p);
or
for(int i = 0; str[i]; i++){
str[i] = tolower(str[i]);
}
Code by Earlz, take a look at this Q&A to look deeper into that.
EDIT : I made a simple program to do this, see if it can help you
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdlib.h>
#include <ctype.h>
void LowerCharacters(char *word, int word_lenth);
int main(void){
char *word = (char *) malloc(10);
bool palindrome = false;
if(word == 0)
{
printf("\nERROR : Out of memory.\n\n");
return 1;
}
printf("\nEnter a word to check if it is palindrome or not : ");
scanf("%s", word);
int word_length = strlen(word);
LowerCharacters(word,word_length);
for(int i = 0; i < word_length; i++)
{
if(word[i] == word[word_length-1-i])
palindrome = true;
else
{
palindrome = false;
break;
}
}
palindrome ? printf("\nThe word %s is palindrome.\n\n", word) : printf("\nThe word %s is not palindrome.\n\n", word);
free(word);
return 0;
}
void LowerCharacters(char *word, int word_length){
for(int i = 0; i < word_length; i++)
word[i] = tolower(word[i]);
}
Input :
Enter a word to check if it is palindrome or not : RadaR
Output :
The word radar is palindrome.
Related
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define FALSE 0
#define TRUE 1
int alphabetic(char *string )
{
int i, valid;
valid = TRUE;
for ( i = 0; i < strlen(string); i++ )
{
if ( toupper ( string[i] ) < 'A' || toupper (string[i] ) > 'Z' )
valid = FALSE;
}
return valid;
}
int main()
{
char c, inputarray[10], temp[10];
int i = 0;
strcpy(temp, inputarray);
printf("%s Please enter string>");
while ( ( c = getchar () ) != '\n')
{
if ( i < 9 )
inputarray[i] = c;
i++;
}
if ( i < 10 )
inputarray[i] = '\0';
else
{
inputarray[9] = '\0';
printf("String too long\n");
return;
}
printf("%s\n",inputarray);
if (! alphabetic (inputarray) )
{
printf("Invalid input");
}
if (strcmp(strrev(inputarray),temp) == 0 )
printf("Palindrome\n");
else
printf("Not palindrome\n");
}
Trying this and still getting 'not palindrome' when input is a palindrome. It says 'stack around inputarray corrupted' when I run the program. Any ideas on how to fix it so reads palindrome and stop the input array being corrupted.
Here is one possible implementation. I've tried to explain in comments as much as I can but feel free toleave a comment if there is something that's not clear.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>
int alphabetic(char *string)
{
int i, valid;
valid = true;
for (i = 0; i < strlen(string); i++)
{
if (toupper(string[i]) < 'A' || toupper(string[i]) > 'Z')
{
valid = false;
// break here we are done;
break;
}
}
return valid;
}
void printArray(char* str)
{
printf("Array = ");
for (int i = 0; i < strlen(str); ++i)
{
printf("%c", str[i]);
}
printf("\n");
}
bool isPalindrome(char* str1, char* str2)
{
bool isValidPalindrome = true;
int length = strlen(str1);
if (length != strlen(str2))
{
printf("Strings must be the same lenth");
isValidPalindrome = false;
}
else
{
--length;
for (int i = length; i >= 0; --i)
{
if (str1[i] != str2[length - i])
{
isValidPalindrome = false;
break;
}
}
}
return isPalindrome;
}
int main()
{
const int length = 10;
char c, inputarray[length], temp[length];
int i = 0;
// Comparing strings that have not been initialized
// produces undefined behavior. Imagine inputArray is equal to:
// inputArray: "my String ... some other unknown stuff"... where does
// the string ends? there is no '\n' in the horizon.
// The stack error you are getting is produced by the statement
// below. I've pusehd this statement below right after inputArray
// has been initialized
// strcpy(temp, inputarray);
// You don't need the format specifier %s unless you
// rewrite your printf statement as printf("%s", "Please enter string");
// for simplicity you can write it as follows
printf("Please enter string: ");
while ((c = getchar()) != '\n')
{
if (i < length - 1)
inputarray[i] = c;
i++;
}
// Pulled the code inside the if to avoid multiple returns // just preference... not needed
if (i < length)
{
inputarray[i] = '\0';
// helper function to print array
printArray(inputarray);
if (!alphabetic(inputarray))
{
printf("Invalid input");
}
// copy the strings here
strcpy(temp, inputarray);
// reverse the string here
strrev(inputarray);
// you will have to roll out your own isPalindrome
// implementation since reversing a string and comparing it
// with itself will always return false e.g.
// inputArray = "hello";
// copy inputArray into temp
// temp = "hello";
// reverse inputArray
// compare strings: "olleh" == "hello" -> false
if (isPalindrome(inputarray, temp) == true)
printf("Palindrome\n");
else
printf("Not palindrome\n");
}
else
{
inputarray[9] = '\0';
printf("String too long\n");
}
return 0;
}
Trying this and still getting 'not palindrome' when input is a palindrome. It says 'stack around inputarray corrupted' when I run the program.
The probable reason for both is that strcpy(temp, inputarray) is called before inputarray is entered. Move this immediately before the if (strcmp(strrev(inputarray),temp) == 0 ), and your program may work. Another error is the %s in printf("%s Please enter string>").
I need to build a program that receives up to 30 chars from the user, and then to play with it.
For example, I need to reverses the sentence and then print it, or to rotate it.
I have been trying to copy the words of the sentence one by one to a matrix of [30][31], but it does not working... any ideas?
I cannot use pointers...
thanks for the help :)
#include <stdio.h>
#include <string.h>
void main(){
int i=0,
j=0,
wrongData=0,
charCounter=0,
word=0,
letter=0;
char st[100],
arr[100]={0},
mat[30][31]={0};
printf("Please, enter your sentence >");
gets(st);
while(i<strlen(st)){
if('A'<=st[i] && st[i]<='Z'){
charCounter++;
arr[j] = st[i];
i++;
j++;
} else if(st[i]==' '){
arr[j] = ' ';
i++;
j++;
while(st[i] == ' '){
i++;
}
} else if(st[i]=='\0'){
arr[j] = '\0';
break;
} else {
puts("ERROR: Incorrect data, try again.");
wrongData=1;
break;
}
if(wrongData==0){
if(charCounter>30){
puts("ERROR: Incorrect data, try again.");
}
}
}
puts(st);
puts(arr);
if(arr[j]==' '){
word++;
}
while(arr[j]!=' ' && letter<32){
strcpy(mat[word],arr);
}
if(arr[j]=='\0'){
mat[word][letter]=arr[j];
}
puts(mat[word]);
}
Taking into account your comment
the problem is that i need to reverse the words not the letters... for
example: if the string is cats hates dogs, i need to get at the end
dogs hates cats
then I think you mean something as the following
#include <stdio.h>
#include <ctype.h>
#include <string.h>
char * reverse_words( char s[] )
{
for ( char *p = s, *q = s; *p; p = q )
{
while ( isspace( ( unsigned char )*p ) ) ++p;
q = p;
while ( *q && !isspace( ( unsigned char )*q ) ) ++q;
for ( size_t i = 0; i < ( q - p ) / 2; i++ )
{
char c = p[i];
p[i] = q[-i-1];
q[-i-1] = c;
}
}
for ( size_t i = 0, n = strlen( s ); i < n / 2; i++ )
{
char c = s[i];
s[i] = s[n-i-1];
s[n-i-1] = c;
}
return s;
}
int main( void )
{
char s[] = "cats hates dogs";
puts( s );
puts( reverse_words( s ) );
return 0;
}
The program output is
cats hates dogs
dogs hates cats
Here is another approach. The idea is to go through the string and record the index where each word starts and ends. Then the words can be printed in reverse order afterwards. (btw - it will also be easy to rotate the words).
#include<stdio.h>
#include <string.h>
int main() {
char st[100] = "here we go again";
int start[30] = { 0 };
int end[30] = { 0 };
int count = 0;
int len = strlen(st);
int i, j;
// Find start and end index of each word
start[0] = 0;
for(i = 0; i < len; ++i)
{
if (st[i] == ' ')
{
end[count] = i;
++count;
start[count] = i + 1;
}
}
end[count] = len;
// Print the words in reverse order
for(i=count; i >= 0; --i)
{
for (j = start[i]; j < end[i]; ++j)
{
printf("%c", st[j]);
}
printf(" ");
}
printf("\n");
return 0;
}
output:
again go we here
fix your approach like this:
#include <stdio.h>
//#include <string.h>
#define MAX_LEN 30
int main(void){
int i, j, n, word;
char st[100], arr[100], mat[MAX_LEN / 2][MAX_LEN + 1];
printf("Please, enter your sentence (up to %d chars and A-Z or space)\n>", MAX_LEN);fflush(stdout);
scanf("%99[^\n]%*c", st);
//validate and reduce of spaces
for(j = i = 0; st[i]; ++i){
if(i > MAX_LEN){
fputs("ERROR: Incorrect data, try again.\n", stderr);
return 1;
}
if('A'<=st[i] && st[i]<='Z'){
arr[j++] = st[i];
} else if(st[i]==' '){
arr[j++] = ' ';
while(st[++i] == ' ')//Skip a continuous space
;
--i;//one back for next loop
} else {
fputs("ERROR: Incorrect data, try again.\n", stderr);
return 1;
}
}
arr[j] = '\0';//st[i]=='\0' never become true in loop
#if DEBUG
puts(st);
puts(arr);
#endif
//split to word
for(word = j = i = 0; arr[i];){
while(arr[i] == ' ')
++i;//skip space
while(arr[i] != ' ' && arr[i] != '\0')
mat[word][j++] = arr[i++];
mat[word++][j] = '\0';
j = 0;
}
#if DEBUG
for(i = 0; i < word; ++i)
puts(mat[i]);
#endif
puts("reverse word");
for(i = 0; i < word; ++i){
if(i)
putchar(' ');
printf("%s", mat[word-1-i]);
}
puts("\nrotate word");
printf("Please, enter number of rotate\n>");fflush(stdout);
scanf("%d", &n);
for(i = 0; i < word; ++i){
if(i)
putchar(' ');
printf("%s", mat[(i+n)%word]);//rotate left
}
}
I used following code..but i am looking for proper condition to put. please help me out with it.
int main(){
int k = 0;
char a[9] = {'\0'}, b[9] = {'\0'};
printf("enter string \n");
gets(a);
int p = strlen(a);
for(int i = p-1; i >= 0; i--){
b[k] = a[i];
k = k+1;
}
for(int j = 0; j < p; j++){
if(a[j] == b[j]){
continue;
}else
printf("not pal");
break;
}
return 0;
}
for(int j=0;j<p;j++)
{
if(a[j]!=b[j])
{
printf("not pal");
return 0;
}
}
printf("string is Palindrome");
return 0;
here is code..
int main(){
int k=0, flag;
char a[9]={'\0'},b[9]={'\0'};
printf("enter string \n");
gets(a);
int p = strlen(a);
for(int i=p-1;i>=0;i--){
b[k]=a[i];
k=k+1;
}
for(int j=0;j<p;j++){
if(a[j]==b[j]){
flag=0;
}else
flag=1;
break;
}
if(flag==0)
printf("yes");
else
printf("no");
return 0;
}
No need to copy the string. The code below uses the size of the whole string to copy the mirrored letter and checks for equality. Complexity is O(N/2) (if that even exists :-p.
#include <stdbool.h>
#include <stddef.h>
#include <stdio.h>
#include <string.h>
bool is_palindrome(const char* s)
{
const size_t len = strlen(s);
size_t i=0;
while(i<len/2-1)
{
if(s[i] != s[len-i-1])
return false;
++i;
}
return true;
}
int main()
{
const char* s1 = "palindrome";
const char* s2 = "palindromemordnilap";
if(is_palindrome(s1))
printf("uhoh");
if(is_palindrome(s2))
printf("yay!");
}
Live demo here.
I'm trying to check whether a string contains a substring in C like:
char *sent = "this is my sample example";
char *word = "sample";
if (/* sentence contains word */) {
/* .. */
}
What is something to use instead of string::find in C++?
if (strstr(sent, word) != NULL) {
/* ... */
}
Note that strstr returns a pointer to the start of the word in sent if the word word is found.
Use strstr for this.
https://cplusplus.com/reference/cstring/strstr
So, you'd write it like..
char *sent = "this is my sample example";
char *word = "sample";
char *pch = strstr(sent, word);
if(pch)
{
...
}
Try to use pointers...
#include <stdio.h>
#include <string.h>
int main()
{
char str[] = "String1 subString1 Strinstrnd subStr ing1subString";
char sub[] = "subString";
char *p1, *p2, *p3;
int i=0,j=0,flag=0;
p1 = str;
p2 = sub;
for(i = 0; i<strlen(str); i++)
{
if(*p1 == *p2)
{
p3 = p1;
for(j = 0;j<strlen(sub);j++)
{
if(*p3 == *p2)
{
p3++;p2++;
}
else
break;
}
p2 = sub;
if(j == strlen(sub))
{
flag = 1;
printf("\nSubstring found at index : %d\n",i);
}
}
p1++;
}
if(flag==0)
{
printf("Substring NOT found");
}
return (0);
}
You can try this one for both finding the presence of the substring and to extract and print it:
#include <stdio.h>
#include <string.h>
int main(void)
{
char mainstring[]="The quick brown fox jumps over the lazy dog";
char substring[20], *ret;
int i=0;
puts("enter the sub string to find");
fgets(substring, sizeof(substring), stdin);
substring[strlen(substring)-1]='\0';
ret=strstr(mainstring,substring);
if(strcmp((ret=strstr(mainstring,substring)),substring))
{
printf("substring is present\t");
}
printf("and the sub string is:::");
for(i=0;i<strlen(substring);i++)
{
printf("%c",*(ret+i));
}
puts("\n");
return 0;
}
And here is how to report the position of the first character off the found substring:
Replace this line in the above code:
printf("%s",substring,"\n");
with:
printf("substring %s was found at position %d \n", substring,((int) (substring - mainstring)));
My own humble (case sensitive) solution:
uint8_t strContains(char* string, char* toFind)
{
uint8_t slen = strlen(string);
uint8_t tFlen = strlen(toFind);
uint8_t found = 0;
if( slen >= tFlen )
{
for(uint8_t s=0, t=0; s<slen; s++)
{
do{
if( string[s] == toFind[t] )
{
if( ++found == tFlen ) return 1;
s++;
t++;
}
else { s -= found; found=0; t=0; }
}while(found);
}
return 0;
}
else return -1;
}
Results
strContains("this is my sample example", "th") // 1
strContains("this is my sample example", "sample") // 1
strContains("this is my sample example", "xam") // 1
strContains("this is my sample example", "ple") // 1
strContains("this is my sample example", "ssample") // 0
strContains("this is my sample example", "samplee") // 0
strContains("this is my sample example", "") // 0
strContains("str", "longer sentence") // -1
strContains("ssssssample", "sample") // 1
strContains("sample", "sample") // 1
Tested on ATmega328P (avr8-gnu-toolchain-3.5.4.1709) ;)
This code implements the logic of how search works (one of the ways) without using any ready-made function:
public int findSubString(char[] original, char[] searchString)
{
int returnCode = 0; //0-not found, -1 -error in imput, 1-found
int counter = 0;
int ctr = 0;
if (original.Length < 1 || (original.Length)<searchString.Length || searchString.Length<1)
{
returnCode = -1;
}
while (ctr <= (original.Length - searchString.Length) && searchString.Length > 0)
{
if ((original[ctr]) == searchString[0])
{
counter = 0;
for (int count = ctr; count < (ctr + searchString.Length); count++)
{
if (original[count] == searchString[counter])
{
counter++;
}
else
{
counter = 0;
break;
}
}
if (counter == (searchString.Length))
{
returnCode = 1;
}
}
ctr++;
}
return returnCode;
}
I believe that I have the simplest answer. You don't need the string.h library in this program, nor the stdbool.h library. Simply using pointers and pointer arithmetic will help you become a better C programmer.
Simply return 0 for False (no substring found), or 1 for True (yes, a substring "sub" is found within the overall string "str"):
#include <stdlib.h>
int is_substr(char *str, char *sub)
{
int num_matches = 0;
int sub_size = 0;
// If there are as many matches as there are characters in sub, then a substring exists.
while (*sub != '\0') {
sub_size++;
sub++;
}
sub = sub - sub_size; // Reset pointer to original place.
while (*str != '\0') {
while (*sub == *str && *sub != '\0') {
num_matches++;
sub++;
str++;
}
if (num_matches == sub_size) {
return 1;
}
num_matches = 0; // Reset counter to 0 whenever a difference is found.
str++;
}
return 0;
}
Using C - No built in functions
string_contains() does all the heavy lifting and returns 1 based index. Rest are driver and helper codes.
Assign a pointer to the main string and the substring, increment substring pointer when matching, stop looping when substring pointer is equal to substring length.
read_line() - A little bonus code for reading the user input without predefining the size of input user should provide.
#include <stdio.h>
#include <stdlib.h>
int string_len(char * string){
int len = 0;
while(*string!='\0'){
len++;
string++;
}
return len;
}
int string_contains(char *string, char *substring){
int start_index = 0;
int string_index=0, substring_index=0;
int substring_len =string_len(substring);
int s_len = string_len(string);
while(substring_index<substring_len && string_index<s_len){
if(*(string+string_index)==*(substring+substring_index)){
substring_index++;
}
string_index++;
if(substring_index==substring_len){
return string_index-substring_len+1;
}
}
return 0;
}
#define INPUT_BUFFER 64
char *read_line(){
int buffer_len = INPUT_BUFFER;
char *input = malloc(buffer_len*sizeof(char));
int c, count=0;
while(1){
c = getchar();
if(c==EOF||c=='\n'){
input[count]='\0';
return input;
}else{
input[count]=c;
count++;
}
if(count==buffer_len){
buffer_len+=INPUT_BUFFER;
input = realloc(input, buffer_len*sizeof(char));
}
}
}
int main(void) {
while(1){
printf("\nEnter the string: ");
char *string = read_line();
printf("Enter the sub-string: ");
char *substring = read_line();
int position = string_contains(string,substring);
if(position){
printf("Found at position: %d\n", position);
}else{
printf("Not Found\n");
}
}
return 0;
}
The same will be achieved with this simpler code: Why use these:
int main(void)
{
char mainstring[]="The quick brown fox jumps over the lazy dog";
char substring[20];
int i=0;
puts("enter the sub string to find");
fgets(substring, sizeof(substring), stdin);
substring[strlen(substring)-1]='\0';
if (strstr(mainstring,substring))
{
printf("substring is present\t");
}
printf("and the sub string is:::");
printf("%s",substring,"\n");
return 0;
}
But the tricky part would be to report at which position in the original string the substring starts...
My code to find out if substring is exist in string or not
// input ( first line -->> string , 2nd lin ->>> no. of queries for substring
following n lines -->> string to check if substring or not..
#include <stdio.h>
int len,len1;
int isSubstring(char *s, char *sub,int i,int j)
{
int ans =0;
for(;i<len,j<len1;i++,j++)
{
if(s[i] != sub[j])
{
ans =1;
break;
}
}
if(j == len1 && ans ==0)
{
return 1;
}
else if(ans==1)
return 0;
return 0;
}
int main(){
char s[100001];
char sub[100001];
scanf("%s", &s);// Reading input from STDIN
int no;
scanf("%d",&no);
int i ,j;
i=0;
j=0;
int ans =0;
len = strlen(s);
while(no--)
{
i=0;
j=0;
ans=0;
scanf("%s",&sub);
len1=strlen(sub);
int value;
for(i=0;i<len;i++)
{
if(s[i]==sub[j])
{
value = isSubstring(s,sub,i,j);
if(value)
{
printf("Yes\n");
ans = 1;
break;
}
}
}
if(ans==0)
printf("No\n");
}
}
#include <stdio.h>
#include <string.h>
int findSubstr(char *inpText, char *pattern);
int main()
{
printf("Hello, World!\n");
char *Text = "This is my sample program";
char *pattern = "sample";
int pos = findSubstr(Text, pattern);
if (pos > -1) {
printf("Found the substring at position %d \n", pos);
}
else
printf("No match found \n");
return 0;
}
int findSubstr(char *inpText, char *pattern) {
int inplen = strlen(inpText);
while (inpText != NULL) {
char *remTxt = inpText;
char *remPat = pattern;
if (strlen(remTxt) < strlen(remPat)) {
/* printf ("length issue remTxt %s \nremPath %s \n", remTxt, remPat); */
return -1;
}
while (*remTxt++ == *remPat++) {
printf("remTxt %s \nremPath %s \n", remTxt, remPat);
if (*remPat == '\0') {
printf ("match found \n");
return inplen - strlen(inpText+1);
}
if (remTxt == NULL) {
return -1;
}
}
remPat = pattern;
inpText++;
}
}
How can I count the number of occurrences in c of each letter (ignoring case) in the string? So that it would print out letter: # number of occurences, I have code to count the occurences of one letter, but how can I count the occurence of each letter in the string?
{
char
int count = 0;
int i;
//int length = strlen(string);
for (i = 0; i < 20; i++)
{
if (string[i] == ch)
{
count++;
}
}
return count;
}
output:
a : 1
b : 0
c : 2
etc...
Let's assume you have a system where char is eight bit and all the characters you're trying to count are encoded using a non-negative number. In this case, you can write:
const char *str = "The quick brown fox jumped over the lazy dog.";
int counts[256] = { 0 };
int i;
size_t len = strlen(str);
for (i = 0; i < len; i++) {
counts[(int)(str[i])]++;
}
for (i = 0; i < 256; i++) {
if ( count[i] != 0) {
printf("The %c. character has %d occurrences.\n", i, counts[i]);
}
}
Note that this will count all the characters in the string. If you are 100% absolutely positively sure that your string will have only letters (no numbers, no whitespace, no punctuation) inside, then 1. asking for "case insensitiveness" starts to make sense, 2. you can reduce the number of entries to the number of characters in the English alphabet (namely 26) and you can write something like this:
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
const char *str = "TheQuickBrownFoxJumpedOverTheLazyDog";
int counts[26] = { 0 };
int i;
size_t len = strlen(str);
for (i = 0; i < len; i++) {
// Just in order that we don't shout ourselves in the foot
char c = str[i];
if (!isalpha(c)) continue;
counts[(int)(tolower(c) - 'a')]++;
}
for (i = 0; i < 26; i++) {
printf("'%c' has %2d occurrences.\n", i + 'a', counts[i]);
}
Like this:
int counts[26];
memset(counts, 0, sizeof(counts));
char *p = string;
while (*p) {
counts[tolower(*p++) - 'a']++;
}
This code assumes that the string is null-terminated, and that it contains only characters a through z or A through Z, inclusive.
To understand how this works, recall that after conversion tolower each letter has a code between a and z, and that the codes are consecutive. As the result, tolower(*p) - 'a' evaluates to a number from 0 to 25, inclusive, representing the letter's sequential number in the alphabet.
This code combines ++ and *p to shorten the program.
One simple possibility would be to make an array of 26 ints, each is a count for a letter a-z:
int alphacount[26] = {0}; //[0] = 'a', [1] = 'b', etc
Then loop through the string and increment the count for each letter:
for(int i = 0; i<strlen(mystring); i++) //for the whole length of the string
if(isalpha(mystring[i]))
alphacount[tolower(mystring[i])-'a']++; //make the letter lower case (if it's not)
//then use it as an offset into the array
//and increment
It's a simple idea that works for A-Z, a-z. If you want to separate by capitals you just need to make the count 52 instead and subtract the correct ASCII offset
#include <stdio.h>
#include <string.h>
void main()
{
printf("PLEASE ENTER A STRING\n");
printf("GIVE ONLY ONE SPACE BETWEEN WORDS\n");
printf("PRESS ENETR WHEN FINISHED\n");
char str[100];
int arr[26]={0};
char ch;
int i;
gets(str);
int n=strlen(str);
for(i=0;i<n;i++)
{
ch=tolower(str[i]);
if(ch>=97 && ch<=122)
{
arr[ch-97]++;
}
}
for(i=97;i<=122;i++)
printf("%c OCCURS %d NUMBER OF TIMES\n",i,arr[i-97]);
return 0;
}
After Accept Answer
A method that meets these specs: (IMO, the other answers do not meet all)
It is practical/efficient when char has a wide range. Example: CHAR_BIT is 16 or 32, so no use of bool Used[1 << CHAR_BIT];
Works for very long strings (use size_t rather than int).
Does not rely on ASCII. ( Use Upper[] )
Defined behavior when a char < 0. is...() functions are defined for EOF and unsigned char
static const char Upper[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
static const char Lower[] = "abcdefghijklmnopqrstuvwxyz";
void LetterOccurrences(size_t *Count, const char *s) {
memset(Count, 0, sizeof *Count * 26);
while (*s) {
unsigned char ch = *s;
if (isalpha(ch)) {
const char *caseset = Upper;
char *p = strchr(caseset, ch);
if (p == NULL) {
caseset = Lower;
p = strchr(caseset, ch);
}
if (p != NULL) {
Count[p - caseset]++;
}
}
}
}
// sample usage
char *s = foo();
size_t Count[26];
LetterOccurrences(Count, s);
for (int i=0; i<26; i++)
printf("%c : %zu\n", Upper[i], Count[i]);
}
You can use the following code.
main()
{
int i = 0,j=0,count[26]={0};
char ch = 97;
char string[100]="Hello how are you buddy ?";
for (i = 0; i < 100; i++)
{
for(j=0;j<26;j++)
{
if (tolower(string[i]) == (ch+j))
{
count[j]++;
}
}
}
for(j=0;j<26;j++)
{
printf("\n%c -> %d",97+j,count[j]);
}
}
Hope this helps.
#include<stdio.h>
#include<string.h>
#define filename "somefile.txt"
int main()
{
FILE *fp;
int count[26] = {0}, i, c;
char ch;
char alpha[27] = "abcdefghijklmnopqrstuwxyz";
fp = fopen(filename,"r");
if(fp == NULL)
printf("file not found\n");
while( (ch = fgetc(fp)) != EOF) {
c = 0;
while(alpha[c] != '\0') {
if(alpha[c] == ch) {
count[c]++;
}
c++;
}
}
for(i = 0; i<26;i++) {
printf("character %c occured %d number of times\n",alpha[i], count[i]);
}
return 0;
}
for (int i=0;i<word.length();i++){
int counter=0;
for (int j=0;j<word.length();j++){
if(word.charAt(i)==word.charAt(j))
counter++;
}// inner for
JOptionPane.showMessageDialog( null,word.charAt(i)+" found "+ counter +" times");
}// outer for
#include<stdio.h>
void frequency_counter(char* str)
{
int count[256] = {0}; //partial initialization
int i;
for(i=0;str[i];i++)
count[str[i]]++;
for(i=0;str[i];i++) {
if(count[str[i]]) {
printf("%c %d \n",str[i],count[str[i]]);
count[str[i]]=0;
}
}
}
void main()
{
char str[] = "The quick brown fox jumped over the lazy dog.";
frequency_counter(str);
}
Here is the C code with User Defined Function:
/* C Program to count the frequency of characters in a given String */
#include <stdio.h>
#include <string.h>
const char letters[] = "abcdefghijklmnopqrstuvwxzy";
void find_frequency(const char *string, int *count);
int main() {
char string[100];
int count[26] = { 0 };
int i;
printf("Input a string: ");
if (!fgets(string, sizeof string, stdin))
return 1;
find_frequency(string, count);
printf("Character Counts\n");
for (i = 0; i < 26; i++) {
printf("%c\t%d\n", letters[i], count[i]);
}
return 0;
}
void find_frequency(const char *string, int *count) {
int i;
for (i = 0; string[i] != '\0'; i++) {
p = strchr(letters, string[i]);
if (p != NULL) {
count[p - letters]++;
}
}
}
Have checked that many of the answered are with static array, what if suppose I have special character in the string and want a solution with dynamic concept. There can be many other possible solutions, it is one of them.
here is the solutions with the Linked List.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct Node {
char data;
int counter;
struct Node* next;
};
void printLinkList(struct Node* head)
{
while (head != NULL) {
printf("\n%c occur %d", head->data, head->counter);
head = head->next;
}
}
int main(void) {
char *str = "!count all the occurances of character in string!";
int i = 0;
char tempChar;
struct Node* head = NULL;
struct Node* node = NULL;
struct Node* first = NULL;
for(i = 0; i < strlen(str); i++)
{
tempChar = str[i];
head = first;
if(head == NULL)
{
node = (struct Node*)malloc(sizeof(struct Node));
node->data = tempChar;
node->counter = 1;
node->next = NULL;
if(first == NULL)
{
first = node;
}
}
else
{
while (head->next != NULL) {
if(head->data == tempChar)
{
head->counter = head->counter + 1;
break;
}
head = head->next;
}
if(head->next == NULL)
{
if(head->data == tempChar)
{
head->counter = head->counter + 1;
}
else
{
node = (struct Node*)malloc(sizeof(struct Node));
node->data = tempChar;
node->counter = 1;
node->next = NULL;
head->next = node;
}
}
}
}
printLinkList(first);
return 0;
}
int charset[256] = {0};
int charcount[256] = {0};
for (i = 0; i < 20; i++)
{
for(int c = 0; c < 256; c++)
{
if(string[i] == charset[c])
{
charcount[c]++;
}
}
}
charcount will store the occurence of any character in the string.
//This is JavaScript Code.
function countWordOccurences()
{
// You can use array of words or a sentence split with space.
var sentence = "The quick brown fox jumped over the lazy dog.";
//var sentenceArray = ['asdf', 'asdf', 'sfd', 'qwr', 'qwr'];
var sentenceArray = sentence.split(' ', 1000);
var output;
var temp;
for(var i = 0; i < sentenceArray.length; i++) {
var k = 1;
for(var j = i + 1; j < sentenceArray.length; j++) {
if(sentenceArray[i] == sentenceArray[j])
k = k + 1;
}
if(k > 1) {
i = i + 1;
output = output + ',' + k + ',' + k;
}
else
output = output + ',' + k;
}
alert(sentenceArray + '\n' + output.slice(10).split(',', 500));
}
You can see it live --> http://jsfiddle.net/rammipr/ahq8nxpf/
//c code for count the occurence of each character in a string.
void main()
{
int i,j; int c[26],count=0; char a[]="shahid";
clrscr();
for(i=0;i<26;i++)
{
count=0;
for(j=0;j<strlen(a);j++)
{
if(a[j]==97+i)
{
count++;
}
}
c[i]=count;
}
for(i=0;i<26;i++)
{
j=97+i;
if(c[i]!=0) { printf("%c of %d times\n",j,c[i]);
}
}
getch();
}
protected void btnSave_Click(object sender, EventArgs e)
{
var FullName = "stackoverflow"
char[] charArray = FullName.ToLower().ToCharArray();
Dictionary<char, int> counter = new Dictionary<char, int>();
int tempVar = 0;
foreach (var item in charArray)
{
if (counter.TryGetValue(item, out tempVar))
{
counter[item] += 1;
}
else
{
counter.Add(item, 1);
}
}
//var numberofchars = "";
foreach (KeyValuePair<char, int> item in counter)
{
if (counter.Count > 0)
{
//Label1.Text=split(item.
}
Response.Write(item.Value + " " + item.Key + "<br />");
// Label1.Text=item.Value + " " + item.Key + "<br />";
spnDisplay.InnerText= item.Value + " " + item.Key + "<br />";
}
}