编码caesar加密码要求密钥两次？

Question

我实现了caesar-cipher算法，但似乎没有捕捉到所有细节，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");
    }

}

输出示例如下：

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-标准输入中的换行符将保留，如果您的格式是%d\n，则需要两次回车。

if子句中的gets(text)删除了换行符-您只需将format更改为%d即可。

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");
    }
}