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I need to write a function that will count words in a string. For the
purpose of this assignment, a "word" is defined to be a sequence
of non-null, non-whitespace characters, separated from other words by
whitespace.
This is what I have so far:
int words(const char sentence[ ]);
int i, length=0, count=0, last=0;
length= strlen(sentence);
for (i=0, i<length, i++)
if (sentence[i] != ' ')
if (last=0)
count++;
else
last=1;
else
last=0;
return count;
I am not sure if it works or not because I can't test it until my whole program is finished and I am not sure it will work, is there a better way of writing this function?
You needed
int words(const char sentence[])
{
}
(note braces).
For loops go with ; instead of ,.
Without any disclaimer, here's what I'd have written:
See it live http://ideone.com/uNgPL
#include <string.h>
#include <stdio.h>
int words(const char sentence[ ])
{
int counted = 0; // result
// state:
const char* it = sentence;
int inword = 0;
do switch(*it) {
case '\0':
case ' ': case '\t': case '\n': case '\r': // TODO others?
if (inword) { inword = 0; counted++; }
break;
default: inword = 1;
} while(*it++);
return counted;
}
int main(int argc, const char *argv[])
{
printf("%d\n", words(""));
printf("%d\n", words("\t"));
printf("%d\n", words(" a castle "));
printf("%d\n", words("my world is a castle"));
}
See the following example, you can follow the approach : count the whitespace between words .
int words(const char *sentence)
{
int count=0,i,len;
char lastC;
len=strlen(sentence);
if(len > 0)
{
lastC = sentence[0];
}
for(i=0; i<=len; i++)
{
if((sentence[i]==' ' || sentence[i]=='\0') && lastC != ' ')
{
count++;
}
lastC = sentence[i];
}
return count;
}
To test :
int main()
{
char str[30] = "a posse ad esse";
printf("Words = %i\n", words(str));
}
Output :
Words = 4
#include <ctype.h> // isspace()
int
nwords(const char *s) {
if (!s) return -1;
int n = 0;
int inword = 0;
for ( ; *s; ++s) {
if (!isspace(*s)) {
if (inword == 0) { // begin word
inword = 1;
++n;
}
}
else if (inword) { // end word
inword = 0;
}
}
return n;
}
bool isWhiteSpace( char c )
{
if( c == ' ' || c == '\t' || c == '\n' )
return true;
return false;
}
int wordCount( char *string )
{
char *s = string;
bool inWord = false;
int i = 0;
while( *s )
{
if( isWhiteSpace(*s))
{
inWord = false;
while( isWhiteSpace(*s) )
s++;
}
else
{
if( !inWord )
{
inWord = true;
i++;
}
s++;
}
}
return i;
}
Here is one of the solutions. It counts words with multiple spaces or just space or space followed by the word.
#include <stdio.h>
int main()
{
char str[80];
int i, w = 0;
printf("Enter a string: ");
scanf("%[^\n]",str);
for (i = 0; str[i] != '\0'; i++)
{
if((str[i]!=' ' && str[i+1]==' ')||(str[i+1]=='\0' && str[i]!=' '))
{
w++;
}
}
printf("The number of words = %d", w );
return 0;
}
I know this is an old thread, but perhaps someone needs a simple solution, just checks for blank space in ascii and compares current char to that while also makign sure first char is not a space, cheers!
int count_words(string text){
int counter = 1;
int len = strlen(text);
for(int i = 0; i < len; i++){
if(text[i] == 32 && i != 0) {
counter++;
}
}
return counter;}
Here is another solution:
#include <string.h>
int words(const char *s)
{
const char *sep = " \t\n\r\v\f";
int word = 0;
size_t len;
s += strspn(s, sep);
while ((len = strcspn(s, sep)) > 0) {
++word;
s += len;
s += strspn(s, sep);
}
return word;
}
#include<stdio.h>
int main()
{
char str[50];
int i, count=1;
printf("Enter a string:\n");
gets(str);
for (i=0; str[i]!='\0'; i++)
{
if(str[i]==' ')
{
count++;
}
}
printf("%i\n",count);
}
#include<stdio.h>
#include<string.h>
int getN(char *);
int main(){
char str[999];
printf("Enter Sentence: "); gets(str);
printf("there are %d words", getN(str));
}
int getN(char *str){
int i = 0, len, count= 0;
len = strlen(str);
if(str[i] >= 'A' && str[i] <= 'z')
count ++;
for (i = 1; i<len; i++)
if((str[i]==' ' || str[i]=='\t' || str[i]=='\n')&& str[i+1] >= 'A' && str[i+1] <= 'z')
count++;
return count;
}
#include <stdio.h>
int wordcount (char *string){
int n = 0;
char *p = string ;
int flag = 0 ;
while(isspace(*p)) p++;
while(*p){
if(!isspace(*p)){
if(flag == 0){
flag = 1 ;
n++;
}
}
else flag = 0;
p++;
}
return n ;
}
int main(int argc, char **argv){
printf("%d\n" , wordcount(" hello world\nNo matter how many newline and spaces"));
return 1 ;
}
I found the posted question after finishing my function for a C class I'm taking. I saw some good ideas from code people have posted above. Here's what I had come up with for an answer. It certainly is not as concise as other's, but it does work. Maybe this will help someone in the future.
My function receives an array of chars in. I then set a pointer to the array to speed up the function if it was scaled up. Next I found the length of the string to loop over. I then use the length of the string as the max for the 'for' loop.
I then check the pointer which is looking at array[0] to see if it is a valid character or punctuation. If pointer is valid then increment to next array index. The word counter is incremented when the first two tests fail. The function then will increment over any number of spaces until the next valid char is found.
The function ends when null '\0' or a new line '\n' character is found. Function will increment count one last time right before it exit to account for the word preceding null or newline. Function returns count to the calling function.
#include <ctype.h>
char wordCount(char array[]) {
char *pointer; //Declare pointer type char
pointer = &array[0]; //Pointer to array
int count; //Holder for word count
count = 0; //Initialize to 0.
long len; //Holder for length of passed sentence
len = strlen(array); //Set len to length of string
for (int i = 0; i < len; i++){
//Is char punctuation?
if (ispunct(*(pointer)) == 1) {
pointer += 1;
continue;
}
//Is the char a valid character?
if (isalpha(*(pointer)) == 1) {
pointer += 1;
continue;
}
//Not a valid char. Increment counter.
count++;
//Look out for those empty spaces. Don't count previous
//word until hitting the end of the spaces.
if (*(pointer) == ' ') {
do {
pointer += 1;
} while (*(pointer) == ' ');
}
//Important, check for end of the string
//or newline characters.
if (*pointer == '\0' || *pointer == '\n') {
count++;
return(count);
}
}
//Redundent return statement.
count++;
return(count);
}
I had this as an assignment...so i know this works.
The function gives you the number of words, average word length, number of lines and number of characters.
To count words, you have to use isspace() to check for whitespaces. if isspace is 0 you know you're not reading whitespace. wordCounter is a just a way to keep track of consecutive letters. Once you get to a whitespace, you reset that counter and increment wordCount. My code below:
Use isspace(c) to
#include <stdio.h>
#include <ctype.h>
int main() {
int lineCount = 0;
double wordCount = 0;
double avgWordLength = 0;
int numLines = 0;
int wordCounter = 0;
double nonSpaceChars = 0;
int numChars = 0;
printf("Please enter text. Use an empty line to stop.\n");
while (1) {
int ic = getchar();
if (ic < 0) //EOF encountered
break;
char c = (char) ic;
if (isspace(c) == 0 ){
wordCounter++;
nonSpaceChars++;
}
if (isspace(c) && wordCounter > 0){
wordCount++;
wordCounter =0;
}
if (c == '\n' && lineCount == 0) //Empty line
{
break;
}
numChars ++;
if (c == '\n') {
numLines ++;
lineCount = 0;
}
else{
lineCount ++;
}
}
avgWordLength = nonSpaceChars/wordCount;
printf("%f\n", nonSpaceChars);
printf("Your text has %d characters and %d lines.\nYour text has %f words, with an average length of %3.2f ", numChars, numLines, wordCount, avgWordLength);
}
Here is one solution. This one will count words correctly even if there are multiple spaces between words, no spaces around interpuncion symbols, etc. For example: I am,My mother is. Elephants ,fly away.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int countWords(char*);
int main() {
char string[1000];
int wordsNum;
printf("Unesi nisku: ");
gets(string); /*dont use this function lightly*/
wordsNum = countWords(string);
printf("Broj reci: %d\n", wordsNum);
return EXIT_SUCCESS;
}
int countWords(char string[]) {
int inWord = 0,
n,
i,
nOfWords = 0;
n = strlen(string);
for (i = 0; i <= n; i++) {
if (isalnum(string[i]))
inWord = 1;
else
if (inWord) {
inWord = 0;
nOfWords++;
}
}
return nOfWords;
}
this is a simpler function to calculate the number of words
int counter_words(char* a){`
// go through chars in a
// if ' ' new word
int words=1;
int i;
for(i=0;i<strlen(a);++i)
{
if(a[i]==' ' && a[i+1] !=0)
{
++words;
}
}
return words;}
My code is supposed to compare 2 strings and returns the common characters in alphabetical order. If there are no common chars, it will return a null string.
However the program is not running.
Code
void strIntersect(char *str1, char *str2, char *str3)
{
int i,j, k;
i = 0;
j = 0;
k = 0;
while(str1[i]!='\0' || str2[j]!='\0')
{
if(strcmp(str1[i],str2[j])>0)
{
str3[k] = str1[i];
k++;
}
else if (strcmp(str2[j],str1[i])>0)
{
str3[k] = str2[j];
k++;
}
i++;
j++;
}
}
Example
Input string 1:abcde
Input string 2:dec
Output: cde
How do I get it to work?
There are quite a few problems with your code
strcmp is not needed for a simple char comparison
Is the 3rd char string allocated by the caller?
Your approach won't work if source strings are either of different sizes or are not alphabetical.
My solution assumes that input is ASCII, and is efficient (used a simple char array with indexes denoting ASCII value of the character).
If a character is found in str1, the char map will have a 1, if it is common, it will have a 2, otherwise, it will have a 0.
void strIntersect(char *str1, char *str2, char *str3)
{
int i=0, j=0, k=0;
char commonCharsMap[128] = { 0 };
while(str1[i] != '\0')
{
commonCharsMap[str1[i++]] = 1;
}
while(str2[j] != '\0')
{
if(commonCharsMap[str2[j]] == 1)
{
commonCharsMap[str2[j++]] = 2;
}
}
for(i=0; i<128; i++)
{
if(commonCharsMap[i] == 2)
{
str3[k++] = i;
}
}
str3[k++] = '\0';
}
int main()
{
char str1[] = "abcde";
char str2[] = "dce";
char str3[30];
strIntersect(str1, str2, str3);
printf("Common chars: %s\n", str3);
return 0;
}
A option is to iterate over the complete second string for each character in the first string
int i = 0;
int k = 0;
while(str1[i] != '\0') {
int j = 0;
while(str2[j] != '\0') {
if (str1[i] == str2[j]) {
str3[k] = str1[i];
k++;
}
j++;
}
i++;
}
I replaced the strcmp because you are comparing single characters not a string
Your if case and Else if case are identical and you are just comparing elements according to your Index. i.e you are comparing first element with first, second with second and so on. This won't provide you solution. I suggest use two for loops. I will provide you code later if you want
#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'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++;
}
}
I need to write a function that will count words in a string. For the
purpose of this assignment, a "word" is defined to be a sequence
of non-null, non-whitespace characters, separated from other words by
whitespace.
This is what I have so far:
int words(const char sentence[ ]);
int i, length=0, count=0, last=0;
length= strlen(sentence);
for (i=0, i<length, i++)
if (sentence[i] != ' ')
if (last=0)
count++;
else
last=1;
else
last=0;
return count;
I am not sure if it works or not because I can't test it until my whole program is finished and I am not sure it will work, is there a better way of writing this function?
You needed
int words(const char sentence[])
{
}
(note braces).
For loops go with ; instead of ,.
Without any disclaimer, here's what I'd have written:
See it live http://ideone.com/uNgPL
#include <string.h>
#include <stdio.h>
int words(const char sentence[ ])
{
int counted = 0; // result
// state:
const char* it = sentence;
int inword = 0;
do switch(*it) {
case '\0':
case ' ': case '\t': case '\n': case '\r': // TODO others?
if (inword) { inword = 0; counted++; }
break;
default: inword = 1;
} while(*it++);
return counted;
}
int main(int argc, const char *argv[])
{
printf("%d\n", words(""));
printf("%d\n", words("\t"));
printf("%d\n", words(" a castle "));
printf("%d\n", words("my world is a castle"));
}
See the following example, you can follow the approach : count the whitespace between words .
int words(const char *sentence)
{
int count=0,i,len;
char lastC;
len=strlen(sentence);
if(len > 0)
{
lastC = sentence[0];
}
for(i=0; i<=len; i++)
{
if((sentence[i]==' ' || sentence[i]=='\0') && lastC != ' ')
{
count++;
}
lastC = sentence[i];
}
return count;
}
To test :
int main()
{
char str[30] = "a posse ad esse";
printf("Words = %i\n", words(str));
}
Output :
Words = 4
#include <ctype.h> // isspace()
int
nwords(const char *s) {
if (!s) return -1;
int n = 0;
int inword = 0;
for ( ; *s; ++s) {
if (!isspace(*s)) {
if (inword == 0) { // begin word
inword = 1;
++n;
}
}
else if (inword) { // end word
inword = 0;
}
}
return n;
}
bool isWhiteSpace( char c )
{
if( c == ' ' || c == '\t' || c == '\n' )
return true;
return false;
}
int wordCount( char *string )
{
char *s = string;
bool inWord = false;
int i = 0;
while( *s )
{
if( isWhiteSpace(*s))
{
inWord = false;
while( isWhiteSpace(*s) )
s++;
}
else
{
if( !inWord )
{
inWord = true;
i++;
}
s++;
}
}
return i;
}
Here is one of the solutions. It counts words with multiple spaces or just space or space followed by the word.
#include <stdio.h>
int main()
{
char str[80];
int i, w = 0;
printf("Enter a string: ");
scanf("%[^\n]",str);
for (i = 0; str[i] != '\0'; i++)
{
if((str[i]!=' ' && str[i+1]==' ')||(str[i+1]=='\0' && str[i]!=' '))
{
w++;
}
}
printf("The number of words = %d", w );
return 0;
}
I know this is an old thread, but perhaps someone needs a simple solution, just checks for blank space in ascii and compares current char to that while also makign sure first char is not a space, cheers!
int count_words(string text){
int counter = 1;
int len = strlen(text);
for(int i = 0; i < len; i++){
if(text[i] == 32 && i != 0) {
counter++;
}
}
return counter;}
Here is another solution:
#include <string.h>
int words(const char *s)
{
const char *sep = " \t\n\r\v\f";
int word = 0;
size_t len;
s += strspn(s, sep);
while ((len = strcspn(s, sep)) > 0) {
++word;
s += len;
s += strspn(s, sep);
}
return word;
}
#include<stdio.h>
int main()
{
char str[50];
int i, count=1;
printf("Enter a string:\n");
gets(str);
for (i=0; str[i]!='\0'; i++)
{
if(str[i]==' ')
{
count++;
}
}
printf("%i\n",count);
}
#include<stdio.h>
#include<string.h>
int getN(char *);
int main(){
char str[999];
printf("Enter Sentence: "); gets(str);
printf("there are %d words", getN(str));
}
int getN(char *str){
int i = 0, len, count= 0;
len = strlen(str);
if(str[i] >= 'A' && str[i] <= 'z')
count ++;
for (i = 1; i<len; i++)
if((str[i]==' ' || str[i]=='\t' || str[i]=='\n')&& str[i+1] >= 'A' && str[i+1] <= 'z')
count++;
return count;
}
#include <stdio.h>
int wordcount (char *string){
int n = 0;
char *p = string ;
int flag = 0 ;
while(isspace(*p)) p++;
while(*p){
if(!isspace(*p)){
if(flag == 0){
flag = 1 ;
n++;
}
}
else flag = 0;
p++;
}
return n ;
}
int main(int argc, char **argv){
printf("%d\n" , wordcount(" hello world\nNo matter how many newline and spaces"));
return 1 ;
}
I found the posted question after finishing my function for a C class I'm taking. I saw some good ideas from code people have posted above. Here's what I had come up with for an answer. It certainly is not as concise as other's, but it does work. Maybe this will help someone in the future.
My function receives an array of chars in. I then set a pointer to the array to speed up the function if it was scaled up. Next I found the length of the string to loop over. I then use the length of the string as the max for the 'for' loop.
I then check the pointer which is looking at array[0] to see if it is a valid character or punctuation. If pointer is valid then increment to next array index. The word counter is incremented when the first two tests fail. The function then will increment over any number of spaces until the next valid char is found.
The function ends when null '\0' or a new line '\n' character is found. Function will increment count one last time right before it exit to account for the word preceding null or newline. Function returns count to the calling function.
#include <ctype.h>
char wordCount(char array[]) {
char *pointer; //Declare pointer type char
pointer = &array[0]; //Pointer to array
int count; //Holder for word count
count = 0; //Initialize to 0.
long len; //Holder for length of passed sentence
len = strlen(array); //Set len to length of string
for (int i = 0; i < len; i++){
//Is char punctuation?
if (ispunct(*(pointer)) == 1) {
pointer += 1;
continue;
}
//Is the char a valid character?
if (isalpha(*(pointer)) == 1) {
pointer += 1;
continue;
}
//Not a valid char. Increment counter.
count++;
//Look out for those empty spaces. Don't count previous
//word until hitting the end of the spaces.
if (*(pointer) == ' ') {
do {
pointer += 1;
} while (*(pointer) == ' ');
}
//Important, check for end of the string
//or newline characters.
if (*pointer == '\0' || *pointer == '\n') {
count++;
return(count);
}
}
//Redundent return statement.
count++;
return(count);
}
I had this as an assignment...so i know this works.
The function gives you the number of words, average word length, number of lines and number of characters.
To count words, you have to use isspace() to check for whitespaces. if isspace is 0 you know you're not reading whitespace. wordCounter is a just a way to keep track of consecutive letters. Once you get to a whitespace, you reset that counter and increment wordCount. My code below:
Use isspace(c) to
#include <stdio.h>
#include <ctype.h>
int main() {
int lineCount = 0;
double wordCount = 0;
double avgWordLength = 0;
int numLines = 0;
int wordCounter = 0;
double nonSpaceChars = 0;
int numChars = 0;
printf("Please enter text. Use an empty line to stop.\n");
while (1) {
int ic = getchar();
if (ic < 0) //EOF encountered
break;
char c = (char) ic;
if (isspace(c) == 0 ){
wordCounter++;
nonSpaceChars++;
}
if (isspace(c) && wordCounter > 0){
wordCount++;
wordCounter =0;
}
if (c == '\n' && lineCount == 0) //Empty line
{
break;
}
numChars ++;
if (c == '\n') {
numLines ++;
lineCount = 0;
}
else{
lineCount ++;
}
}
avgWordLength = nonSpaceChars/wordCount;
printf("%f\n", nonSpaceChars);
printf("Your text has %d characters and %d lines.\nYour text has %f words, with an average length of %3.2f ", numChars, numLines, wordCount, avgWordLength);
}
Here is one solution. This one will count words correctly even if there are multiple spaces between words, no spaces around interpuncion symbols, etc. For example: I am,My mother is. Elephants ,fly away.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int countWords(char*);
int main() {
char string[1000];
int wordsNum;
printf("Unesi nisku: ");
gets(string); /*dont use this function lightly*/
wordsNum = countWords(string);
printf("Broj reci: %d\n", wordsNum);
return EXIT_SUCCESS;
}
int countWords(char string[]) {
int inWord = 0,
n,
i,
nOfWords = 0;
n = strlen(string);
for (i = 0; i <= n; i++) {
if (isalnum(string[i]))
inWord = 1;
else
if (inWord) {
inWord = 0;
nOfWords++;
}
}
return nOfWords;
}
this is a simpler function to calculate the number of words
int counter_words(char* a){`
// go through chars in a
// if ' ' new word
int words=1;
int i;
for(i=0;i<strlen(a);++i)
{
if(a[i]==' ' && a[i+1] !=0)
{
++words;
}
}
return words;}