我对这个问题有些困惑,因为我可以解决该问题并提交并获得了充分的信誉,但是当我在循环之前打印初始变量时,该代码只是单词。这段代码有效:
int main(int argc, string argv[]) {
// c = (p + k) % 26, where c is result text, p is input and k
// is key
//considers if arg count is two
if (argc == 2) {
int n = strlen(argv[1]);
int check = 0;
if (isdigit(argv[1][0])) {
for (int i = 1; i < n; i++) {
if (isdigit(argv[1][i]) || argv[1][i] == '0') {
check++;
} else {
check--;
}
}
}
// verifies all characters are numeric
if (check != n - 1) {
printf("Usage: ./caesar key\n");
return 1;
}
} else {
printf("Usage: ./caesar key\n");
// returning 1 identifies an error and exits the program
return 1;
}
int key = atoi(argv[1]);
string plaintext = get_string("plaintext: ");
printf("%i\n", key);
printf("%s\n", plaintext);
int m = strlen(plaintext);
printf("%i\n", m);
char ciphertext[m];
int usekey = (key % 26);
printf("%i\n", key);
// NEED to figure out how to handle wrap around
// need to understand ASCII
for (int i = 0; i < m; i++) {
int c = plaintext[i];
//encrypts upper case letters
if (c >= 65 && c <= 90) {
//incorporates wrap around for uppercase
if (c + usekey >= 90) {
int val = 90 - c;
int key2 = usekey - val;
char cipher = 64 + key2;
ciphertext[i] = cipher;
}
//considers if key works fine
else {
char cipher = c + usekey;
ciphertext[i] = cipher;
}
}
//encrypts lower case letters
else if (c >= 97 && c <= 122) {
//incorporates wrap around for lowercase
if (c + usekey >= 122) {
int val = 122 - c;
int key2 = usekey - val;
char cipher = 96 + key2;
ciphertext[i] = cipher;
} else {
char cipher = c + usekey;
ciphertext[i] = cipher;
}
} else {
//encrypts punctuation
ciphertext[i] = c;
}
printf("*\n");
}
printf("ciphertext: %s\n", ciphertext);
}
但是,此代码不起作用(使用1作为密钥将a
加密为b
,使用12作为密钥加密world, say hello!
作为iadxp, emk tqxxa!
)。正确答案后,它会随机打印不同的字符,我不知道为什么。
int main(int argc, string argv[]) {
// c = (p + k) % 26, where c is result text, p is input and k
// is key
//considers if arg count is two
if (argc == 2) {
int n = strlen(argv[1]);
int check = 0;
if (isdigit(argv[1][0])) {
for (int i = 1; i < n; i++) {
if (isdigit(argv[1][i]) || argv[1][i] == '0') {
check++;
} else {
check--;
}
}
}
// verifies all characters are numeric
if (check != n - 1) {
printf("Usage: ./caesar key\n");
return 1;
}
} else {
printf("Usage: ./caesar key\n");
// returning 1 identifies an error and exits the program
return 1;
}
int key = atoi(argv[1]);
string plaintext = get_string("plaintext: ");
int m = strlen(plaintext);
char ciphertext[m];
int usekey = (key % 26);
// NEED to figure out how to handle wrap around
// need to understand ASCII
for (int i = 0; i < m; i++) {
int c = plaintext[i];
//encrypts upper case letters
if (c >= 65 && c <= 90) {
//incorporates wrap around for uppercase
if (c + usekey >= 90) {
int val = 90 - c;
int key2 = usekey - val;
char cipher = 64 + key2;
ciphertext[i] = cipher;
}
//considers if key works fine
else {
char cipher = c + usekey;
ciphertext[i] = cipher;
}
}
//encrypts lower case letters
else if (c >= 97 && c <= 122) {
//incorporates wrap around for lowercase
if (c + usekey >= 122) {
int val = 122 - c;
int key2 = usekey - val;
char cipher = 96 + key2;
ciphertext[i] = cipher;
} else {
char cipher = c + usekey;
ciphertext[i] = cipher;
}
}
//encrypts punctuation
else {
ciphertext[i] = c;
}
}
printf("ciphertext: %s\n", ciphertext);
}
答案 0 :(得分:0)
我认为您的条件是否不应该按预期进行。您可以打印“ argv [1] [i]”并查看问题。这是我的代码可能会帮助您。
bool isNumber(char number[])
{
int i = 0;
for (; number[i] != 0; i++)
{
if (!isdigit(number[i])) //check if there is something that is not digit
{
return false;
}
}
return true;
}
int main(int argc, string argv[])
{
if (argc == 2 && isNumber(argv[1]) == 1)
{
int k = atoi(argv[1]);
string plainText, chipherText;
plainText = get_string("plaintext: ");
printf("ciphertext: ");
for (int i = 0, n = strlen(plainText) ; i < n; i++)
{
// checking if it is lowercase 97 = a to 112 = z and if it + 13 characters along.
if (plainText[i] >= 'a' && plainText[i] <= 'z')
{
printf("%c", (((plainText[i] - 'a') + k) % 26) + 'a'); // print out lowercase with key
} // if it it between uppercase A and Z
else if (plainText[i] >= 'A' && plainText[i] <= 'Z')
{
printf("%c", (((plainText[i] - 'A') + k) % 26) + 'A'); // print out uppercase with key
}
else
{
printf("%c", plainText[i]);
}
}
printf("\n");
return 0;
}
else if (argc != 2 || isNumber(argv[1]) == 0)
{
printf("Error\n");
return 1;
}
}
答案 1 :(得分:0)
您为m
分配了cyphertext
个字节,这对于空终止符是不够的,您也没有设置空终止符,导致随机字符出现在加密输出之后。这实际上是未定义的行为,因此任何事情都可能发生,包括程序崩溃。
您不需要存储加密的文本,只需一次输出一个字节。
另外,请勿使用65
和90
等ASCII值,而应使用更具可读性的字符常量'A'
和'Z'
。
这是简化版:
#include <stdio.h>
#include <stdlib.h>
#include <cs50.h>
int main(int argc, string argv[]) {
if (argc != 2) {
printf("Usage: ./caesar key\n");
// returning 1 identifies an error and exits the program
return 1;
}
char *p;
int key = strtol(argv[1], &p, 10);
if (*p || p == argv[1]) {
printf("caesar: invalid argument: %s\n", argv[1]);
return 1;
}
string plaintext = get_string("plaintext: ");
// assuming ASCII
for (size_t i = 0; plaintext[i]; i++) {
int c = plaintext[i];
if (c >= 'A' && c <= 'Z') {
c = 'A' + (c - 'A' + key) % 26;
} else
if (c >= 'a' && c <= 'z') {
c = 'a' + (c - 'a' + key) % 26;
}
putchar(c);
}
putchar('\n');
free(plaintext);
return 0;
}