So I want to create a function using C to find the longest repeated non overlapping substring in a given string. For example: input banana. Output: an.
I was thinking using comparison of the array of the string and checking for repeats. Is that a viable approach? How would I be able to compare substrings with the rest of the strings. I want to avoid using suffix trees if possible
#include <stdio.h>
#include <string.h>
void stringcheck(char a[],int len, int s1, int s2)
{
int i=s1+1;
int j=s2+1;
if(j<=len&&a[i]==a[j])
{
printf("%c",a[i]);
stringcheck(a,len,i,j);
}
}
void dupcheck(char a[], int len, int start)
{
for(int i=start;i<len-1;i++)
{
for(int j=i+1;j<=len;j++)
{
if(a[i]==a[j])
{
printf("%c",a[i]);
stringcheck(a,len,i,j);
i=len;
}
}
}
}
int main()
{
char input[99];
scanf("%s",input);
int start=0;
int len =strlen(input);
dupcheck(input,len,start);
return 0;
}
Yes, this is a valid approach.
You can compare the string - character by character, that way no need to truly save a substring.
You can see a dynamic solution using c++ taking that approach here: https://www.geeksforgeeks.org/longest-repeating-and-non-overlapping-substring/
This solution can be converted to c without many changes.
Another variant if the option is to save the substring by its' indexes.
You can then compare it against the string, and save the max substring, however this will take O(n^3) when the above solution does it in O(n^2).
edit: I converted the solution to c:
#include <stdio.h>
#include <string.h>
void longestRepeatedSubstring(char * str, char * res)
{
int n = strlen(str);
int LCSRe[n+1][n+1];
int res_length = 0; // To store length of result
int i, j, index = 0;
// Setting all to 0
memset(LCSRe, 0, sizeof(LCSRe));
// building table in bottom-up manner
for (i=1; i<=n; i++)
{
for (j=i+1; j<=n; j++)
{
// (j-i) > LCSRe[i-1][j-1] to remove
// overlapping
if (str[i-1] == str[j-1] &&
LCSRe[i-1][j-1] < (j - i))
{
LCSRe[i][j] = LCSRe[i-1][j-1] + 1;
// updating maximum length of the
// substring and updating the finishing
// index of the suffix
if (LCSRe[i][j] > res_length)
{
res_length = LCSRe[i][j];
index = (i>index) ? i : index;
}
}
else
LCSRe[i][j] = 0;
}
}
// If we have non-empty result, then insert all
// characters from first character to last
// character of string
j=0;
if (res_length > 0) {
for (i = index - res_length + 1; i <= index; i++) {
res[j] = str[i-1];
j++;
}
}
res[j]=0;
}
// Driver program to test the above function
int main()
{
char str[] = "banana";
char res[20];
longestRepeatedSubstring(str, res);
printf("%s",res);
return 0;
}
Related
I try to split one string to 3-gram strings. But turns out that the resulting substrings were always messy. The length and char ** input... are needed, since I will use them as args later for python calling the funxtion.
This is the function I wrote.
struct strArrIntArr getSearchArr(char* input, int length) {
struct strArrIntArr nameIndArr;
// flag of same bit
int same;
// flag/index of identical strings
int flag = 0;
// how many identical strings
int num = 0;
// array of split strings
char** nameArr = (char **)malloc(sizeof(char *) * (length - 2));
if ( nameArr == NULL ) exit(0);
// numbers of every split string
int* valueArr = (int* )malloc(sizeof(int) * (length-2));
if ( valueArr == NULL ) exit(0);
// loop length of search string -2 times (3-gram)
for(int i = 0; i<length-2; i++){
if(flag==0){
nameArr[i - num] = (char *)malloc(sizeof(char) * 3);
if ( nameArr[i - num] == NULL ) exit(0);
printf("----i------------%d------\n", i);
printf("----i-num--------%d------\n", i-num);
}
flag = 0;
// compare splitting string with existing split strings,
// if a string exists, it would not be stored
for(int k=0; k<i-num; k++){
same = 0;
for(int j=0; j<3; j++){
if(input[i + j] == nameArr[k][j]){
same ++;
}
}
// identical strings found, if all the three bits are the same
if(same == 3){
flag = k;
num++;
break;
}
}
// if the current split string doesn't exist yet
// put current split string to array
if(flag == 0){
for(int j=0; j<3; j++){
nameArr[i-num][j] = input[i + j];
valueArr[i-num] = 1;
}
}else{
valueArr[flag]++;
}
printf("-----string----%s\n", nameArr[i-num]);
}
// number of N-gram strings
nameIndArr.length = length- 2- num;
// array of N-gram strings
nameIndArr.charArr = nameArr;
nameIndArr.intArr = valueArr;
return nameIndArr;
}
To call the function:
int main(int argc, const char * argv[]) {
int length = 30;
char* input = (char *)malloc(sizeof(char) * length);
input = "googleapis.com.wncln.wncln.org";
// split the search string into N-gram strings
// and count the numbers of every split string
struct strArrIntArr nameIndArr = getSearchArr(input, length);
}
Below is the result. The strings from 17 are messy.
----i------------0------
----i-num--------0------
-----string----goo
----i------------1------
----i-num--------1------
-----string----oog
----i------------2------
----i-num--------2------
-----string----ogl
----i------------3------
----i-num--------3------
-----string----gle
----i------------4------
----i-num--------4------
-----string----lea
----i------------5------
----i-num--------5------
-----string----eap
----i------------6------
----i-num--------6------
-----string----api
----i------------7------
----i-num--------7------
-----string----pis
----i------------8------
----i-num--------8------
-----string----is.
----i------------9------
----i-num--------9------
-----string----s.c
----i------------10------
----i-num--------10------
-----string----.co
----i------------11------
----i-num--------11------
-----string----com
----i------------12------
----i-num--------12------
-----string----om.
----i------------13------
----i-num--------13------
-----string----m.w
----i------------14------
----i-num--------14------
-----string----.wn
----i------------15------
----i-num--------15------
-----string----wnc
---i------------16------
----i-num--------16------
-----string----ncl
----i------------17------
----i-num--------17------
-----string----clnsole
----i------------18------
----i-num--------18------
-----string----ln.=C:
----i------------19------
----i-num--------19------
-----string----n.wgram 馻绚s
----i------------20------
----i-num--------20------
-----string----n.wgram 馻绚s
-----string----n.wgram 馻绚s
-----string----n.wgram 馻绚s
-----string----n.wgram 馻绚s
-----string----n.wgram 馻绚s
-----string----n.oiles(騛窑=
----i------------26------
----i-num--------21------
-----string----.orSModu鯽蓼t
----i------------27------
----i-num--------22------
-----string----org
under win10, codeblocks 17.12, gcc 8.1.0
You are making life complicated for you in several places:
Don't count backwards: Instead of making num the count of duplicates, make it the count of unique trigraphs.
Scope variable definitions in functions as closely as possible. You have several uninitialized variables. You have declared them at the start of the function, but you need them only in local blocks.
Initialize as soon as you allocate. In your code, you use a flag to determine whather to create a new string. The code to allocate he string and to initialize it are in different blocks. Those blocks have the same flag as condition, but the flag is updated in between. This could lead to asynchronities, even to bugs when you try to initialize memory that wasn't allocated.
It's probably better to keep the strings and their counts together in a struct. If anything, this will help you with sorting later. This also offers some simplification: Instead of allocating chunks of 3 bytes, keep a char array of four bytes in the struct, so that all entries can be properly null-terminated. Those don't need to be allocated separately.
Here's an alternative implementation:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
struct tri {
char str[4]; // trigraph: 3 chars and NUL
int count; // count of occurrences
};
struct stat {
struct tri *tri; // list of trigraphs with counts
int size; // number of trigraphs
};
/*
* Find string 'key' in list of trigraphs. Return the index
* or in the array or -1 if it isn't found.
*/
int find_trigraph(const struct tri *tri, int n, const char *key)
{
for (int i = 0; i < n; i++) {
int j = 0;
while (j < 3 && tri[i].str[j] == key[j]) j++;
if (j == 3) return i;
}
return -1;
}
/*
* Create an array of trigraphs from the input string.
*/
struct stat getSearchArr(char* input, int length)
{
int num = 0;
struct tri *tri = malloc(sizeof(*tri) * (length - 2));
for(int i = 0; i < length - 2; i++) {
int index = find_trigraph(tri, num, input + i);
if (index < 0) {
snprintf(tri[num].str, 4, "%.3s", input + i); // see [1]
tri[num].count = 1;
num++;
} else {
tri[index].count++;
}
}
for(int i = 0; i < num; i++) {
printf("#%d %s: %d\n", i, tri[i].str, tri[i].count);
}
struct stat stat = { tri, num };
return stat;
}
/*
* Driver code
*/
int main(void)
{
char *input = "googleapis.com.wncln.wncln.org";
int length = strlen(input);
struct stat stat = getSearchArr(input, length);
// ... do stuff with stat ...
free(stat.tri);
return 0;
}
Footnote 1: I find that snprintf(str, n, "%.*s", len, str + offset) is useful for copying substrings: The result will not overflow the buffer and it will be null-terminated. There really ought to be a stanard function for this, but strcpy may overflow and strncpy may leave the buffer unterminated.
This answer tries to fix the existing code instead of proposing alternative/better solutions.
After fixing the output
printf("-----string----%s\n", nameArr[i-num]);
in the question, there is still another important problem.
You want to store 3 characters in nameArr[i-num] and allocate space for 3 characters. Later you print is as a string in the code shown above. This requires a trailing '\0' after the 3 characters, so you have to allocate memory for 4 characters and either append a '\0' or initialize the allocated memory with 0. Using calloc instead of malloc would automatically initialize the memory to 0.
Here is a modified version of the source code
I also changed the initialization of the string value and its length in main() to avoid the memory leak.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct strArrIntArr {
int length;
char **charArr;
int *intArr;
};
struct strArrIntArr getSearchArr(char* input, int length) {
struct strArrIntArr nameIndArr;
// flag of same bit
int same;
// flag/index of identical strings
int flag = 0;
// how many identical strings
int num = 0;
// array of split strings
char** nameArr = (char **)malloc(sizeof(char *) * (length - 2));
if ( nameArr == NULL ) exit(0);
// numbers of every split string
int* valueArr = (int* )malloc(sizeof(int) * (length-2));
if ( valueArr == NULL ) exit(0);
// loop length of search string -2 times (3-gram)
for(int i = 0; i<length-2; i++){
if(flag==0){
nameArr[i - num] = (char *)malloc(sizeof(char) * 4);
if ( nameArr[i - num] == NULL ) exit(0);
printf("----i------------%d------\n", i);
printf("----i-num--------%d------\n", i-num);
}
flag = 0;
// compare splitting string with existing split strings,
// if a string exists, it would not be stored
for(int k=0; k<i-num; k++){
same = 0;
for(int j=0; j<3; j++){
if(input[i + j] == nameArr[k][j]){
same ++;
}
}
// identical strings found, if all the three bits are the same
if(same == 3){
flag = 1;
num++;
break;
}
}
// if the current split string doesn't exist yet
// put current split string to array
if(flag == 0){
for(int j=0; j<3; j++){
nameArr[i-num][j] = input[i + j];
valueArr[i-num] = 1;
}
nameArr[i-num][3] = '\0';
}else{
valueArr[flag]++;
}
printf("-----string----%s\n", nameArr[i-num]);
}
// number of N-gram strings
nameIndArr.length = length- 2- num;
// array of N-gram strings
nameIndArr.charArr = nameArr;
nameIndArr.intArr = valueArr;
return nameIndArr;
}
int main(int argc, const char * argv[]) {
int length;
char* input = strdup("googleapis.com.wncln.wncln.org");
length = strlen(input);
// split the search string into N-gram strings
// and count the numbers of every split string
struct strArrIntArr nameIndArr = getSearchArr(input, length);
}
This other answer contains more improvements which I personally would prefer over the modified original solution.
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'm trying to understand how the return value of a function works, through the following program that has been given to me,
It goes like this :
Write a function that given an array of character v and its dim, return the capital letter that more often is followed by its next letter in the alphabetical order.
And the example goes like : if I have the string "B T M N M P S T M N" the function will return M (because two times is followed by N).
I thought the following thing to create the function:
I'm gonna consider the character inserted into the array like integer thank to the ASCII code so I'm gonna create an int function that returns an integer but I'm going to print like a char; that what I was hoping to do,
And I think I did, because with the string BTMNMPSTMN the function prints M, but for example with the string 'ABDPE' the function returns P; that's not what I wanted, because should return 'A'.
I think I'm misunderstanding something in my code or into the returning value of the functions.
Any help would be appreciated,
The code goes like this:
#include <stdio.h>
int maxvolte(char a[],int DIM) {
int trovato;
for(int j=0;j<DIM-1;j++) {
if (a[j]- a[j+1]==-1) {
trovato=a[j];
}
}
return trovato;
}
int main()
{
int dim;
scanf("%d",&dim);
char v[dim];
scanf("%s",v);
printf("%c",maxvolte(v,dim));
return 0;
}
P.S
I was unable to insert the value of the array using in a for scanf("%c,&v[i]) or getchar() because the program stops almost immediately due to the intepretation of '\n' a character, so I tried with strings, the result was achieved but I'd like to understand or at least have an example on how to store an array of character properly.
Any help or tip would be appreciated.
There are a few things, I think you did not get it right.
First you need to consider that there are multiple pairs of characters satisfying a[j] - a[j+1] == -1
.
Second you assume any input will generate a valid answer. That could be no such pair at all, for example, ACE as input.
Here is my fix based on your code and it does not address the second issue but you can take it as a starting point.
#include <stdio.h>
#include <assert.h>
int maxvolte(char a[],int DIM) {
int count[26] = {0};
for(int j=0;j<DIM-1;j++) {
if (a[j] - a[j+1]==-1) {
int index = a[j] - 'A'; // assume all input are valid, namely only A..Z letters are allowed
++count[index];
}
}
int max = -1;
int index = -1;
for (int i = 0; i < 26; ++i) {
if (count[i] > max) {
max = count[i];
index = i;
}
}
assert (max != -1);
return index + 'A';
}
int main()
{
int dim;
scanf("%d",&dim);
char v[dim];
scanf("%s",v);
printf("answer is %c\n",maxvolte(v,dim));
return 0;
}
#include <stdio.h>
int maxvolte(char a[],int DIM) {
int hold;
int freq;
int max =0 ;
int result;
int i,j;
for(int j=0; j<DIM; j++) {
hold = a[j];
freq = 0;
if(a[j]-a[j+1] == -1) {
freq++;
}
for(i=j+1; i<DIM-1; i++) { //search another couple
if(hold==a[i]) {
if(a[i]-a[i+1] == -1) {
freq++;
}
}
}
if(freq>max) {
result = hold;
max=freq;
}
}
return result;
}
int main()
{
char v[] = "ABDPE";
int dim = sizeof(v) / sizeof(v[0]);
printf("\nresult : %c", maxvolte(v,dim));
return 0;
}
I have written a C program to find out the number of similar characters between two strings. If a character is repeated again it shouldn't count it.
Like if you give an input of
everest
every
The output should be
3
Because the four letters "ever" are identical, but the repeated "e" does not increase the count.
For the input
apothecary
panther
the output should be 6, because of "apther", not counting the second "a".
My code seems like a bulk one for a short process. My code is
#include<stdio.h>
#include <stdlib.h>
int main()
{
char firstString[100], secondString[100], similarChar[100], uniqueChar[100] = {0};
fgets(firstString, 100, stdin);
fgets(secondString, 100, stdin);
int firstStringLength = strlen(firstString) - 1, secondStringLength = strlen(secondString) - 1, counter, counter1, count = 0, uniqueElem, uniqueCtr = 0;
for(counter = 0; counter < firstStringLength; counter++) {
for(counter1 = 0; counter1 < secondStringLength; counter1++) {
if(firstString[counter] == secondString[counter1]){
similarChar[count] = firstString[counter];
count++;
break;
}
}
}
for(counter = 0; counter < strlen(similarChar); counter++) {
uniqueElem = 0;
for(counter1 = 0; counter1 < counter; counter1++) {
if(similarChar[counter] == uniqueChar[counter1]) {
uniqueElem++;
}
}
if(uniqueElem == 0) {
uniqueChar[uniqueCtr++] = similarChar[counter];
}
}
if(strlen(uniqueChar) > 1) {
printf("%d\n", strlen(uniqueChar));
printf("%s", uniqueChar);
} else {
printf("%d",0);
}
}
Can someone please provide me some suggestions or code for shortening this function?
You should have 2 Arrays to keep a count of the number of occurrences of each aplhabet.
int arrayCount1[26],arrayCount2[26];
Loop through strings and store the occurrences.
Now for counting the similar number of characters use:
for( int i = 0 ; i < 26 ; i++ ){
similarCharacters = similarCharacters + min( arrayCount1[26], arrayCount2[26] )
}
There is a simple way to go. Take an array and map the ascii code as an index to that array. Say int arr[256]={0};
Now whatever character you see in string-1 mark 1 for that. arr[string[i]]=1; Marking what characters appeared in the first string.
Now again when looping through the characters of string-2 increase the value of arr[string2[i]]++ only if arr[i] is 1. Now we are tallying that yes this characters appeared here also.
Now check how many positions of the array contains 2. That is the answer.
int arr[256]={0};
for(counter = 0; counter < firstStringLength; counter++)
arr[firstString[counter]]=1;
for(counter = 0; counter < secondStringLength; counter++)
if(arr[secondString[counter]]==1)
arr[secondString[counter]]++;
int ans = 0;
for(int i = 0; i < 256; i++)
ans += (arr[i]==2);
Here is a simplified approach to achieve your goal. You should create an array to hold the characters that has been seen for the first time.
Then, you'll have to make two loops. The first is unconditional, while the second is conditional; That condition is dependent on a variable that you have to create, which checks weather the end of one of the strings has been reached.
Ofcourse, the checking for the end of the other string should be within the first unconditional loop. You can make use of the strchr() function to count the common characters without repetition:
#include <stdio.h>
#include <string.h>
int foo(const char *s1, const char *s2);
int main(void)
{
printf("count: %d\n", foo("everest", "every"));
printf("count: %d\n", foo("apothecary", "panther"));
printf("count: %d\n", foo("abacus", "abracadabra"));
return 0;
}
int foo(const char *s1, const char *s2)
{
int condition = 0;
int count = 0;
size_t n = 0;
char buf[256] = { 0 };
// part 1
while (s2[n])
{
if (strchr(s1, s2[n]) && !strchr(buf, s2[n]))
{
buf[count++] = s2[n];
}
if (!s1[n]) {
condition = 1;
}
n++;
}
// part 2
if (!condition ) {
while (s1[n]) {
if (strchr(s2, s1[n]) && !strchr(buf, s1[n]))
{
buf[count++] = s1[n];
}
n++;
}
}
return count;
}
NOTE: You should check for buffer overflow, and you should use a dynamic approach to reallocate memory accordingly, but this is a demo.
I have a question regarding an issue with a program in C I am making. I am going to write two different strings next to each other in two columns. I haven't found clear answers to my question since they almost always give examples of numbers with a known length or amount.
I have two strings, with a maximum length of 1500 characters, but to me unknown length. Let's for the sake of learning given them these values:
char string1[] = "The independent country is not only self-governed nation with own authorities.";
char string2[] = "This status needs the international diplomatic recognition of sovereignty.";
I want to write them next to each other, with a column width of twenty characters. I have set the difference between the columns to a regular 'tab'. Like this:
The independent coun This status needs th
try is not only self e international dipl
-governed nation wit omatic recognition o
h own authorities. f sovereignty.
I have tried with the following code but it isn't effective since I can't figure out how to adapt it to the length of the strings. It also just adapted to write five rows. I also get the below error.
Could someone please give me an example of how this could be done, and maybe with a pre-defined c-function in order to avoid using the for-loops.
void display_columns(char *string1, char *string2);
int main()
{
char string1[] = "The independent country is not only self-governed nation with own authorities.";
char string2[] = "This status needs the international diplomatic recognition of sovereignty.";
display_columns(string1,string2);
}
void display_columns(char *string1, char *string2)
{
int i,j;
for(i=0;i<5;i++)
{
for(j=0+20*i;j<20+20*i;j++)
{
printf("%c",string1[j]);
}
printf("\t");
for(j=0+20*i;j<20+20*i;j++)
{
printf("%c",string2[j]);
}
}
}
I guess this is more generic way to do it.
void print_line(char *str, int *counter) {
for (int i = 0; i < 20; i++) {
if (str[*counter] != '\0') {
printf("%c", str[*counter]);
*counter += 1;
}
else { printf(" "); }
}
}
void display_columns(char *string1, char *string2)
{
int counter = 0, counter2 = 0;
while (1) {
print_line(string1, &counter);
printf("\t");
print_line(string2, &counter2);
printf("\n");
if (string1[counter] == '\0' && string2[counter2] == '\0') {
break;
}
}
}
To print a single character, use:
printf("%c",string1[j]);
or
putchar(string1[j]);
This is the reason for the warnings and segmentation fault.
With this fix, the program somewhat works, you just have to print a newline as the last part of the loop:
for(i=0;i<5;i++)
{
for(j=0+20*i;j<20+20*i;j++)
{
putchar(string1[j]);
}
printf("\t");
for(j=0+20*i;j<20+20*i;j++)
{
putchar(string2[j]);
}
putchar('\n');
}
Update: For the function to work with strings of variable lengths, try this:
void display_columns(char *string1, char *string2)
{
int i,j;
int len1 = strlen(string1);
int len2 = strlen(string2);
int maxlen = (len1 > len2) ? len1 : len2;
int numloops = (maxlen + 20 - 1) / 20;
for(i=0; i<numloops; i++)
{
for(j=0+20*i;j<20+20*i;j++)
{
if (j < len1)
putchar(string1[j]);
else
putchar(' '); // Fill with spaces for correct alignment
}
printf("\t");
for(j=0+20*i;j<20+20*i;j++)
{
if (j < len2)
putchar(string2[j]);
else
break; // Just exit from the loop for the right side
}
putchar('\n');
}
}