编码凯撒加密代码要求输入两次密钥？

Question

我实现了 caesar-cipher 算法，但似乎我没有掌握所有细节，因为当我运行程序时，它要求输入两次密钥！

#include <stdio.h>
#include <string.h>
#define SIZE 1024
char text[SIZE];
int text_2[SIZE];
int key;
int encoder(char text[SIZE]){
        printf("plaintext: \n");
        gets(text);
        for (int i = 0; i < strlen(text); ++i) {
            text_2[i] = (int) text[i];
            text_2[i] = text_2[i] + key;
            text[i] = (char) text_2[i];

        }
    return 0;
}

int main() {
    printf("enter the key: \n");
    scanf("%d\n",&key);
    if(key < 26 && key > 0){
        printf("nice!, now enter a word to encrypt it\n");
        gets(text); //this step is necessary to pass text onto encoder function.
        encoder(text);
        puts(text);
    } else{
       printf("Yeuch!!\n");
    }

}

输出示例为：

enter the key: 
2 //I press 2 and nothing happens, then it asks for it again, hence why I have two 2's
2
nice!, now enter a word to encrypt it
plaintext: 
a
c

Process finished with exit code 0

Answer 1

%d 忽略换行符 - stdin 中的换行符将保留，如果您的格式是 %d\n，则需要输入两次。

if 子句中的

和 gets(text) 删除换行符 - 您只需将格式更改为 %d.

解决方案：

int main() {
    printf("enter the key: \n");
    scanf("%d", &key);
    if (key < 26 && key > 0) {
        printf("nice!, now enter a word to encrypt it\n");
        gets(text); // this step is necessary to pass text onto encoder
                    // function.
        encoder(text);
        puts(text);
    } else {
        printf("Yeuch!!\n");
    }
}

编码凯撒加密代码要求输入两次密钥？

Coding caesar encryption code asks for key twice?

c

algorithm

cs50

caesar-cipher