CS50 凯撒 output 作为 aaaaa

Question

For some reason when I change the copy of string s that is (string c), string s also gets changed and for loop j keeps on looping until it reached position zero in string alphabet which is 'a'.

int key = atoi(keys);

    string s = get_string("plaintext: ");
    string c = s;

    int length = strlen(s);

    string alphabet = "abcdefghijklmnopqrstuvwxyz";
    string ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    int position = 0;

    for (int i = 0; i < length; i++)
    {
        printf("Before Conversion: %c\n",s[i]);

        for (int j = 0; j < 26; j++)
        {

            if(s[i] == alphabet[j])
            {
                position = (j + key) % 26;
                c[i] = alphabet[position];
                printf("Ascii: %i + Key: %i.. Position = %i\n",j,key,position);

                position = 0;
                printf("Not Converted: %c\n",s[i]);
                printf("After conversion: %c\n\n",c[i]);
            }
            if(s[i] == ALPHABET[j])
            {
                position = (j + key) % 26;
                c[i] = ALPHABET[position];
                printf("Ascii: %i + Key: %i.. Position = %i\n",j,key,position);

                position = 0;
                printf("After conversion: %c\n\n",c[i]);
            }
        }
    }

    printf("ciphertext: %s\n",c);

Answer 1

Change:

string s = get_string("plaintext: ");
string c = s;

to

string s = get_string("plaintext: ");

string c = malloc(strlen(s) + 1);
strcpy(c, s);

That will make it work. The biggest problem here is that the course CS50 is using

typedef char* string;

to hide the complexity of pointers for new students, but in reality it just causes confusion. In order to fully understand why the above solution works, you need to understand pointers and dynamic allocation.