I just want to reverse a string using for loop and array. Don't want to use any predefined function. I used the following code but its near to nothing. Please share some good suggestions.
int main(){
char a[]="this is a man";
char b[30];
int p= sizeof(a)/sizeof(a[0]);
for(int i=p-1;i>0;i--){
for(int j=0;j<p;j++){
b[j]=a[i];
}
}
printf("array is %s",b);
return 0;
}
#include<stdio.h>
int main(){
char str[] = "str to rev";
char revstr[12]={'\0'};
int i, j;
int length = strlen(str);
j = 0;
for(i = length-1; i>=0; i--){
revstr[j] = str[i];
j = j + 1;
}
printf("%s", revstr);
return 0;
}
1) In your first for loop, you have to reach 0 (i>=0)
for(int i=p-1;i>=0;i--){
2) The a[p-1] contains the null termination('\0') of your string a[]. And the null termination should not be included in the array reverse procedure. So in your first loop you should start from p-2 and not from p-1.
And after finishing the reversing you have to add a '\0' (null terminator) at the end of your b array
b[j]='\0'; // add this
printf("array is %s",b);
return 0;
3) And as said in the other answers, you have to use only one loop and not 2 loops.
int i,j;
for (i=p-2, j=0; i>=0; i--,j++) {
b[j]=a[i];
}
b[j]='\0';
printf("array is %s",b);
Using while loop::
void main()
{
char str[100],temp;
int i,j=0;
printf("nEnter the string :");
gets(str);
i=0;
j=strlen(str)-1;
while(i<j)
{
temp=str[i];
str[i]=str[j];
str[j]=temp;
i++;
j--;
}
printf("nReverse string is :%s",str);
return(0);
}
Using for loop::
void StrRev(char *str)
{
int i, len, endpos;
len = strlen(str);
endpos = len-1;
for(i = 0; i < len / 2; i++)
{
char temp = str[i];
str[i] = str[endpos - i];
str[endpos - i] = temp ;
}
}
Related
I'm learning C and I've a problem with this school homework.
I have to make function which get two strings from user as parameters. The function removes all spaces from the first string and returns the "cleaned" strings as the other parameter.
The main function ask three strings, uses function to remove spaces and prints "cleaned" strings.
My code doesn't work as it should? What goes wrong?
#include <stdio.h>
void removeSpaces(char *, char *);
int main(){
int i, j;
char string[101], strings[1][101];
for(i = 0; i <= 2; i++){
fgets(string, 100, stdin);
for(j = 0; string[j] != '\0'; j++){
strings[i][j] = string[j];
}
strings[i][j] = '\0';
removeSpaces(strings[i], strings[i]);
}
for(i = 0; i <= 0; i++){
for(j = 0; j <= 101; j++){
printf("%c", strings[i][j]);
}
}
}
void removeSpaces(char *string1, char *string2){
int i, j;
for(i = 0; string1[i] != '\0'; i++){
if(string1[i] != ' '){
string2[i] = string1[j];
j++;
}
}
string2[i] = '\0';
}
You have to be more careful when writing code. There are several things wrong:
In removeSpaces(), you never initialize j. So it can be anything.
You are also mixing up i and j inside removeSpaces(). i should only be used to index string1, and j only for string2.
strings[1][101] is only one string, not 3. But the first for-loop in main() runs 3 times.
You don't have to print strings character by character, just printf("%s", strings[i]) or fputs(strings[i], stdout).
I'm not sure why you used a two-dimensional array strings here. You only need two strings. Renaming the variables can also help you avoid getting confused. Consider:
#include <stdio.h>
static void removeSpaces(const char *input, char *output) {
int i, o;
for(i = 0, o = 0; input[i] != '\0'; i++) {
if(input[i] != ' ') {
output[o] = input[i];
o++;
}
}
output[o] = '\0';
}
int main() {
char input[100], output[100];
fgets(input, sizeof input, stdin);
removeSpaces(input, output);
fputs(output, stdout);
}
how to code a user defined function that searches and replaces a character occurrences of any of the character contained in another string with a character string.
Cannot used any string variable in the code, has to be a user defined function.
Thanks
This is what i have tried so far
#define _CRT_SECURE_NO_WARNINGS
#include
#include
void s1();
void s2();
int main(void)
{
int i=0;
s1();
s2();
printf("c = {'$'} ");
}//main
void s1(){
int i = 0;
while (i <= 40){
printf("%c", (rand() % 25) + 'A');
i++;
}
}
void s2(){
char s2[20];
printf("\nEnter a string of minimum 2 and maximum 20 characters= ");
gets(s2);
puts(s2);
}
/*
I just need to make another function that searches s1 and replaces any occurrence of any of the character contained is s2 with a character that can be anything(e.g. '$')
*/
//If I have understood your question then this should be answer
char *replace(char [] a, char b[], int lower, int upper){
char c[100];
int j = 0;
for(int i = 0; i < lower; i++){
c[j] = a[i];
j++;
}
for(int i = 0; i < strlen(b); i++){
c[j] = b[i];
j++;
}
for(int i = upper; i < strlen(a); i++){
c[j] = a[i];
j++;
}
c[j] = '\0'
for(int i = 0; i < strlen(c); i++){
a[i]= c[i];
}
a[i] = '\0';
return a;
}
C program for removal of duplicate characters from the given string. It uses the O(n2) can we do it in O(n) order. Please comment on this program.
int main()
{
char a[100],b[100],temp='\0';
int i,n,j,count=0,p=0,k=0;
printf("ENTRE THE STRING \n");
scanf("%s",a);
n = strlen(a);
i=0;
while(i < n)
{
count=0;
temp = a[i];
for(j = i ; j < n ; j++ )
{
if(temp==a[j])
{
count++;
}
}
if(count<2)
{
b[k] = temp;
k++;
}
i++;
}
b[k]='\0';
printf("THE RESULTED STRING IS \n");
for(p = 0 ; p < k ; p++)
printf("%c ",b[p]);
printf("\n");
return 0;
}
You can create a O(n) algorithm for this.
Steps:
Create another array bucket[] with size 255. (Should adjust all the characters)
Initialise every element in bucket[] to 0.
Run a loop and increment the bucket[] at the index a[i].
Now, run another loop through the bucket[], if bucket[i] > 0, append the (char) i to the b[] array.
Code:
#include <stdio.h>
#include <string.h>
int main()
{
char a[100], b[100];
int bucket[256] = {0};
int i;
printf("Enter the string:");
scanf("%s",a);
int n = strlen(a);
for(i = 0; i < n; ++i)
{
//Incrementing the character count of each character.
bucket[a[i]]++;
}
//Keep track of the index where the next character is to be appended.
int b_pos = 0;
for (i = 0; i < 256; ++i)
{
//Character occurs in a[], we don't care if it occurs once
//or twice, we just need one instance of it.
if (bucket[i] > 0)
{
b[b_pos] = (char) i;
b_pos++;
}
}
b[b_pos] = '\0';
printf("Modified string : %s",b);
}
Take a look at this:
int main()
{
char a[100],b[100];
int i,n,j,count=0,k=0;
printf("ENTRE THE STRING \n");
scanf("%s",a);
n = strlen(a);
b[0] = a[0];
k = 1;
for(i=1;i<n;i++)
{
for(j=0;j<i;j++)
{
if(a[i] == b[j])
{
count = 1;
break;
}
}
if(count == 0)
{
b[k] = a[i];
k++;
}
else
{
count = 0;
}
}
b[k] = 0;
printf("RESULT %s",b);
return 0;
}
Can someone advise why the loop in the main dies after the fifth iteration never completing
it's intended goal of reducing the character array down to 1 final element? I've gotten it this
far and am completely consumed as their should be 11 iterations as returned by the call
size_t strlen( char const *str )
{
int length = 0;
while (*str++ !='\0')
{
length += 1;
}
return length;
}
void abracadabra( char *word )
{
int i, c;
int len = strlen(word)-1;
for (i = 0; i <= len; i++)
{
putchar(*word);
putchar(' ');
*(word++);
}
}
int main()
{
char word[250];
int i, j;
printf ("enter your word:\n");
scanf ("%[^\n]s", &word);
for (i = 0; i <= strlen(word)-1; i++)
{
abracadabra(word);
putchar('\0');
printf("\n");
for (j = 0; j <= i; j++)
{
putchar('\0');
}
word[strlen(word) - 1] = '\0';
}
word[strlen(word)-1] = '\0';
printf("\n");
system("pause");
return 0;
}
Each time you execute the outer for loop in main the size of the string decreases by 1. The counter i is also increasing by 1 each time. This causes you to run the loop half of the times that you intend to.
int size = strlen(word);
for (i = 0; i < size; i++) {
\\same inner code
}
Subbing the above code for the outer for loop in main resolves the issue.
The is mistake in using the variable i and strlen in for loop. i is keep increasing and word length is decreasing. So in the mid, loop is terminated due to i>strlen (word)
int main()
{
char word[250];
int i, j;
printf ("enter your word:\n");
scanf ("%[^\n]s", &word);
// for (i = 0; i <= strlen(word)-1; i++)
while ( strlen(word) )
{
abracadabra(word);
putchar('\0');
printf("\n");
word[strlen(word) - 1] = '\0';
}
word[strlen(word)-1] = '\0';
printf("\n");
system("pause");
return 0;
}
I've nearly finished my anagram solver program where I input two strings and get the result of whether they are anagrams of each other. For this example i'm using 'Payment received' and 'Every cent paid me'.
The problem i'm getting is when I output the letterCount arrays, letterCount1 is incorrect (it doesn't think there is a character 'd' but there is.) but letterCount2 is correct.
Can anyone see a problem with this because i'm completely baffled?
#include <stdio.h>
#include <string.h>
int checkAnagram(char string1[], char string2[])
{
int i;
int count = 0, count2 = 0;
int letterCount1[26] = {0};
int letterCount2[26] = {0};
for(i = 0; i < strlen(string1); i++)
{
if(!isspace(string1[i]))
{
string1[i] = tolower(string1[i]);
count++;
}
}
for(i = 0; i < strlen(string2); i++)
{
if(!isspace(string2[i]))
{
string2[i] = tolower(string2[i]);
count2++;
}
}
if(count == count2)
{
for(i = 0; i < count; i++)
{
if(string1[i] >='a' && string1[i] <= 'z')
{
letterCount1[string1[i] - 'a'] ++;
}
if(string2[i] >='a' && string2[i] <= 'z')
{
letterCount2[string2[i] - 'a'] ++;
}
}
printf("%s\n", string1);
for(i = 0; i < 26; i++)
{
printf("%d ", letterCount1[i]);
printf("%d ", letterCount2[i]);
}
}
}
main()
{
char string1[100];
char string2[100];
gets(string1);
gets(string2);
if(checkAnagram(string1, string2) == 1)
{
printf("%s", "Yes");
} else
{
printf("%s", "No");
}
}
That's because your count holds the count of non-space characters, but you keep the strings with the spaces.
For example, the string "hello world" has 11 characters, but if you run it through the loops your count will be 10 (you don't count the space). However, when you later go over the strings and count the appearance of each letter, you will go over the first 10 characters, therefore completely ignoring the last character - a 'd'.
To fix it, you need to go over all characters of the string, and only count the alphanumeric ones.
I fixed it for you:
#include <stdio.h>
#include <string.h>
int checkAnagram(char string1[], char string2[])
{
int i;
int count = 0, count2 = 0;
int letterCount1[26] = {0};
int letterCount2[26] = {0};
int len1 = strlen(string1);
int len2 = strlen(string2);
for(i = 0; i < len1; i++)
{
if(!isspace(string1[i]))
{
string1[i] = tolower(string1[i]);
count++;
}
}
for(i = 0; i < len2; i++)
{
if(!isspace(string2[i]))
{
string2[i] = tolower(string2[i]);
count2++;
}
}
if(count == count2)
{
for (i=0; i<len1; i++)
if (!isspace(string1[i]))
letterCount1[string1[i]-'a']++;
for (i=0; i<len2; i++)
if (!isspace(string2[i]))
letterCount2[string2[i]-'a']++;
int flag = 1;
for(i = 0; flag && i < 26; i++)
if (letterCount1[i] != letterCount2[i])
flag = 0;
return flag;
}
return 0;
}
main()
{
char string1[100];
char string2[100];
gets(string1);
gets(string2);
if(checkAnagram(string1, string2) == 1)
{
printf("%s", "Yes");
} else
{
printf("%s", "No");
}
}
First, don't calculate an string's length inside a loop. I extracted them into len1 and len2 variables.
Second, your loop was wrong! You shouldn't go up to count, you should go up to that string's length.
Third, you didn't return anything from checkAnagram function.