I want to input an array of integers without giving spaces.
For ex:- 032146548 ,each integer should be stored in array distinctly ,
i.e a[0]=0,a[1]=3,a[2]=2 and so on.
How can i do this ?
I think it's clearer to say "each digit", since it's not at all obvious how many "integers" the character sequence 032146548 represents (the common practice is "one") once you know it's supposed to be several.
The simplest way is to just read it in as a string of digits, then convert each digit to its integer counterpart by subtracting '0':
char line[12];
unsigned int a[10];
if(fgets(line, sizeof line, stdin) != NULL)
{
const size_t digits = strlen(line) - 1;
for(size_t i = 0; i < sizeof a; ++i)
{
if(i < digits && isdigit((unsigned int) line[i]))
a[i] = line[i] - '0';
else
a[i] = 0;
}
}
Use this if you are reading from file,
int i=0;
while(scanf("%1d",&a[i])==1)
{
i++;
}
Use this if you know how many inputs are there,
for(int i=0;i<inputLength;i++)
{
scanf("%1d",&a[i]);
}
#include <stdio.h>
int main(){
int a[16];
int i, j, stat;
char ch[2] ={0};
for(i=0;i<16;++i){
if(1!=(stat=scanf("%1d%1[^0-9]", &a[i], ch))){
if(stat==2)
++i;
break;
}
}
for(j=0;j<i;++j)
printf("%d ", a[j]);
printf("\n");
return 0;
}
Related
I want to compare the integers in a string with integers (0-9) and I wrote this -
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
char num[100];
int count = 0;
scanf("%s", num);
int len = strlen(num);
for (int i = 0; i <= 9; i++)
{
for (int j = 0; j <= len; j++)
{
if (i == (num[j] - '0'))
{
count++;
}
}
printf("%d ", count);
count = 0;
}
return 0;
}
No problems with this (works in most cases but it is failing in few cases). So can you please give me alternate and best idea to do this?
Thanks in advance
Complete pic -
The root cause is not in char comparison, but in the under-allocated buffer:
char num[100];
The assignment constraint is:
1 <= len(num) <= 1000
After increasing the buffer size, all the tests pass.
Besides a too small input buffer (i.e. 100 instead of 1001), I think your approach is too complex.
Instead of a nested loop, I'll suggest an array to count the frequency, i.e. an array with 10 elements so that you have a counter for each digit.
int main() {
char num[1001]; // 1000 chars + 1 zero termination
int count[10] = {0}; // Array of 10 zero initialized counters, one for each digit
scanf("%1000s", num); // At max accept 1000 chars input
char* p = num;
while (*p)
{
if (isdigit(*p) ++count[*p - '0'];
++p;
}
for (int i = 0; i < 10; ++i) printf("%d ", count[i]);
puts("");
return 0;
}
If you don't want to use isdigit you can instead do:
if (*p >= '0' && *p <= '9') ++count[*p - '0'];
I'm quite new to C and I'm wondering why in the code below, the char I'm comparing to each letter of the string word is showing that it's equal everytime.
For example
If I've inputted the word
apple
and I'm looking for any repeating char in "apple" my function. I pass in to the function each char of apple such as a, p, p etc. It should return 1 when I pass in p since it's repeated, but instead, for every char of apple, my function says a == word[0], a == word[1] even though word[1] for "apple" is 'p'.
I know char is ASCII, so each char has a number value, but I'm not sure why this is not working. Perhaps, I'm using the pointer *word in the functions arguments incorrectly?
My code is below for my function, rpt_letter:
int rpt_letter(char *word, char c)
{
int i;
int count = 0;
i = 0;
printf("This is the WORD %s\n", word);
while(count < 2)
{
if(word[i] == c)
{
count++;
printf("the count is %d\n the char is %c and the string is %c\n", count, c, word[i]);
}
i++;
}
if (count<2)
{
// printf("letter %c was not found in the array. \n", c);
return 0;
}
else
{
//printf("letter %c was found at index %d in the array.\n", c, mid);
repeats[rpt_counter] = c;
rpt_counter++;
return 1;
}
return 0;
}
I'll include the main method just in case -- but I believe the main method is working well
int main(void)
{
//! showArray(list, cursors=[ia, ib, mid])
//int n = 51;
char word[51];
scanf("%s", word);
//length of string
for (n=0; word[n] != '\0'; n++); //calculate length of String
printf("Length of the string: %i\n", n);
int count = 0;
//sort words
int i;
char swap = ' ';
for(int k = 0; k < n; k++)
{
for (i=0; i<n-1; i++)
{
//if prev char bigger then next char
if (word[i] > word[i+1])
{
//make swap = prev char
swap = word[i];
//switch prev char with next char
word[i] = word[i+1];
//make next letter char
word[i+1] = swap;
}
}
}
printf("%s\n", word);
for (i=0; i<n-1; i++)
{
int rpt = rpt_letter(word, word[i]);
if(rpt == 1)
{
count++;
}
}
printf("%d", count);
return 0;
}
I've tried a number of things such as using the operator !=, also <, > but it gives me the same result that each word[ia] == c.
You are getting this issue because in your code rpt_letter() the while loop has a terminating condition count >= 2. Now consider input apple and character a. As a appears in apple only once, the count after traversing the whole word remains 1. But the loop doesn't terminate. So, the index i becomes greater than the length of string and tries to check the character appearing after that.
The loop terminates eventually when it gets another a this way. You need to add a check for the terminating null character in your loop so that it doesn't cross the length of the string .
Change the while loop condition to something like -
while((count < 2) && (word[i] != '\0'))
Here is my code. I need to find out the number of times a given word(a short string) occurs in a sentence(a long string).
Sample Input: the
the cat sat on the mat
Sample Output: 2
For some reason the string compare function is not working and my output is coming as zero. Kindly ignore the comments in the code as they have been put to debug the code.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(){
char word[50];
gets(word);
int len = strlen(word);
//printf("%d",len);
char nword[len];
char s[100];
strcpy(nword,word);
puts(nword);
printf("\n");
gets(s);
//printf("%d",strlen(s));
char a[50][50];
int i,j,k;
j = 0;
for(i=0;i<strlen(s);i++)
{
a[i][j] = s[i];
printf("%c",a[i][j]);
if(s[i] == ' ')
{
j++;
printf("\n");
}
}
printf("%d",j);
k = j;
//printf("\nk assigned\n");
j = 0;
//printf("j equal to zero\n");
int count = 0;
int temp = 0;
//printf("count initialized.\n");
for(i=0;i<k;i++)
{
if(strcmp(a[i],nword) == 0)
count++;
}
printf("\n%d",count);
return 0;
}
Your main problem is with this loop for numerous reasons
int i,j,k;
j = 0;
for(i=0;i<strlen(s);i++)
{
a[i][j] = s[i];
printf("%c",a[i][j]);
if(s[i] == ' ')
{
j++;
printf("\n");
}
}
Firstly you've got your indexes into a backwards - a[i][j] means the i-th string and the j-th character, but since you're incrementing j for each word you want it the other way around - a[j][i].
Secondly you can't use i for both indexing into s and a. Think about what happens when you are building the second string. In your example input the second word starts when i is 4 so the first character will be stored as a[1][4]=s[4] which leaves a[1][0] to a[1][3] uninitialised. So you have to use a 3rd variable to track where you are in the other string.
When you hit a space, you don't want to add it to your word as it won't match later on. You also need to add in a null-terminator character to the end of each string or else your code won't know where the end of the string is.
Putting the above together gives you something like this:
int i,j,k;
k = j = 0;
for(i=0;i<strlen(s);i++)
{
if(s[i] == ' ')
{
a[j][k] = '\0';
j++;
k=0;
printf("\n");
}
else
{
a[j][k] = s[i];
printf("%c",a[j][k]);
k++;
}
}
a[j][k]='\0';
The problem is that a is a two-dimentional array and you reference it as a one dimention. Maby you use a 2-dimentional array to represent i=line, j=character. If you keep this idea then you'll have to do this:
j=0;
for(i=0;i<k;i++)
{
if(strcmp(a[i][j],nword) == 0)
count++;
j++;
}
But then it will be difficult to detect words that are split in half. I'd recommend keeping a as a one dimentional array. Copy the contents of s[i] serially and when you want to distinguish lines use the \r\n operator.
I think you use your 2-dimensional array wrong. a[0][j] should be the first word from s[i]. But what you are doing is a[i][0] = s[i] which makes no sense to me.
Best regards
I would implement this using the functions strtok() and strcmp():
int main(void)
{
char word[] = "the"; /* the word you want to count*/
char sample[] = "the cat sat on the mat"; /* the string in which you want to count*/
char delimiters[] = " ,;.";
int counter;
char* currentWordPtr;
/* tokenize the string */
currentWordPtr = strtok(sample, delimiters);
while(currentWordPtr != NULL)
{
if(strcmp(word, currentWordPtr) == 0)
{
counter++;
}
/* get the next token (word) */
currentWordPtr = strtok(NULL, delimiters);
}
printf("Number of occurences of \"%s\" is %i\n", word, counter);
return 0;
}
I am trying to get the most frequent characters from an array.
Here's my code
#include <stdio.h>
int main(void)
{
int c[1000];
char input[] = "abcdab";
int i;
for(i=0; input[i]; i++)
{
c[input[i]]++;
}
int j = 0;
char str = 0;
for(i=0; i<256; i++)
{
if(c[i] > j)
{
j = c[i];
str = i;
}
}
printf("%c\n", str);
return 0;
}
It returns 'a'
But I want to get 'a' and 'b' since they are the most frequent characters in the array.
Any help would be appreciated, thank you.
You are passing through the entire array looking for a maximum, and remembering the first one. With the solution you have, you need an additional loop:
for(i=0; i<256; i++){ // Look for all maximums
if(c[i] == j) // If it is the maximum
{
printf("%c\n", i); // print the character
}
}
Note that your array c is not initialized to all zeroes, so it is purely by chance (not really) that the code is working. If you want c to be all zeroes, you need to declare it as int c[1000] = {0}; or to call memset on it.
i'm developing a little function to display the most frequent character in a (char) array.
This is what I've accomplished so far, but I think i'm on the wrong way.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char test[10] = "ciaociaoci";
max_caratt(test, 10);
}
int max_caratt(char input[], int size)
{
int i;
char max[300];
max[0] = input[0];
for (i=0; i<size; i++)
{
if(strncmp(input,input[i],1) == 1)
{
printf("occourrence found");
max[i] = input[i];
}
}
}
Any help?
Actually, the correct code is this.
It's just a corrected version of IntermediateHacker's below snippet.
void main()
{
int array[255] = {0}; // initialize all elements to 0
char str[] = "thequickbrownfoxjumpedoverthelazydog";
int i, max, index;
for(i = 0; str[i] != 0; i++)
{
++array[str[i]];
}
// Find the letter that was used the most
max = array[0];
index = 0;
for(i = 0; str[i] != 0; i++)
{
if( array[str[i]] > max)
{
max = array[str[i]];
index = i;
}
}
printf("The max character is: %c \n", str[index]);
}
The easiest way to find the most common character is to create an int array of 255 and just increment the arraly element that corresponds to the character. For example: if the charcter is 'A', then increment the 'A'th element (if you look at any ascii table you will see that the letter 'A' has a decimal value of 65)
int array[255] = {0}; // initialize all elements to 0
char str[] = "The quick brown fox jumped over the lazy dog.";
int i, max, index;
// Now count all the letters in the sentence
for(i = 0; str[i] != 0; i++)
{
++array[str[i]];
}
// Find the letter that was used the most
max = array[0];
index = 0;
for(i = 0; str[i] != 0; i++)
{
if( array[i] > max)
{
max = array[i];
index = i;
}
}
printf("The max character is: %c \n", (char)index);
You're passing a (almost) string and a char to strncmp(). strncmp() takes two strings (and an integer). Your program shouldn't even compile!
Suggestion: increase the warning level of your compiler and mind the warnings.
You may want to look at strchr() ...
Assuming an input array of 0-127, the following should get you the most common character in a single pass through the string. Note, if you want to worry about negative numbers, shift everything up by +127 as needed...
char mostCommonChar(char *str) {
/* we are making the assumption that the string passed in has values
* between 0 and 127.
*/
int cnt[128], max = 0;
char *idx = str;
/* clear counts */
memset((void *)cnt, 0, sizeof(int) * 128);
/* collect info */
while(*idx) {
cnt[*idx]++;
if(cnt[*idx] > cnt[max]) {
max = *idx;
}
idx++;
}
/* we know the max */
return max;
}
If you don't need to preserve the input array, you could sort the input array first, then find the longest contiguous run of a single character. This approach is slower, but uses less space.
I made a working version using structs. It works fine, I guess, but I think there's a MUCH better way to write this algorithm.
#include <stdio.h>
#include <stdlib.h>
struct alphabet {
char letter;
int times;
};
typedef struct alphabet Alphabet;
void main() {
char string[300];
gets(string);
Alphabet Alph[300];
int i=0, j=0;
while (i<=strlen(string)) {
while(j<=300) {
if(string[i] != Alph[j].letter) {
Alph[i].letter = string[i];
Alph[i].times = 1;
}
else {
Alph[j].times++;
}
j++;
}
j=0;
i++;
}
int y,max=0;
char letter_max[0];
for (y=0; y<strlen(string); y++) {
printf("Letter: %c, Times: %d \n", Alph[y].letter, Alph[y].times);
if(Alph[y].times>max) {
max=Alph[y].times;
letter_max[0]=Alph[y].letter;
}
}
printf("\n\n\t\tMost frequent letter: %c - %d times \n\n", letter_max[0], max);
}
I saw you all creating big arrays and "complex" stuff so here I have easy and simple code xD
char most_used_char (char s[]) {
int i; //array's index
int v; //auxiliary index for counting characters
char c_aux; //auxiliary character
int sum = 0; //auxiliary character's occurrence
char c_max; //most used character
int max = 0; //most used character's occurrence
for (i = 0; s[i]; i++) {
c_aux = s[i];
for (v = 0; s[v]; v++)
if (c_aux == s[v]) sum++; /* responsible cycle for counting
character occurrence */
if (sum > max) { //checks if new character is the most used
max = sum;
c_max = c_aux;
}
sum = 0; /* reset counting variable so it can counts new
characters occurrence */
}
return c_max; //this is the most used character!
}