c - 即使 char 数组以 null 终止并经过检查， printf 也会吐出垃圾

Question

这个问题已经有答案了:

printf outputting garbage instead of specific characters (2 个回答)

已关闭 5 年前。

我有一个作业，我们要制作一个凯撒(rot)密码，但不使用数组符号(a[b])。看起来它应该可以工作，但是即使在第 41 行和第 51 行执行了检查，它也会在字符串末尾喷射出随机垃圾

/* lab 08 */
/* exercise for loops and pointers */
/* assume that plain text which needs to be encrypted has
 * only capital letters */

#include <stdio.h>
#include <stdlib.h>


int Caesar_encrypt(char *p, char *s, int enckey);
int Caesar_decrypt(char *p, char *s, int enckey);

int main(void){
    char A[]="THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
    char B[50], /* stores encrypted output string   */
    C[50];  /* stores decrypted output using string B as source */
    int enckey,
    statuse, statusd; /* they store return values from the functions */

    printf("Plaintext for encryption is : %s \n", A);

    printf("Input numerical key for Caesar's cypher : ");
    scanf("%d", &enckey );  // COMPLETE THIS

    putchar('\n');
    printf("You entered encryption key %d \n", enckey);
    /* encrypt by Caesar's cypher */
    statuse= Caesar_encrypt(A, B, enckey);

    printf("Ciphertext is : %s \n", B);

    /* decrypt by Caesar's cypher */
    statusd = Caesar_decrypt(B, C, enckey);
    printf("Decrypted text is: %s \n", C);
    exit (0);
}

int Caesar_encrypt(char *p, char *s, int enckey) {  // COMPLETE THE INPUT PARAMETERS
    enckey %= 26;
    int i;
    for (i = 0; *(p+i); i++) {
        if (*(p+i) != ' ') *(s+i) = (*(p+i) - 'A' + enckey) % 26 + 'A';
        else *(s+i) = ' ';
    }
    return 0;
}

int Caesar_decrypt(char *p, char *s, int enckey) {  // COMPLETE THE INPUT PARAMETERS
    enckey %= 26;
    int i;
    for (i = 0; *(p+i); i++) {
        if (*(p+i) != ' ') *(s+i) = (*(p+i) - 'A' + 26 - enckey) % 26 + 'A';
        else  *(s+i) = ' ';
    }
    return 0;
}

示例输出:

Plaintext for encryption is : THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG 
Input numerical key for Caesar's cypher : 1

You entered encryption key 1 
Ciphertext is : UIF RVJDL CSPXO GPY KVNQT PWFS UIF MBAZ EPH8???THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG 
Decrypted text is: THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG47;Q9=@SGD PTHBJ AQNVM ENW ITLOR NUDQ SGD KZYX CNF8???THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG 

Answer 1

您不会在加密或解密函数中以零终止结果字符串。

您需要在加密函数中的循环之后添加 *(p+i) = 0; 或者最好添加 p[i] = 0;，并且s[i] = 0; 在解密函数中的循环之后。