如何找到数组中某个元素的总数（C）

Question

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

Answer 1

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;

  }

Answer 2

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);
 }

Answer 3

<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 printf s 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 (eg #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!

Answer 4

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 )