i am trying to write a program which reverses a entire string and also may print reversing each word of the string more than one time.
For example my string is:
"2
Fox jumps over the lazy dog."
for this, the output should be:
.god .god yzal yzal eht eht revo revo spmuj spmuj xoF xoF.
I am trying to store each word in a 2d array and then reverse each word but i am not able to do that.
here is my code. kindly help
Note: how do we provide EOF in console
#include <stdio.h>
#include <stdlib.h>
int main() {
char string[100][100];
char ch;
int i = 0, j = 0, l = 0, count = 0, x = 0, y = 0;
while ((ch = getchar()) != EOF) {
string[i][l++] = ch;
if (ch == ' ') {
string[i][l] = '\n';
i++;
l = 0;
count++;
}
}
for (x = 0; x <= count; x++) {
int length = strlen(string[x]) - 1;
for (y = length; y >= 0; --y)
printf("%s", string[y]);
}
return 0;
}
Here are a few changes to your code check the comments for explanation.
int main()
{
char string[100][100];
char ch;
int i=0,j=0, l=0, count=0, x=0, y=0;
while((ch=getchar())!=EOF)
{
string[i][l++] = ch;
if(ch==' ')
{
string[i][l] = '\0'; //make the string null terminated otherwise you cant work with its length
i++;
l=0;
count++;
}
}
string[i][l]='\0'; //make the last string null terminated
for(x=count; x>=0; x--) //read from last counter
{
int length = strlen(string[x])-1;
for(y=length; y>=0; --y)
{
printf("%c", string[x][y]); //print by each character and not string.
}
}
return 0;
}
Corrections in your code:
C strings are null terminated, but you are terminating your strings with newline \n character, which is wrong.
You are storing the whitespace with the string, reversing will be difficult in this case.
Your print statement won't reverse the string, print it character by character.
For the output that you need, you can consider this code.
#include <stdio.h>
#include <stdlib.h>
int main() {
char string[100][100];
char ch;
int i = 0, j = 0, l = 0, count = 0, x = 0, y = 0;
while ((ch = getchar()) != EOF) {
if (ch == ' ') {
string[i][l] = '\0'; /// Null terminate the string
i++;
l = 0;
count++;
}
else
string[i][l++] = ch; /// Don't add whitespace to the string
}
string[i][l] = '\0';
for (x = count; x >= 0; x--) {
int length = strlen(string[x]) - 1;
for (y = length; y >= 0; --y)
printf("%c", string[x][y]); /// Print the string in reverse
printf(" ");
for (y = length; y >= 0; --y)
printf("%c", string[x][y]); /// Twice
printf(" ");
}
return 0;
}
Input
2 Fox jumps over the lazy dog.
Output
.god .god yzal yzal eht eht revo revo spmuj spmuj xoF xoF 2 2
See http://ideone.com/qaIoW9
If I understand you want to duplicate each reversed word N times as specified by the first number in the string, then something like the following would work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *strrevdup (char* str);
int main (int argc, char **argv) {
if (argc < 2 ) {
fprintf (stderr, "error: insufficient input, usage: %s \"# string\"\n",
argv[0]);
return 1;
}
char *str = strdup (argv[1]);
char *p = str;
char *rev = NULL;
int mult = 0;
int i = 0;
while (*p && *p != ' ') p++;
*p = 0;
mult = atoi (str);
*p = ' ';
if (!mult) return 1;
while (*p && *p == ' ') p++;
rev = strrevdup (p);
char *r = rev;
printf ("\n the reversed string with duplicated words '%d' times is:\n\n", mult);
for (p = strtok (r, " "); p; p = strtok (NULL, " \n"))
for (i = 0; i < mult; i++)
printf (" %s", p);
printf ("\n\n");
free (str);
free (rev);
return 0;
}
/** strrevdup - reverse duplicate string, swaps src & dest each iteration.
* Takes valid string, duplicates and reverses, original is preserved.
* Returns pointer to reversed string on success, NULL otherwise.
* Requires string.h, caller is responsible for freeing memory allocated.
*/
char *strrevdup (char* str)
{
if (!str) {
printf ("%s() error: invalid string\n", __func__);
return NULL;
}
char *rstr = strdup (str);
char *begin = rstr;
char *end = rstr + strlen (rstr) - 1;
char tmp;
while (end > begin){
tmp=*end;
*end-- = *begin;
*begin++ = tmp;
}
return rstr;
}
Output
$ ./bin/strrevndup "2 Fox jumps over the lazy dog."
the reversed string with duplicated words '2' times is:
.god .god yzal yzal eht eht revo revo spmuj spmuj xoF xoF
you may try this code although it will reverse and print words in different lines.. you may try few more things to get the desired answer.
` #include <stdio.h>
#include <stdlib.h>
#include<string.h>
int main()
{
char string[1024][1024];
char ch;
int t,z;
scanf("%d",&t);
z=t;
int i=0, l=0, count=0, x=0, y=0;
getchar();
while((ch=getchar())!=EOF)
{
if(ch=='\n')
{
i++;
l=0;
count++;
string[i][l++] = ch;
i++;
l=0;
}
else if(ch==' ')
{
i++;
l=0;
count++;
}
else{
string[i][l++] = ch;
}
}
for(x=count+1; x>=0; x--)
{
if(string[x][0]=='\n')
{
printf("\n");
}
else{
char *rev=strrev(string[x]);
while(t--)
printf("%s ",rev);
t=z;
}
}
return 0;
}`
Related
I have program to remove the similar words from string but this program only removing at once word not a repeating words.
For example input:
sabunkerasmaskera kera
and should an output:
sabunmas
This my code:
#include <stdio.h>
#include <string.h>
void remove(char x[100], char y[100][100], char words[100]) {
int i = 0, j = 0, k = 0;
for (i = 0; x[i] != '\0'; i++) {
if (x[i] == ' ') {
y[k][j] = '\0';
k++;
j = 0;
} else {
y[k][j] = x[i];
j++;
}
}
y[k][j] = '\0';
j = 0;
for (i = 0; i < k + 1; i++) {
if (strcmp(y[i], kata) == 0) {
y[i][j] = '\0';
}
}
j = 0;
for (i = 0; i < k + 1; i++) {
if (y[i][j] == '\0')
continue;
else
printf("%s ", y[i]);
}
printf ("\n");
}
int main() {
char x[100], y[100][100], kata[100];
printf ("Enter word:\n");
gets(x);
printf("Enter word to remove:\n");
gets(words);
remove(x, y, words);
return 0;
}
My program output its:
sabunkerasmaskerara
and that should not be the case. Maybe I need your opinion to fixed this program and also I need help to make it better.
Your solution does not work because it uses strcmp to compare the string portions, which only works if the substring is at the end of the string, as this makes it null-terminated.
You should instead use strstr to locate the matches and use memmove to shift the string contents.
There are other issues in your code:
do not use gets()
y is unnecessary for this task.
words is not defined
Here is a modified version:
#include <stdio.h>
#include <string.h>
char *remove_all(char *str, const char *word) {
size_t len = strlen(word);
if (len != 0) {
char *p = str;
while ((p = strstr(p, word)) != NULL) {
memmove(p, p + len, strlen(p + len) + 1);
}
}
return str;
}
int main() {
char str[100], word[100];
printf ("Enter string:\n");
if (!fgets(str, sizeof str, stdin))
return 1;
printf("Enter word to remove:\n");
if (!fgets(word, sizeof word, stdin))
return 1;
word[strcspn(word, "\n")] = '\0'; // strip the trailing newline if any
remove_all(str, word);
fputs(str, stdout);
return 0;
}
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;}
What I need to write:
1.Get a main string from user.
2.Get a subString from a user.
Every match of the subString in the main string, change its letters to uppercase.
Do not use string's functions like strstr.
For example:
main string: abcdeffghfhkfff
sub string: ff
outut: abcdeFFghfhkFFf
Problem: Well, I'm having troubles to continue writing the code after I found one match. for example after I found the first 'f' in the main string, how can I continue check if the second 'f' is adjacent to the found 'f', if not, then try to find another 'f' and check subsequent matches of the subarray until we've found that the length of the substring matches the number of subsequent matches in the string? Here's what I've tried, and in writing the logic of the for loop in 'replaceSubstring' function
#include <stdio.h>
#include <stdlib.h>
#include<conio.h>
#include <string.h>
#define N 101
void replaceSubstring(char *str, char *subStr);
void main()
{
char str[N], subStr[N];
while (strlen(str) != 0 || strlen(subStr) != 0)
{
str[0] = 0;
printf("Enter text: ");
gets(str);
printf("Enter substring: ");
scanf("%s", subStr);
replaceSubstring(str, subStr);
}
}
void replaceSubstring(char *str, char *SubStr)
{
int i, count = 0, j = 0, k = 0;
for (i = 0; i <= strlen(str); i++)
{
if (str[i] == SubStr[k])
{
k++;
count++;
if (count == strlen(SubStr))
{
str[i] -= 32;
}
}
}
puts(str);
getchar();
}
You can use strstr() function to do this more easly, like this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 101
void replaceSubstring(char *str, char *subStr);
void main()
{
char str[N], subStr[N];
while (strlen(str) != 0 || strlen(subStr) != 0)
{
str[0] = 0;
printf("Enter text: ");
gets(str);
printf("Enter substring: ");
scanf("%s", subStr);
replaceSubstring(str, subStr);
}
}
void replaceSubstring(char *str, char *SubStr)
{
int i;
char *tmp;
while((tmp = strstr(str, SubStr)) != NULL)
{
for (i = 0; i < strlen(SubStr); i++)
{
tmp[i] -= 32;
}
}
puts(str);
getchar();
}
Here another version of replaceSubstring() function without using strstr() function:
void replaceSubstring(char *str, char *SubStr)
{
int i = 0, found = 1, j = 0, k = 0;
while (i < strlen(str))
{
if (str[i] == SubStr[0])
{
found = 1;
for(k = 0; k < strlen(SubStr); k++)
{
if(str[i+k] != SubStr[k])
{
found = 0;
break;
}
}
if(found)
{
for(k = 0; k < strlen(SubStr); k++)
{
str[i+k] -= 32;
}
i += strlen(SubStr);
}
else
i++;
}
else
i++;
}
puts(str);
getchar();
}
To solve this without using strstr(), i would do something like this:
void replaceSubstring(char *str, char *SubStr)
{
int i = 0, equals = 0, j = 0, k = 0;
for(i=0;i<strlen(str);i++){
j = i;
equals = 1;
k=0;
while(k<strlen(SubStr)&&(equals == 1)){
if(SubStr[k] != str[j]){
equals = 0;
}
k++;
j++;
}
if(equals == 1){
for(j=i;j<i+k;j++){
str[j] -= 32;
}
}
}
puts(str);
getchar();
}
I'm pretty sure this works correctly.
input: abcdeffghfhkfff
substring: ff
output: abcdeFFghfhkFFf
Here is a demonstrative program that shows how the function can be written
#include <stdio.h>
#include <string.h>
#include <ctype.h>
char * replaceSubstring( char *s1, const char *s2 )
{
char *p = s1;
size_t n = strlen( s2 );
while ( ( p = strstr( p, s2 ) ) != NULL )
{
for ( size_t i = 0; i < n; ++i, ++p ) *p = toupper( ( unsigned char )*p );
}
return s1;
}
int main( void )
{
char s[] = "abcdeffghfhkfff";
puts( s );
puts( replaceSubstring( s, "ff" ) );
}
Its output is
abcdeffghfhkfff
abcdeFFghfhkFFf
Take into account that according to the C Standard function main without parameters shall be declared like`
int main( void )
Also it is a bad idea to use "magic" numbers like 32 like in this statement
tmp[i] -= 32;
For example if in the environment there are used EBCDIC characters then this statement will be simply wrong.
Moreover even for ASCII characters this statement is invalid because it is not necessary that original characters are in lower case.
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;}
#include <stdio.h>
int main(void)
{
int i,j;
int wordstart = -1;
int wordend = -1;
char words[]= "this is a test";
char temp;
// Reverse each word
for (i = 0; i < strlen(words); ++i)
{
wordstart = -1;
wordend = -1;
if(words[i] != ' ')
wordstart = i;
for (j = wordstart; j < strlen(words); ++j)
{
if(words[j] == ' ')
{
wordend = j - 1;
break;
}
}
if(wordend == -1)
wordend = strlen(words);
for (j = wordstart ; j <= (wordend - wordstart) / 2; ++j)
{
temp = words[j];
words[j] = words[wordend - (j - wordstart)];
words[wordend - (j - wordstart)] = temp;
}
i = wordend;
printf("reversed string is %s:", words);
}
}
I tried in this way but i am getting this output:
siht is a test
my expected output is:
test a is this
I would appreciate if some one could come with a different approach for which time complexity is very less or correct me if it is the right approach. Thanks
Perhaps this belongs on the code review site instead?
Your approach seems very efficient to me (except that I would only call strlen(words) once and save the result in a register).
Two possible bugs look like:
wordend = strlen(words);
should be
wordend = strlen(words)-1;
and
for(j = wordstart ; j <= (wordend - wordstart) / 2 ; ++j) {
should be
for(j = wordstart ; j <= (wordend + wordstart) / 2 ; ++j) {
Final code looks like (with some extra {}):
#include <stdio.h>
int main(int argc,char *argv[])
{
int i,j;
char words[]= "this is a test";
int L=strlen(words);
// Reverse each word
for(i = 0; i < L; ++i) {
int wordstart = -1;
int wordend = -1;
if(words[i] != ' ')
{
wordstart = i;
for(j = wordstart; j < L; ++j) {
if(words[j] == ' ') {
wordend = j - 1;
break;
}
}
if(wordend == -1)
wordend = L-1;
for(j = wordstart ; j <= (wordend + wordstart) / 2 ; ++j) {
char temp = words[j];
words[j] = words[wordend - (j - wordstart)];
words[wordend - (j - wordstart)] = temp;
}
i = wordend;
}
}
printf("reversed string is %s:",words);
return 0;
}
You can create a double linked list as a base data structure. Then, iterate through the words and insert them in the list as you find them.
When you reach the end of the sentence, simply traverse the list backwards and print the words as you go through them
Simply we can just use a n*1 2D character array tailored to suit our needs!!!
#include <stdlib.h>
int main()
{
char s[20][20];
int i=0, length=-1;
for(i=0;;i++)
{
scanf("%s",s[i]);
length++;
if(getchar()=='\n')
break;
}
for(i=length;i>=0;i--)
printf("%s ",s[i]);
return 0;
}
Start tokenizing the line from the last character and continue to the first character. Keep one pointer anchored at the base of the current word, and another pointed which will decrease while a word start is not found. When you find a word start while scanning like this, print from the word start pointer to the word end anchor. Update the word end anchor to the previous character of the current word start char.
You might want to skip the blankspace characters while scanning.
UPDATE
This is a quick implementation:
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#define MAX_BUF 256
void show_string (char *str, int i, int n)
{
while (i <= n)
{
printf ("%c", str[i]);
i++;
}
}
int main (void)
{
char str[MAX_BUF];
int end_anchor, start_ptr;
int state;
printf ("\nEnter a string: ");
scanf (" %[^\n]", str);
start_ptr = strlen (str) - 1;
end_anchor = start_ptr;
state = 0;
while (start_ptr >= -1)
{
switch (state)
{
case 0:
if ((!isspace (str[start_ptr]) && (start_ptr >= 0)))
{
start_ptr--;
}
else
{
state = 1;
}
break;
case 1:
show_string (str, start_ptr + 1, end_anchor);
state = 2;
start_ptr--;
printf (" ");
break;
case 2:
if (!isspace (str[start_ptr]))
{
state = 0;
end_anchor = start_ptr;
}
else
{
start_ptr--;
}
break;
}
}
printf ("\n");
return 0;
}
The end_anchor points to each end word, and the start_ptr finds the start of the word of which the end is held by end_anchor. When we find a word start (by blankspace characters or start_ptr = -1), we print all the characters from start_ptr + 1 to end_anchor. The + 1 is because of the implementation: start_ptr points to the blankspace character, and the print routine will print all the characters from i to n. Once we have detected one blank space we print it and we skip adjacent blankspaces (in case 2) and preserve only one which is manually printed. Once a non blankspace is detected, we have got another word end, for which we set the end_anchor to this index in the case 2, and set state = 0 , so that we can search for the word start again.
if(words[i] != ' ')
wordstart = i;
This statement what about the else part? if words[i] == ' ', and wordstart remains -1.
So maybe try to use:
while (words[i] && words[i] == ' ') ++i;
if (!words[i])
break;
wordstart = i;
Then you should output the result out of the i loop.
Finally, if you want to get the result you expected, you should reverse the whole sentence once more, with the way you used in the loop.
I would use write function similar to strrchr for finding last occurence of ' ', if its found print word that follows, rewrite this ' ' with '\0' and repeat it in loop till no more words are found. At the end I would print the content of this string again because there is most likely no ' ' before the first word.
I would write own function instead of strrchr because strrchr calculates the lenght of the given string, which is redundant in this case. This length doesn't have to be calculated more than once.
Here's the code:
char* findLastWord(char* str, int* len)
{
int i;
for (i = *len - 1; i >= 0; --i)
{
if (str[i] == ' ')
{
str[i] = '\0';
if (i < *len - 1)
{
*len = i - 1;
return &str[i + 1];
}
}
}
return NULL;
}
int main (int argc, char *argv[])
{
char str[] = " one two three four five six ";
int len = strlen(str);
char* lastWord = findLastWord(str, &len);
while (lastWord != NULL)
{
printf("%s\n", lastWord);
lastWord = findLastWord(str, &len);
}
if (len > 1)
printf("%s\n", str);
return 0;
}
output:
six
five
four
three
two
one
Hope this helps ;)
#include<stdio.h>
#include<string.h>
void reverse(char *str, size_t len)
{
char tmp;
size_t beg, end;
if (len <=1) return;
for (beg=0,end=len; beg < --end ; beg++) {
tmp = str[beg];
str[beg] = str[end];
str[end] = tmp;
}
}
int main(void)
{
char sentence[] = "one two three four five";
size_t pos, len;
printf("Before:%s\n",sentence);
for (pos = len= 0; sentence[pos]; pos += len) {
pos += strspn( sentence+pos, " \t\n" );
len = strcspn( sentence+pos, " \t\n" );
reverse ( sentence + pos, len );
}
reverse ( sentence , pos );
printf("After:%s\n",sentence);
return 0;
}
#include <iostream>
#include <string>
using namespace std;
char* stringrev(char s[], int len)
{
char *s1 = (char*)malloc(len+1);
int i=0;
while (len>0)
{
s1[i++] = s[--len];
}
s1[i++] = '\0';
return s1;
}
void sentrev(char s[], int len)
{
int i=0; int j=0;
char *r = (char*)malloc(len+1);
while(1)
{
if(s[j] == ' ' || s[j] == '\0')
{
r = stringrev(s+i, j-i);
i = j+1;
cout<<r<<" ";
}
if (s[j] == '\0')
break;
j++;
}
}
int main()
{
char *s = "this is a test";
char *r = NULL;
int len = strlen(s);
cout<<len<<endl;
r = stringrev(s, len);
cout<<r<<endl;
sentrev(r, len);
return 0;
}
The above code snap reverse the sentence, using char *r
and printing cout<
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main()
{
char st[50], rst[50];
printf("Enter the sentence...\n");
gets(st);
int len=strlen(st), p;
int j=-1,k;
p=len;
for(int i=(len-1); i>=0; i--)
{
//searching for space or beginning
if(st[i]==' ')
{
//reversing and storing each word except the first word
for(k=i+1;k<p;k++)
{
//printf("%c",st[k]);
rst[++j]=st[k];
}
j++;
rst[j]=' ';
printf("\n");
p=i;
}
else if(i==0)
{
//for first word
for(k=i;k<p;k++)
{
//printf("%c",st[k]);
rst[++j]=st[k];
}
}
}
printf("Now reversing the sentence...\n");
puts(rst);
return 0;
}
Use a main for loop to traverse till the end of the sentence:
Copy the letters in a string until you find a space.
now call add#beginning function and in that function add the string each time you pass a string to the linked list.
print the contents of the linked list with a space inbetween to get the expected output
My code,just traverse from the last and if you find a space print the characters before it,now change the end to space-1;This will print till the second word,finally just print the first word using a single for loop.Comment for alter approach.
Program:
#include<stdio.h>
int main()
{
char str[200];
int i,j,k;
scanf("%[^\n]s",&str);
for(i=0;str[i]!='\0';i++);
i=i-1;
for(j=i;j>=0;j--)
{
if((str[j])==' ')
{
for(k=j+1;k<=i;k++)
{
printf("%c",str[k]);
}
i=j-1;
printf(" ");
}
}
for(k=0;k<=i;k++)
{
printf("%c",str[k]);
}
}
using stack
#include <iostream>
#include <stdio.h>
#include <stack>
int main()
{
std::stack<string> st;
char *words= "this is a test";
char * temp = (char *)calloc(1, sizeof(*temp));
int size1= strlen(words);
int k2=0;
int k3=0;
for(int i=0;i<=size1;i++)
{
temp[k2] = words[i];
k2++;
if(words[i] == ' ')
{
k3++;
if(k3==1)
temp[k2-1]='\0';
temp[k2]='\0';
st.push(temp);
k2=0;
}
if(words[i] == '\0')
{
temp[k2]='\0';
st.push(temp);
k2=0;
break;
}
}
while (!st.empty())
{
printf("%s",st.top().c_str());
st.pop();
}