I was working on a program for my intro to C class (xtra credit assignment) and can't figure out how to discard duplicates numbers on an array. The problem asks to only print non duplicates; so I able to print the first number, compare the following and print if different, I discard the next if a duplicate, but the thing is I've only figured out how to compare the one number it following one, I figured I could do another for loop inside the for loop, but I'm getting super confused and just can't figure it out. I've already submitted my code last week, I've just been working on this trying to figure it out for myself so any help/guidance would be greatly appreciated.
"EDIT: Here's the problem: Use a single-subscripted array to solve the following problem. Read in 20 numbers, each of which is between 10 and 100, inclusive. As each number is read, print it only if it's not a duplicate of a number already read. Provide for the worst case in which all 20 numbers are different. Use the smallest possible array to solve this problem"
Thanks in advance, and any advice on how I'm writing my program would also be appreciated as I'm a total noob, and trying to be a good programmer with as little bad habits as possible.
#include <stdio.h>
#include <stdlib.h>
#define AS 20
void findDuplicate (int af[], int fAS);
int main(){
int a[AS], i , j, k;
int last = 0;
printf("Enter %d numbers between 10 and 100:\n", AS);
for (i = 0; i < AS; i++){
scanf("%d",&a[i] );
if (a[i] >= 10 && a[i] <= 100 ){
continue;
} else {
printf("You must enter values between 10 - 100\n");
i = i -1;
}
}
findDuplicate(a, AS);
system ("pause");
return 0;
}
void findDuplicate (int af[], int fAS){
int c;
printf("You entered ");
for (c=0; c < fAS; c++){
if (af[c] != af[c+1]){
printf("%d ", af[c]);
}
continue;
}
printf("\n");
}
You should first define an Array which can save as many variables as you want ..
Lets say you are comparing for 10-100 which means 91 possible different digits.
so , define array with the size of 91. and then do the scanning in for loop for 91 times to find out if you have that variable entered previously. If not then save it and display it ,else discard it.
Related
I'm trying to create a complete C program to read ten alphabets and display them on the screen. I shall also have to find the number of a certain element and print it on the screen.
#include <stdio.h>
#include <conio.h>
void listAlpha( char ch)
{
printf(" %c", ch);
}
int readAlpha(){
char arr[10];
int count = 1, iterator = 0;
for(int iterator=0; iterator<10; iterator++){
printf("\nAlphabet %d:", count);
scanf(" %c", &arr[iterator]);
count++;
}
printf("-----------------------------------------");
printf("List of alphabets: ");
for (int x=0; x<10; x++)
{
/* I’m passing each element one by one using subscript*/
listAlpha(arr[x]);
}
printf("%c",arr);
return 0;
}
int findTotal(){
}
int main(){
readAlpha();
}
The code should be added in the findTotal() element. The output is expected as below.
Output:
List of alphabets : C C C A B C B A C C //I've worked out this part.
Total alphabet A: 2
Total alphabet B: 2
Total alphabet C: 6
Alphabet with highest hit is C
I use an array to count the number of the existence of each character,
I did this code but the display of number of each character is repeated in the loop
int main()
{
char arr[100];
printf("Give a text :");
gets(arr);
int k=strlen(arr);
for(int iterator=0; iterator<k; iterator++)
{
printf("[%c]",arr[iterator]);
}
int T[k];
for(int i=0;i<k;i++)
{
T[i]=arr[i];
}
int cpt1=0;
char d;
for(int i=0;i<k;i++)
{int cpt=0;
for(int j=0;j<k;j++)
{
if(T[i]==T[j])
{
cpt++;
}
}
if(cpt>cpt1)
{
cpt1=cpt;
d=T[i];
}
printf("\nTotal alphabet %c : %d \n",T[i],cpt);
}
printf("\nAlphabet with highest hit is : %c\n",d,cpt1);
}
There is no way to get the number of elements You write in an array.
Array in C is just a space in the memory.
C does not know what elements are actual data.
But there are common ways to solve this problem in C:
as mentioned above, create an array with one extra element and, fill the element after the last actual element with zero ('\0'). Zero means the end of the actual data. It is right if you do not wish to use '\0' among characters to be processed. It is similar to null-terminated strings in C.
add the variable to store the number of elements in an array. It is similar to Pascal-strings.
#include <stdio.h>
#include <string.h>
#define ARRAY_SIZE 10
char array[ARRAY_SIZE + 1];
int array_len(char * inp_arr) {
int ret_val = 0;
while (inp_arr[ret_val] != '\0')
++ret_val;
return ret_val;
}
float array_with_level[ARRAY_SIZE];
int array_with_level_level;
int main() {
array[0] = '\0';
memcpy(array, "hello!\0", 7); // 7'th element is 0
printf("array with 0 at the end\n");
printf("%s, length is %d\n", array, array_len(array));
array_with_level_level = 0;
const int fill_level = 5;
int iter;
for (iter = 0; iter < fill_level; ++iter) {
array_with_level[iter] = iter*iter/2.0;
}
array_with_level_level = iter;
printf("array with length in the dedicated variable\n");
for (int i1 = 0; i1 < array_with_level_level; ++i1)
printf("%02d:%02.2f ", i1, array_with_level[i1]);
printf(", length is %d", array_with_level_level);
return 0;
}
<conio.h> is a non-standard header. I assume you're using Turbo C/C++ because it's part of your course. Turbo C/C++ is a terrible implementation (in 2020) and the only known reason to use it is because your lecturer made you!
However everything you actually use here is standard. I believe you can remove it.
printf("%c",arr); doesn't make sense. arr will be passed as a pointer (to the first character in the array) but %c expects a character value. I'm not sure what you want that line to do but it doesn't look useful - you've listed the array in the for-loop.
I suggest you remove it. If you do don't worry about a \0. You only need that if you want to treat arr as a string but in the code you're handling it quite validly as an array of 10 characters without calling any functions that expect a string. That's when it needs to contain a 0 terminator.
Also add return 0; to the end of main(). It means 'execution successful' and is required to be conformant.
With those 3 changes an input of ABCDEFGHIJ produces:
Alphabet 1:
Alphabet 2:
Alphabet 3:
Alphabet 4:
Alphabet 5:
Alphabet 6:
Alphabet 7:
Alphabet 8:
Alphabet 9:
Alphabet 10:-----------------------------------------List of alphabets: A B C D E F G H I J
It's not pretty but that's what you asked for and it at least shows you've successfully read in the letters. You may want to tidy it up...
Remove printf("\nAlphabet %d:", count); and insert printf("\nAlphabet %d: %c", count,arr[iterator]); after scanf(" %c", &arr[iterator]);.
Put a newline before and after the line of minus signs (printf("\n-----------------------------------------\n"); and it looks better to me.
But that's just cosmetics. It's up to you.
There's a number of ways to find the most frequent character. But at this level I recommend a simple nested loop.
Here's a function that finds the most common character (rather than the count of the most common character) and if there's a tie (two characters with the same count) it returns the one that appears first.
char findCommonest(const char* arr){
char commonest='#'; //Arbitrary Bad value!
int high_count=0;
for(int ch=0;ch<10;++ch){
const char counting=arr[ch];
int count=0;
for(int c=0;c<10;++c){
if(arr[c]==counting){
++count;
}
}
if(count>high_count){
high_count=count;
commonest=counting;
}
}
return commonest;
}
It's not very efficient and you might like to put some printfs in to see why!
But I think it's at your level of expertise to understand. Eventually.
Here's a version that unit-tests that function. Never write code without a unit test battery of some kind. It might look like chore but it'll help debug your code.
https://ideone.com/DVy7Cn
Footnote: I've made minimal changes to your code. There's comments with some good advice that you shouldn't hardcode the array size as 10 and certainly not litter the code with that value (e.g. #define ALPHABET_LIST_SIZE (10) at the top).
I have used const but that may be something you haven't yet met. If you don't understand it and don't want to learn it, remove it.
The terms of your course will forbid plagiarism. You may not cut and paste my code into yours. You are obliged to understand the ideas and implement it yourself. My code is very inefficient. You might want to do something about that!
The only run-time problem I see in your code is this statement:
printf("%c",arr);
Is wrong. At this point in your program, arr is an array of char, not a single char as expected by the format specifier %c. For this to work, the printf() needs to be expanded to:
printf("%c%c%c%c%c%c%c%c%c%c\n",
arr[0],arr[1],arr[2],arr[3],arr[4],
arr[5],arr[6],arr[7],arr[8],arr[9]);
Or: treat arr as a string rather than just a char array. Declare arr as `char arr[11] = {0};//extra space for null termination
printf("%s\n", arr);//to print the string
Regarding this part of your stated objective:
"I shall also have to find the number of a certain element and print it on the screen. I'm new to this. Please help me out."
The steps below are offered to modify the following work
int findTotal(){
}
Change prototype to:
int FindTotal(char *arr);
count each occurrence of unique element in array (How to reference)
Adapt above reference to use printf and formatting to match your stated output. (How to reference)
I made a program and then it didn't worked. As in the other post someone advised me to trying to debug my programs, I learned it and debugged this one. Probably it has some basic errors of writting but that's because I've changed a lot of thing recently to understand what's happening. In third time when I input a value on screen on that loop, it changes my var "i" to that value instead of keeping that number in my array "grade".
First I tried to make it all in one loop, the first one, but as always it didn't help much, and then i wrote the code by this manner as you'll see
#include <stdio.h>
#include <stdlib.h>
int main()
{
int j=0,sum=0,i=0;
int grade[]={0};
for(;j<100;j++){
printf("Type a grade:\t");
scanf("%d",&grade[j]);
if(grade[j]<10||grade[j]>20){
break;
}
}
for(;i<j;i++){
sum=sum+grade[i];
}
float average=sum/j;
printf("The average is: %.2f\n",average);
system("pause");
return 0;
}
The exercicise says that you need to read "x" grades from a student and it needs to be between 10 and 20, if the number is out of this range it stops the loop.After I just need to calculate the average os these grades.I don't really know if my var average is being calculated correctly, cause I didn't could reach over there because of my problem. If you input 11, 12 and 13 it should give to a sum of 36, but gaves me 26, i don't know how.
Erik, you should define your array in a coherent way. To allow the necessary number of elements, try defining a numeric constant. You could use it for both define the number of iterations of your cycle and the size of your grade array. You can also avoid a new cycle to calculate the sum of the array, you can do this operation while reading the grades, using only one for loop. Try this way:
#include <stdio.h>
#include <stdlib.h>
#define MAX_GRADES 100
int main()
{
int j,sum=0,i;
float average;
int grade[MAX_GRADES];
for(j = 0 ; j < MAX_GRADES; j++)
{
printf("Type a grade:\t");
scanf("%d",&i);
if ( (i<10) || (i>20) )
break;
grade[j] = i;
sum += i;
}
if (j > 0)
average = (float)sum/j;
else
average = 0;
printf("The average is: %d, %d, %.2f\n",sum, j, average);
system("pause");
return 0;
}
Currently learning C - and I have no clue where I'm going wrong in this code:
#include <stdio.h>
int main()
{
char alphabet[20];
int i;
for (int i = 0; i > 20; i++)
{
printf("Enter in a letter:\n");
scanf("%s", alphabet[i]);
if (alphabet[i] == alphabet[i+1])
{
printf("Duplicate Letters");
};
return 0;
}
}
The program that I am asked to make for class — I'm required to create a 1D array, add validation for alphabetical letters and duplicate letters, as well as creating a function for sorting the letters and specifying the number of times each letter was put in.
As much as I've been able to attempt coding is:
Create a 1D array to read 20 alphabetical letters
Add validation for duplicate letters and printf 'Duplicate Letters'
but every time I try, the program terminates at 'Enter in a letter:' or it won't execute.
Where did I go wrong?
For background: I work mainly on Windows 7 because that's what the school has — using MinGW as my compiler — but for working at home I use MacOS using Terminal as the compiler.
for (int i = 0; i > 20; i++)
You're telling the computer here to initialize i to 0, and then, while i is greater than 20, do the loop. However, since i starts at 0, it will never be greater than 20.
for (int i = 0; i < 20; i++)
And, yes, as comments have pointed out, your use of scanf is incorrect. Lacking a better C reference for it, check out http://www.cplusplus.com/reference/cstdio/scanf/ for descriptions of its arguments.
Hey this is my first post here. i have been assigned with an excercise to count the most frequent word in c programming language. first and foremost i need to read a number which tells me how many words i will have to read. then i need to use calloc with max element size 50. after that i read the strings. my original idea was to create a one-dimensional array which im gonna later sort alphabetically and then counting and printing the most frequent word would be easy. but after some hours of research i found out i need to use a two-dimensional array and things went out of control. ive been studying computer science for 3 months now and this exercise seems tough. do u have any other suggestions?. the example was this:
10
hello
world
goodbye
world
thanks
for
all
hello
the
fish
hello
my code so far is
int main()
{
int i, n, j, temp;
int *a;
printf("Eisagete to plhthos twn leksewn:");
scanf("%d",&n);
a = (int*)calloc(n,50);
printf("Eisagete tis %d lekseis:\n",n);
for( i=0 ; i < n ; i++ )
{
scanf("%d",&a[i]);
}
for (i = 0 ; i < ( n - 1 ); i++)
{
for (j = 0 ; j < n - i - 1; j++)
{
if (a[j] > a[j+1])
{
temp = a[j];
a[j] = a[j+1];
a[j+1] = temp;
}
}
}
dont mind the printfs they are in greek and they are just there to make it look better. i also want to point out that this version is used for integers and not for strings just to start off.
im currently trying a linear search but im not sure if it will help
As you point out, the code you show is related to reading and sorting integers; it is only loosely related to the word counting problem.
How would you count the occurrences of each number? You'd have to
Read the next number;
If you already have a tally for the number, you add one to the tally for that number;
If you've not seen the number before, you create a tally for it and set its count to one.
When all the numbers are read, you search through the set of tallies, looking for the one with the largest count.
Record the number and count of the first entry.
For each subsequent entry:
If the count is larger than the current maximum, record the new maximum count and the entry.
Print the information about the number with the largest count and what that count is.
Replace numbers with words and the general outline will be very similar. You might allocate storage for each string (distinct word) separately.
It is easy to count the number of distinct words or the total number of words as you go. Note that you do not need to store all the words; you only need to store the distinct words. And the count at the front of the list is computer science education gone astray; you don't need the count to make it work (but you probably have to live with it being in the data; the simplest thing is to ignore the first line of input since it really doesn't help very much at all). The next simplest thing is to note that unless they're fibbing to you, the maximum number of distinct words will be the number specified, so you can pre-allocate all the space you need in one fell swoop.
Very simple. Store your words in a 2D array the loop through it and each time you take a word loop through it again starting from the current index and check if there is an equal. ech time the child loop ends check if the number of occurrences is bigger than the last maximum.
#include <stdio.h>
int main()
{
int i, j, occurrence=0, maximum = 0;
char *index_max = NULL;
char wl[10][10] = {"hello","world","goodbye","world","thanks","for","all","hello","the","world"};
for (i=0; i<10; i++){
occurrence = 0;
for (j=i; j<10; j++){
if (!strcmp(*(wl+i), *(wl+j))){
occurrence++;
}
}
if (occurrence>maximum){
maximum = occurrence;
index_max = *(wl+i);
}
}
if (index_max != NULL){
printf("The most frequent word is \"%s\" with %d occurrences.\n", index_max, maximum);
}
return 0;
}
So I have only ever programmed in c++, but I have to do a small homework that requires the use of c. The problem I encountered is where I need a loop to read in numbers separated by spaces from the user (like: 1 5 6 7 3 42 5) and then take those numbers and fill an array.
the code I wrote is this:
int i, input, array[10];
for(i = 0; i < 10; i++){
scanf("%d", &input);
array[i] = input;
}
EDIT: added array definition.
any suggestions or hints would be very highly appreciated.
Irrespective of whatever is wrong here, you should quickly learn to NEVER write code that does not check the return value from any API call that you make. scanf returns a value, and you have to be interested in what it says. If the call fails, your logic is different, yes?
Perhaps in this case it would tell you what's going wrong. The docs are here.
Returns the number of fields
successfully converted and assigned;
the return value does not include
fields that were read but not
assigned. A return value of 0
indicates that no fields were
assigned.
This code working good.
If your numbers is less than 10, then you must know how many numbers is before you start reading this numbers, or last number must be something like 0 to terminate output then you can do while(true) loop, but for dynamically solution you must read all line into string and then using sscanf to reading numbers from this string.
You need the right #include and a proper main. The following works for me
#include <stdio.h>
int main(void) {
/* YOUR CODE begin */
int i, input, array[10];
for (i = 0; i < 10; i++) {
scanf("%d", &input);
array[i] = input;
}
/* end of YOUR CODE */
return 0;
}
i'm not a c programmer but i can suggest an algorithm which is to use scanf("%s",&str) to read all the input into a char[] array then loop over it and test using an if statment if the current char is a space, if it is then add the preceeding number to the array