为什么显示随机数？ C程序读取字符串字母

Question

So I have to create a program that reads the user input and shows how many times each letter appears in that string, and also how many non-letters but my code for alphabets is showing random numbers..

#include <stdio.h>
#include <string.h>

#define SIZE 100

void readInput (char string[]);
void Calc(char string[], int letters[]);

void
readInput (char string[])
{
    printf ("Enter a String:\n");
    fgets (string, SIZE, stdin);

    string[strlen (string) - 1] = '\0';
}

void
Calc(char string[], int letters[])
{
    int c = 0, x;
    while (string[c] != '\0')
    {
        if (string[c] >= 'a' && string[c] <= 'z')
        {
            x = string[c] - 'a';
            letters[x]++;
        }

        c++;
    }

    for (c = 0; c < 26; c++)
        printf ("%c occurs %d times in the entered string.\n", c + 'a', letters[c]);
}

int
main ()
{
    char string[SIZE];
    int letters[26];

    readInput (string);
    Calc(string, letters);
    return 0;
}

This is the output

I'm new to strings I've googled examples but can't seem to find whats wrong with my code and no idea how I will include the non-letters part.

Answer 1

The contents of letters are not initialised . Formally the behaviour of your program is indeterminate .

Sort that by writing int letters[26] = {0}; Doing that sets all elements to zero, which is what you want in this case.

Answer 2

letters[] is uniinitialized. Solution int letters[26]={0} .

You are reading an uninitialized value which is indeterminate. And the results doesn't conform to as you expect it to be.

Here you want to initialize the elements with 0 denoting that you haven't seen any charcaters yet. (automatic storage duration).

A better way to overwrite the '\\n' would be

string[strcspn(string, "\n")] = 0;

Answer 3

您应该使用所有0值初始化letters