I know how to print a single string vertically.
char test[100] = "test";
int i;
for(i=0;i<strlen(test);i++){
printf("%c\n",test[i]);
}
Which will give me:
t
e
s
t
But how can I print an array of strings vertically? For example:
char listOfTest[2][10] = {"testing1","quizzing"};
So it can return:
tq
eu
si
tz
iz
gi
1g
Simply loop through the first string and print each character at index i in the first string and the second string till you reach the null terminator of first string
NOTE: this only work when string 1 and string2 are equal in length and will need modification for other test cases
#include <stdio.h>
int main(void)
{
char listOfTest[2][10] = {"testing1","quizzing"};
int i = 0;
//loop through string 1 till NULL is reach
while (listOfTest[0][i])
{
//prints char at index i in string 1 and 2
printf("%c%c\n", listOfTest[0][i], listOfTest[1][i]);
//increment the index value
i++;
}
return (0);
}
Simply print a character from both strings.
(Better to test for the null character rather than repeatedly call strlen().)
for(i = 0; listOfTest[0][i] && listOfTest[1][i]; i++) {
printf("%c%c\n", listOfTest[0][i], listOfTest[1][i]);
}
To extend to n strings ...
size_t num_of_strings = sizeof listOfTest/sizeof listOfTest[0];
bool done = false;
for (size_t i = 0; listOfTest[0][i] && !done; i++) {
for (size_t n = 0; n < num_of_strings; n++) {
if (listOfTest[n][i] == '\0') {
done = true;
break;
}
printf("%c", listOfTest[n][i]);
}
printf("\n");
}
Related
This program, tokenizes a user input string, removes extra spaces and saves each word into a 2D array and then print the tokens
EXAMPLE:
input: " Hello world string house and car"
output and EXPECTED output:
token[0]: Hello
token[1]: world
token[2]: string
token[3]: house
token[4]: and
token[5]: car
THE PROBLEM:
the problem is that I achieved this by using strlen() function when printing the tokens(code located at the very bottom), I am not supposed to use any other library than stdio.h and stdlib.h, since strlen() function is defined in string.h i tried to use sizeof(arr) / sizeof(arr[0]); but it does not work as I want, the result using sizeof is :
token[0]: Hello
token[1]: world
token[2]: string
token[3]: house
token[4]: and
token[5]: car
�oken[6]: ��
token[7]: �
token[8]: ����
token[9]: �
token[10]:
I WOULD LIKE TO HAVE THE EXPECTED OUTPUT WITHOUT USING STRLEN()
#include<stdio.h>
#include <stdlib.h>
#define TRUE 1
char tokenize(char *str, char array[10][20])
{
int n = 0, i, j = 0;
for(i = 0; TRUE; i++)//infinite loop until is the end of the string '\0'
{
if(str[i] != ' '){
//position 1, char 1
array[n][j++] = str[i];// if, it is not space, we save the character
}
else{
array[n][j++] = '\0';//end of the first word
n++;// position for next new word
j=0;// start writting char at position 0
}
if(str[i] == '\0')
break;
}
return 0;
}
//removes extra spaces
char* find_word_start(char* str){
/*also removes all extra spaces*/
char *result = (char*) malloc(sizeof(char) *1000);
int c = 0, d = 0;
// no space at beginning
while(str[c] ==' ') {
c++;
}
while(str[c] != '\0'){ // till end of sentence
result[d++] = str[c++]; //take non-space characters
if(str[c]==' ') { // take one space between words
result[d++] = str[c++];
}
while(str[c]==' ') { //
c++;
}
}
result[d-1] = '\0';
//print or return char?
return result;
free(result);
}
int main()
{
char str[]=" Hello world string dudes and dudas ";
//words, and chars in each word
char arr[10][20];
//call the method to tokenize the string
tokenize(find_word_start(str),arr);
int row = sizeof(arr) / sizeof(arr[0]);
/*----------------------------------------------------------------------*/
/*----------------------------------------------------------------------*/
for(int i = 0;i <= strlen(arr);i++)
/*----------------------------------------------------------------------*/
/*----------------------------------------------------------------------*/
printf("token[%d]: %s\n", i, arr[i]);
return 0;
}
Your code using strlen() may appear the work in this instance but it is not correct.
strlen(arr) makes no semantic sense because arr is not a string. It happens in this case to return 5 because arr has the same address as arr[0], then you kludged it to work for the 6 word output by using the test i <= strlen(arr) in the for loop. The two values strlen(arr) and the number of strings stored in arr are not related.
The expression sizeof(arr) / sizeof(arr[0]) determines the run-time constant number arrays within the array of arrays arr (i.e. 10), not the number of valid strings assigned. It is your code's responsibility to keep track of that either with a sentinel value such as an empty string, or by maintaining a count of strings assigned.
I suggest you change tokenize to return the number of strings (currently it is inexplicably defined to return a char, but in fact only ever rather uselessly returns zero):
int tokenize( char* str, char array[][20] )
{
...
return n ;
}
Then:
int rows = tokenize( find_word_start(str), arr ) ;
for( int i = 0; i < rows; i++ )
{
printf( "token[%d]: %s\n", i, arr[i] ) ;
}
How can I complete the function canArrangeWords() ?
Question : Given a set of words check if we can arrange them in a list such that the last letter of any word and first letter of another word are same. The input function canArrangeWords shall contain an integer num and array of words arr. num denotes the number of word in the list (1<=num<=100). arr shall contain words consisting of lower case letters between 'a' - 'z' only . return 1 if words can be arranged in that fashion and -1 if cannot.
Input : 4 pot ten nice eye
output : 1
input : 3 fox owl pond
output: -1
Please help me complete this program .
**
#include<stdio.h>
#include<string.h>
int canArrangewords(int,char [100][100]);
void main(){
int n ,count=0 , i ;
char arrayS[100][100];
scanf("%d",&n);
for (i = 0; i < n; ++i)
{
scanf("%s",arrayS[i]);
}
for(i=0;i<n;i++)
{
printf("%s",arrayS[i]);
printf("\n");
}
printf("%c\n",arrayS[2][4]);
canArrangewords(n , arrayS);
}
int canArrangewords(int n,char arrayS[100][100]){
int i , j ;
for ( i = 0; i < n; i++)
{
for ( j = i+1 ; j < strlen(arrayS[j+1]); i++)
{
int flag = strlen(arrayS[j+1]) - 1;
int temp = strcmp(arrayS[i][0],arrayS[j][flag]);
}
}
}
}
Well, first of all think of the way you can reach that answer.
If you only need to know if they can or can not be arranged and you do not have to do so your self you can use an empty array of int array[26] for each letter a-z.
The rule is that from all the first and last letters for all the words only two MAY appear an odd amount of times - the first letter of first word in list and the last letter in the last word in the list, the rest MUST appear an even amount of times. I would add a check to make sure the letters are lowercase as well. good luck!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MINASCII 97
#define LETTERS 26
void UpdateArray(char letter, int* arr)
{
if(arr[letter - MINASCII] == 0)
{
++arr[letter - MINASCII];
}
else
{
--arr[letter - MINASCII];/*for each second time same letter is seen reduce back to zero */
}
}
int canArrangewords(int wordNum, char* wordArr[])
{
int arr[LETTERS] = {0};
int i = 0;
int count = 0 ;
char first;
char last;
char* string;
for (i= 0; i< wordNum; ++i)
{
string = wordArr[i];
first = string[0];
last = string[strlen(string)-1];
UpdateArray(first, &arr[0]);
UpdateArray(last, &arr[0]);
}
for(i = 0; i< LETTERS; ++i)
{
count+=arr[i];
}
if(count == 2 || count == 0)/*either once each or twice -see word1 example in main*/
{
return 1;
}
return -1;
}
int main()
{
int i = 0;
char* words[] = {"pot", "ten", "nice", "eye"};
char* words1[] = {"pot", "ten", "nip"};
char* words2[] = {"fox", "owl", "pond"};
i = canArrangewords(4,words);
printf("%d\n", i);
i = canArrangewords(3,words1);
printf("%d\n", i);
i = canArrangewords(3,words2);
printf("%d\n", i);
return 0;
}
Change your array of words into an array of pointers to words. Then you can simply exchange the pointers.
To speed things up, instead of a pointer to a word, have it point to a structure:
struct WORD {
char *firstchar; // begin of word
char *lastchar; // last char of word
} *words[100]; // array of 100 pointers to words
To read the words:
char buf[100];
for (i = 0; i < n; ++i)
{
scanf("%s",buf);
int len= strlen(buf);
words[i]= malloc(sizeof(struct WORDS));
words[i]->firstchar= malloc(len+1);
strcpy(words[i]->firstchar, buf);
words[i]->lastchar= words[i]->firstchar + len-1;
}
Now compare and sort:
if (*words[i]->lastchar == *words[j]->firstchar) {
struct WORDS *tmp= words[i+1];
words[i+1]= words[j];
words[j]= tmp;
}
Do this in a loop, a kind of bubble sort. I leave that to you.
I made a code and my target is to put spacewhere the input word was found in a sentence.
i neet to replece the small word with space
like:
Three witches watched three watches
tch
output:
Three wi es wa ed three wa es
I made this code:
#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
for (i = 0; i < B; i++)
{
for(j=0;j<S;j++)
{
if(small[j]!=big[i])
{
j=0;
break;
}
if(small[j]=='\0')
{
while (i-(j-1)!=i)
{
i = i - j;
big[i] = '\n';
i++;
}
}
}
}
puts(big);
}
First of all, in your exemple you work with newline '\n' and not with space.
Consider this simple example:
#include<stdio.h>
#define S 8
#define B 50
void main() {
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int cpt = 0;
int smallSize = 0;
// loop to retrieve smallSize
for (i = 0; i < S; i++)
{
if (small[i] != '\0')
smallSize++;
}
// main loop
for (i = 0; i < B; i++)
{
// stop if we hit the end of the string
if (big[i] == '\0')
break;
// increment the cpt and small index while the content of big and small are equal
if (big[i] == small[j])
{
cpt++;
j++;
}
// we didn't found the full small word
else
{
j = 0;
cpt = 0;
}
// test if we found the full word, if yes replace char in big by space
if (cpt == smallSize)
{
for (int k = 0; k < smallSize; k++)
{
big[i-k] = ' ';
}
j = 0;
cpt = 0;
}
}
puts(big);
}
You need first to retrieve the real size of the small array.
Once done, next step is to look inside "big" if there is the word small inside. If we find it, then replace all those char by spaces.
If you want to replace the whole small word with a single space, then you'll need to adapt this example !
I hope this help !
A possible way is to use to pointers to the string, one for reading and one for writing. This will allow to replace an arbitrary number of chars (the ones from small) with a single space. And you do not really want to nest loops but une only one to process every char from big.
Last but not least, void main() should never be used except in stand alone environment (kernel or embedded development). Code could become:
#include <stdio.h>
#define S 8
#define B 50
int main() { // void main is deprecated...
char small[S] = {"ol"};
char big[B] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0; // pointer to written back big
for (i = 0; i < B; i++)
{
if (big[i] == 0) break; // do not process beyond end of string
if(small[j]!=big[i])
{
for(int l=0; l<j; l++) big[k++] = small[l]; // copy an eventual partial small
big[k++] = big[i]; // copy the incoming character
j=0; // reset pointer to small
continue;
}
else if(small[++j] == 0) // reached end of small
{
big[k++] = ' '; // replace chars from small with a single space
j = 0; // reset pointer to small
}
}
big[k] = '\0';
puts(big);
return 0;
}
or even better (no need for fixed sizes of strings):
#include <stdio.h>
int main() { // void main is deprecated...
char small[] = {"ol"};
char big[] = {"my older gradmom see my older sister"};
int i = 0, j = 0;
int k = 0; // pointer to written back big
for (i = 0; i < sizeof(big); i++)
{
if(small[j]!=big[i])
...
In C strings are terminated with a null character '\0'. Your code defines a somehow random number at the beginning (B and S) and iterates over that much characters instead of the exact number of characters, the strings actually contain. You can use the fact that the string is terminated by testing the content of the string in a while loop.
i = 0;
while (str[i]) {
...
i = i + 1;
}
If you prefer for loops you can write it also as a for loop.
for (i = 0; str[i]; i++) {
...
}
Your code does not move the contents of the remaining string to the left. If you replace two characters ol with one character , you have to move the remaining characters to the left by one character. Otherwise you would have a hole in the string.
#include <stdio.h>
int main() {
char small[] = "ol";
char big[] = "my older gradmom see my older sister";
int s; // index, which loops through the small string
int b; // index, which loops through the big string
int m; // index, which loops through the characters to be modified
// The following loops through the big string up to the terminating
// null character in the big string.
b = 0;
while (big[b]) {
// The following loops through the small string up to the
// terminating null character, if the character in the small
// string matches the corresponding character in the big string.
s = 0;
while (small[s] && big[b+s] == small[s]) {
// In case of a match, continue with the next character in the
// small string.
s = s + 1;
}
// If we are at the end of the small string, we found in the
// big string.
if (small[s] == '\0') {
// Now we have to modify the big string. The modification
// starts at the current position in the big string.
m = b;
// First we have to put the space at the current position in the
// big string.
big[m] = ' ';
// And next the rest of the big string has to be moved left. The
// rest of the big string starts, where the match has ended.
while (big[b+s]) {
m = m + 1;
big[m] = big[b+s];
s = s + 1;
}
// Finally the big string has to be terminated by a null
// character.
big[m+1] = '\0';
}
// Continue at next character in big string.
b = b + 1;
}
puts(big);
return 0;
}
I wrote a function that finds the longest string in a 2d array, it works, partially. My problem is that it takes the first longest string that it finds without checking the other ones.
For example, the following list of strings:
eke
em
ekeke
eme
e
ememeememe
emem
ekekee
eooeeeeefe
eede
My function catches "ekeke" (the third string from the list) as the longest instead of "ememeememe ".
Here is my function:
void length(char str[][MAX])
{
int i = 0;
for(i = 1; i < LEN; i++)
{
if(strlen(str[i]) > strlen(str[i-1]))
{
if(strlen(str[i]) > strlen(str[i+1]))
{
printf("%s", str[i]);
break;
}
}
}
}
LEN is a constant, his value is 10.
MAX is a constant, his value is 50.
The strings are given by the user.
Thanks.
You are only comparing the previous and next strings. You need to check the lengths of all the strings.
void length(char str[][MAX])
{
size_t longest = strlen(str[0]);
szie_t j = 0;
for(size_t i = 1; i < LEN; i++)
{
size_t len = strlen(str[i]);
if(longest < len)
{
longest = len;
j = i;
}
}
printf("%s", str[j]);
}
I am assuming you have at least 1 string and handle corner cases (if user inputs less than LEN strings etc -- depends on how you fill the str with strings).
I wrote a function that finds the longest string in a 2d array, it works, partially. My problem is that it takes the first longest string that it finds without checking the other ones.
For example, the following list of strings:
eke
em
ekeke
eme
e
ememeememe
emem
ekekee
eooeeeeefe
eede
My function catches "ekeke" (the third string from the list) as the longest instead of "ememeememe ".
Here is my function:
void length(char str[][MAX])
{
int i = 0;
for(i = 1; i < LEN; i++)
{
if(strlen(str[i]) > strlen(str[i-1]))
{
if(strlen(str[i]) > strlen(str[i+1]))
{
printf("%s", str[i]);
break;
}
}
}
}
LEN is a constant, his value is 10.
MAX is a constant, his value is 50.
The strings are given by the user.
Thanks.
You are only comparing the previous and next strings. You need to check the lengths of all the strings.
void length(char str[][MAX])
{
size_t longest = strlen(str[0]);
szie_t j = 0;
for(size_t i = 1; i < LEN; i++)
{
size_t len = strlen(str[i]);
if(longest < len)
{
longest = len;
j = i;
}
}
printf("%s", str[j]);
}
I am assuming you have at least 1 string and handle corner cases (if user inputs less than LEN strings etc -- depends on how you fill the str with strings).