Objective:-To print A3B5 as shown below..
AAABBBBB
I have achieved this through following code:-
#include <stdio.h>
#include <conio.h>
void main(){
char char1[2],*ptr_char,c;
int number[2],*ptr_number,i,j=0,k=0;
ptr_char = char1;
ptr_number = number;
printf("Enter character string\n");
for(i=0;i<2;i++)
scanf("%c",&char1[i]);
printf("Enter number array\n");
for(i=0;i<2;i++)
scanf("%d",&number[i]);
for(i=0;i<ptr_number[j];i++)
printf("%c",ptr_char[k]);
j++;
k++;
for(i=0;i<ptr_number[j];i++)
printf("%c",ptr_char[k]);
getch();
}
///////////////////////////////////////////////////////////
Now i have to take a single array in which my A3B5 can be taken instead of taking two array.
But i am unable to take array like that.
I want to take A3B5 in single array .
Is this possible?
You can say:
char csrc[]= "A3B5" ;
The issue here is that 3 and 5 are not numbers, they are characters. So you will have to convert from "3" to 3. You have several options for that, including sscanf and atoi. As others have suggested, read in the entire string at once, and it will be in the char array, and then decompose it into the separate parts you want.
check this code
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int count = 0, i = 0;
char temp = 0;
char ip[16] = {0};
char filter[8] = {0};
int num[8] = {0};
printf("Enter input :\t");
scanf("%s", ip);
for (count = 0, i =0; count < strlen(ip); count++, i++)
{
filter[i] = ip[count]; //take out the character to print
count++;
temp = ip[count]; //read the number of print iteration as a character
num[i] = atoi(&temp); // convert "n" [charater] to n [int]
}
for (count = 0; count < strlen(ip)/2 ; count++)
{ //print loop
for (i = 0; i < num[count]; i++)
printf("%c\t", filter[count]);
printf("\n\n");
}
return 0;
}
Sample run and o/p
[sourav#localhost test]$ ./a.out
Enter input : S2G5
S S
G G G G G
[sourav#localhost test]$
Related
So, I was writing this code for counting the digit frequency i.e. the number of times the digits from 0-9 has appeared in a user inputted string(alphanumeric). So, I took the string, converted into integer and tried to store the frequency in "count" and print it but when I run the code, count is never getting incremented and the output comes all 0s. Would be grateful if anyone points out in which part my logic went wrong.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
// takes string input
char *s;
s = malloc(1024 * sizeof(char));
scanf("%[^\n]", s);
s = realloc(s, strlen(s) + 1);
//turns the string to int
int x = atoi(s);
int temp = x, len = 0;
//calculates string length
while (x != 0) {
x = x / 10;
len++;
}
x = temp;
//parses through the string and matches digits with each number
for (int j = 0; j < 10; j++){
int count = 0;
for(int i = 0; i < len; i++){
if(x % 10 == j){
count++;
}
x = x / 10;
}
x = temp;
printf("%d ", count);
}
return 0;
}
To write a correct and reasonable digit-counting program:
Do not allocate any buffer for this.
Create an array to count the number of times each digit occurs. The array should have ten elements, one for each digit.
Initialize the array to zero in each element.
In a loop, read one character at a time.
Leave the loop when the read routine (such as getchar) indicates end-of-file or a problem, or, if desired, returns a new-line or other character you wish to use as an end-of-input indication.
Inside the loop, check whether the character read is a digit. If the character read is a digit, increment the corresponding element of the array.
After the loop, execute a new loop to iterate through the digits.
Inside that loop, for each digit, print the count from the array element for that digit.
Your approach is way to complicated for a very easy task. This will do:
void numberOfDigits(const char *s, int hist[10]) {
while(*s) {
if(isdigit(*s))
hist[*s - '0']++;
s++;
}
}
It can be used like this:
int main(void) {
char buf[1024];
int hist[10];
fgets(buf, sizeof buf, stdin);
numberOfDigits(s, hist);
for(int i=0; i<10; i++)
printf("Digit %d occurs %d times\n", i, hist[i]);
}
This can also be quite easily achieved without a buffer if desired:
int ch;
int hist[10];
while((ch = getchar()) != EOF) {
if(isdigit(ch))
hist[ch - '0']++;
}
#include <stdio.h>
int main(void) {
int input = 1223330;
int freq[10] = {0};
input = abs(input);
while(input)
{
freq[input%10]++;
input /= 10;
}
for(int i=0; i<10; ++i)
{
printf("%d: %.*s\n", i, freq[i], "*************************************************");
}
return 0;
}
Output:
Success #stdin #stdout 0s 5668KB
0: *
1: *
2: **
3: ***
4:
5:
6:
7:
8:
9:
This app is currently limited by the size of an int (approximately 9 or 10 digits).
You can update it to use a long long easily, which will get you to about 19 digits.
I am making a program which requires the user to input an argument (argv[1]) where the argument is every letter of the alphabet rearranged however the user likes it. Examples of valid input is "YTNSHKVEFXRBAUQZCLWDMIPGJO" and "JTREKYAVOGDXPSNCUIZLFBMWHQ". Examples of invalid input would then be "VCHPRZGJVTLSKFBDQWAXEUYMOI" and "ABCDEFGHIJKLMNOPQRSTUYYYYY" since there are duplicates of 'V' and 'Y' in the respective examples.
What I know so far is, that you can loop through the whole argument like the following
for (int j = 0, n = strlen(argv[1]); j < n; j++)
{
//Place something in here...
}
However, I do not quite know if this would be the right way to go when looking for duplicates? Furthermore, I want the answer to be as simple as possible, right know time and cpu usage is not a priority, so "the best" algorithm is not necessarily the one I am looking for.
Try it.
#include <stdio.h>
#include <stddef.h>
int main()
{
char * input = "ABCC";
/*
*For any character, its value must be locate in 0 ~ 255, so we just
*need to check counter of corresponding index whether greater than zero.
*/
size_t ascii[256] = {0, };
char * cursor = input;
char c = '\0';
while((c=*cursor++))
{
if(ascii[c] == 0)
++ascii[c];
else
{
printf("Find %c has existed.\n", c);
break;
}
}
return 0;
}
assuming that all your letters are capital letters you can use a hash table to make this algorithm work in O(n) time complexity.
#include<stdio.h>
#include<string.h>
int main(int argc, char** argv){
int arr[50]={};
for (int j = 0, n = strlen(argv[1]); j < n; j++)
{
arr[argv[1][j]-'A']++;
}
printf("duplicate letters: ");
for(int i=0;i<'Z'-'A'+1;i++){
if(arr[i]>=2)printf("%c ",i+'A');
}
}
here we make an array arr initialized to zeros. this array will keep count of the occurrences of every letter.
and then we look for letters that appeared 2 or more times those are the duplicated letters.
Also using that same array you can check if all the letters occured at least once to check if it is a permutation
I don't know if this is the type of code that you are looking for, but here's what I did. It looks for the duplicates in the given set of strings.
#include <stdio.h>
#include <stdlib.h>
#define max 50
int main() {
char stringArg[max];
int dupliCount = 0;
printf("Enter A string: ");
scanf("%s",stringArg);
system("cls");
int length = strlen(stringArg);
for(int i=0; i<length; i++){
for(int j=i+1; j<length; j++){
if(stringArg[i] == stringArg[j]){
dupliCount +=1;
}
}
}
if(dupliCount > 0)
printf("Invalid Input");
printf("Valid Input");
}
This code snippet is used to count the duplicate letters in the array.
If you want to remove the letters, Also this code snippet is helpful .Set null when the same characters are in the same letter.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int count=0;
int array[]={'e','d','w','f','b','e'};
for(int i=0;i<array.Lenth;i++)
{
for(int j=1;j<array.Length;j++)
{
if(array[i]==array[j])
{
count++;
}
}
}
printf("The dublicate letter count is : %d",count);
}
Problem
I have made sorting program which is similiar to other found at
https://beginnersbook.com/2015/02/c-program-to-sort-set-of-strings-in-alphabetical-order/
but program which i made is not working.
I think both are same but my program giving me waste output.
Also i want to know in other program count is set to 5 for example and it should take 6 input starting from 0 but it is getting only 5,How?
My Program
#include <string.h>
#include <stdio.h>
int main() {
char str[4][10],temp[10];
int i,j;
printf("Enter strings one by one : \n");
for(i=0;i<5;i++)
scanf("%s",str[i]);
for(i=0;i<5;i++)
for(j=i+1;j<5;j++)
if(strcmp(str[i],str[j])>0){
strcpy(temp,str[i]);
strcpy(str[i],str[j]);
strcpy(str[j],temp);
}
printf("\nSorted List : ");
for(i=0;i<5;i++)
printf("\n%s",str[i]);
printf("\n\n");
return 0;
}
Use qsort().
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int pstrcmp( const void* a, const void* b )
{
return strcmp( *(const char**)a, *(const char**)b );
}
int main()
{
const char* xs[] =
{
"Korra",
"Zhu Li",
"Asami",
"Mako",
"Bolin",
"Tenzin",
"Varrick",
};
const size_t N = sizeof(xs) / sizeof(xs[0]);
puts( "(unsorted)" );
for (int n = 0; n < N; n++)
puts( xs[ n ] );
// Do the thing!
qsort( xs, N, sizeof(xs[0]), pstrcmp );
puts( "\n(sorted)" );
for (int n = 0; n < N; n++)
puts( xs[ n ] );
}
Please don’t use bubble sort. In C, you really do not have to write your own sorting algorithm outside of specialized needs.
Here is a program which will sort and print your inputted strings. Answering a little late, but just in case others have a similar question.
// This program will sort strings into either ascending or descending order
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_SIZE 1000
#define EQUAL 0
#define ASCENDING 0
#define DESCENDING 1
// Function prototypes
void swap_str(char str1[], char str2[]);
void sort_strings(char str[MAX_SIZE][MAX_SIZE], int len, int order_type);
void print_2d_array(char str[MAX_SIZE][MAX_SIZE], int len);
int main(void) {
int order_type;
char str[MAX_SIZE][MAX_SIZE];
// User selecting the order type
printf("-----------[Order Type]-----------\n");
printf("Sort your strings in ascending or descending order?\n");
printf("0 = Ascending | 1 = descending\n");
scanf("%d", &order_type);
if (order_type != ASCENDING && order_type != DESCENDING) {
printf("Please enter 0 or 1\n");
exit(1);
}
// User inputting their strings
printf("---------[Enter Strings]----------\n");
printf("Enter Strings one by one.\n");
printf("Max Strings: %d | Max String Len: %d\n", MAX_SIZE, MAX_SIZE);
int i = 0;
while ((i < MAX_SIZE) && (scanf("%s", str[i]) == 1)) i++;
if (i == MAX_SIZE) printf("You reached the maximum strings allowed\n");
// Program output of the sorted strings
printf("---------[Sorted Strings]---------\n");
sort_strings(str, i, ASCENDING);
print_2d_array(str, i);
return 0;
}
// Swaps two strings (assuming memory allocation is already correct)
void swap_str(char str1[], char str2[]) {
char temp[MAX_SIZE];
strcpy(temp, str1);
strcpy(str1, str2);
strcpy(str2, temp);
}
// Will sort a 2D array in either descending or ascending order,
// depending on the flag you give it
void sort_strings(char str[MAX_SIZE][MAX_SIZE], int len, int order_type) {
int i = 0;
while (i < len) {
int j = 0;
while (j < len) {
if ((order_type == ASCENDING) &&
(strcmp(str[i], str[j]) < EQUAL)) {
swap_str(str[i], str[j]);
} else if ((order_type == DESCENDING) &&
(strcmp(str[i], str[j]) > EQUAL)) {
swap_str(str[i], str[j]);
}
j++;
}
i++;
}
}
// Will print out all the strings 2d array
void print_2d_array(char str[MAX_SIZE][MAX_SIZE], int len) {
int i = 0;
while (i < len) {
printf("%s\n", str[i]);
i++;
}
}
Example (Ascending order):
-----------[Order Type]-----------
Sort your strings in ascending or descending order?
0 = Ascending | 1 = descending
0
---------[Enter Strings]----------
Enter Strings one by one.
Max Strings: 1000 | Max String Len: 1000
Mango
Watermelon
Apple
Banana
Orange
// I pressed CTRL+D here (Linux) or CTRL+Z then enter (on Windows). Essentially triggering EOF. If you typed the MAX_SIZE it would automatically stop.
---------[Sorted Strings]---------
Apple
Banana
Mango
Orange
Watermelon
it should take 6 input starting from 0 but it is getting only 5,How?
This loop
for(i=0;i<5;i++)
scanf("%s",str[i]);
execute for i being 0, 1, 2, 3, 4 so it loops 5 times.
If you want 6 loops do
for(i=0;i<=5;i++)
^
Notice
or
for(i=0;i<6;i++)
^
Notice
Also notice this line
char str[6][10],temp[10];
^
Notice
so that you reserve memory for 6 strings
This is not an answer, but some criticism of the code you refer to:
#include<stdio.h>
#include<string.h>
int main(){
int i,j,count;
char str[25][25],temp[25];
puts("How many strings u are going to enter?: ");
scanf("%d",&count); // (1)
puts("Enter Strings one by one: ");
for(i=0;i<=count;i++) // (2)
gets(str[i]);
for(i=0;i<=count;i++)
for(j=i+1;j<=count;j++){
if(strcmp(str[i],str[j])>0){
strcpy(temp,str[i]);
strcpy(str[i],str[j]);
strcpy(str[j],temp);
}
}
printf("Order of Sorted Strings:"); // (3)
for(i=0;i<=count;i++)
puts(str[i]);
return 0;
}
And the criticism:
(1) scanf("%d",&count); reads a number into count, and returns after that. It does not consume the line break(!)
(2) this loop does not print anything, just reads. However if you put
for(i=0;i<=count;i++){
printf("%d:",i);
gets(str[i]);
}
in its place, you will suddenly see that it asks for names 0...5, just skips the 0 automatically. That is where the line break is consumed, it reads an empty string. You can also make it appear, if instead of putting 5 into the initial question, you put 5 anmoloo7.
(3) in the printout the names appear below the title Order of Sorted Strings. But there is no linebreak in that printf. The thing is that the empty string is "smaller" than any other string, so it gets to the front of the list, and that is printed there first. If you do the 'trick' of appending a name after the initial number, the output will look different, there will be 6 names, and one of them appended directly to the title.
Plus there is the thing what you probably get from your compiler too: gets is a deadly function, forget its existence and use fgets with stdin as appears in other answers.
I have this sample that I made:
#include <stdio.h>
#include <string.h>
void main()
{
char str[100],ch;
int i,j,l;
printf("\n\nSort a string array in ascending order :\n");
printf("--------------------------------------------\n");
printf("Input the string : ");
fgets(str, sizeof str, stdin);
l=strlen(str);
/* sorting process */
for(i=1;i<l;i++)
for(j=0;j<l-i;j++)
if(str[j]>str[j+1])
{
ch=str[j];
str[j] = str[j+1];
str[j+1]=ch;
}
printf("After sorting the string appears like : \n");
printf("%s\n\n",str);
}
#include <stdio.h>
#include <string.h>
#define STRING_MAX_WIDTH 255
void sort_strings(char str_arr[][STRING_MAX_WIDTH],int len){
char temp[STRING_MAX_WIDTH];
for(int i=0;i<len-1;i++){
if(strcmp(&str_arr[i][0],&str_arr[i+1][0])>0){
strcpy(temp, str_arr[i+1]);
strcpy(str_arr[i+1],str_arr[i]);
strcpy(str_arr[i],temp);
sort_strings(str_arr, len);
}
}
}
int main(){
char str_arr[][STRING_MAX_WIDTH] = {"Test", "Fine", "Verb", "Ven", "Zoo Keeper", "Annie"};
int len = sizeof(str_arr)/STRING_MAX_WIDTH;
sort_strings(str_arr, len);
for(int i=0;i<len;i++){
printf("%s\r\n", str_arr[i]);
}
}
Hi how would you count the number of occurences in the given word like shown below because with the program I have right now it doesn't seem to room correctly.
#include <stdio.h>
int main(void)
{
char a;
char lang[] = "pneumonoultramicroscopicsilicovolcanoconiosis";
char i = 0;
char count = 0;
printf("pneumonoultramicroscopicsilicovolcanoconiosis\n");
printf("\nEnter the letter you want to find the number of\n");
scanf("%c", &lang);
for (i = 0; i <= 46; i++)
if (a == lang[i]) {
count++;
}
printf("Number of %c is %d..\n", a, count);
return 0;
Your scanf is the problem.
Try:
scanf("%c",&a);
declare count as int instead of char
also change scanf to take input a
#include <stdio.h>
int main(void)
{
char a;
char lang[] = "pneumonoultramicroscopicsilicovolcanoconiosis";
char i = 0;
int count = 0;
printf("pneumonoultramicroscopicsilicovolcanoconiosis\n");
printf("\nEnter the letter you want to find the number of\n");
scanf("%c", &a);
for (i = 0; i <= 46; i++)
if (a == lang[i]) {
count++;
}
printf("Number of %c is %d..\n", a, count);
return 0;
}
You probably shouldn't hard code the max length of the string (46) if at all possible in case you are given a longer string, but assuming it's an assigned assignment that is set it shouldn't be a problem.
i and count should also be ints if possible for a bigger size count. And &lang should be &a since lang is already assigned while a is your checker.
iI'm trying to understand how to properly use arrays in C and tried to write a simple program, which is supposed to take a 5-integers array and eliminate zeroes to the left. This is my attempt:
#include <stdio.h>
int main() {
int seq[5];
int i;
int cor[5];
int counter;
printf("Type the 5 numbers: ");
scanf("%s", &seq);
for (i=0; i<5; i++){
if (seq[i] != 0) {
for (counter=0; counter<5-i; counter++){
cor[counter]=seq[i+counter];
}
break;
}
}
printf("%s", cor);
return 0;
}
The idea was that when something such as 00101 was entered, the program would look at each entry and check whether it is 0. If it isn't, at the position i, it would write a new array in which the 0-th position is assigned the value of the original array at i, the 1-th position would be assigned the value at i+1 and so on, and then it would print this new array, which should have no useless zeroes to the left. But it just prints the original array. What is wrong? Sorry for the begginner's question.
You have two ways to do what you want.
using chars :
#include <stdio.h>
int main() {
char seq[6]; /* reserve one place for terminating null */
int i;
char cor[5];
int counter;
printf("Type the 5 digits : ");
scanf("%5s", seq); /* limit to 5 chars and seq is already an array : ne & */
for (i=0; i<5; i++){
/* should control seq[i] >= '0' && seq[i] <= '9' */
if (seq[i] != '0') {
for (counter=0; counter<6-i; counter++){ /* also copy terminating null */
cor[counter]=seq[i+counter];
}
break;
}
}
printf("%s", cor);
return 0;
}
but user could give you chars that are not digits.
using ints :
#include <stdio.h>
int main() {
int seq[5];
int i;
int cor[5];
int counter;
printf("Type the 5 numbers: ");
i = scanf("%d%d%d%d%d", seq, seq+1, seq+2, seq+3, seq+4); /* seq+i = &(seq[i]) */
/* should control i == 5 */
for (i=0; i<5; i++){
if (seq[i] != 0) {
for (counter=0; counter<5-i; counter++){
cor[counter]=seq[i+counter];
}
break;
}
}
printf("%d%d%d%d%d", cor[0], cor[1], cor[2], cor[3], cor[4]);
return 0;
}
You cannot store a string in an integer array. You have to store strings in character arrays.