CS50 如何修复错误：使用未声明的标识符“i”？

Question

I'm working on my PSet2 Caesar Problem. After finishing my code here is the mistakes I got. Any advice how to fix them? Really appreciated.

caesar.c:46:10: error: use of undeclared identifier 'i'
    for (i = 0; i < strlen(plaintext); i++)

         ^
caesar.c:46:17: error: use of undeclared identifier 'i'
    for (i = 0; i < strlen(plaintext); i++)

                ^
caesar.c:46:40: error: use of undeclared identifier 'i'
    for (i = 0; i < strlen(plaintext); i++)

                                       ^
caesar.c:48:31: error: use of undeclared identifier 'i'
        if (isupper(plaintext[i]))

                              ^
caesar.c:50:39: error: use of undeclared identifier 'i'
            printf("%c", (((plaintext[i] - 65) + k) %26) + 65);

                                      ^
caesar.c:52:36: error: use of undeclared identifier 'i'
        else if (islower(plaintext[i]))

                                   ^
caesar.c:54:39: error: use of undeclared identifier 'i'
            printf("%c", (((plaintext[i] - 97) + k) %26) + 97);

                                      ^
caesar.c:58:36: error: use of undeclared identifier 'i'
            printf("%c", plaintext[i]);

#include <cs50.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, string argv[])
{
    //Check that there is one command-line argument
    
    if (argc != 2)
    {
        printf ("Usage: ./caesar key\n");
        return 1;
    }
    else printf ("Success!\n");
    
    //Define the key 
    
    string key = argv[1];
    
    //Check if input is a digit
    
    for (int i = 0; i < strlen(argv[1]); i++)
    {
        if (!isdigit(argv[1][i]))
        {
            printf ("Usage: ./caesar key\n");
            return 1;
        }
        else printf ("Success!\n%s\n", key);
    }
    
    //Get plain text from user
    
    string plaintext = get_string("Plaintext: ");
    
    //Define key
    
    int k = atoi(key);
    printf("ciphertext: ");
    
    //Obtain ciphertext
    
    for (i = 0; i < strlen(plaintext); i++)
    {
        if (isupper(plaintext[i]))
        {
            printf("%c", (((plaintext[i] - 65) + k) %26) + 65);
        }
        else if (islower(plaintext[i]))
        {
            printf("%c", (((plaintext[i] - 97) + k) %26) + 97);
        }
        else
        {
            printf("%c", plaintext[i]);
        }
    }
    printf("\n");
}

Answer 1

"use of undeclared identifier 'i'" means that you have not declared the identifier i before using it - the compiler doesn't recognize the name.

In this case the code should be for (int i = 0; ...

Answer 2

Good to see you're doing cs50. You have missed a little, yet important thing. Since you have not declared the variable i outside any loop, thus, it is only available to that loop. So, when you are using it anywhere outside the loop, it's actually undeclared variable, and that's what is happening here. So, you need to declare i in the last for loop too. I've attached the modified code.

#include <cs50.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, string argv[])
{
//Check that there is one command-line argument

if (argc != 2)
{
    printf ("Usage: ./caesar key\n");
    return 1;
}
else printf ("Success!\n");

//Define the key 

string key = argv[1];

//Check if input is a digit

for (int i = 0; i < strlen(argv[1]); i++)
{
    if (!isdigit(argv[1][i]))
    {
        printf ("Usage: ./caesar key\n");
        return 1;
    }
    else printf ("Success!\n%s\n", key);
}

//Get plain text from user

string plaintext = get_string("Plaintext: ");

//Define key

int k = atoi(key);
printf("ciphertext: ");

//Obtain ciphertext

for (int i = 0; i < strlen(plaintext); i++) // declared i
{
    if (isupper(plaintext[i]))
    {
        printf("%c", (((plaintext[i] - 65) + k) %26) + 65);
    }
    else if (islower(plaintext[i]))
    {
        printf("%c", (((plaintext[i] - 97) + k) %26) + 97);
    }
    else
    {
        printf("%c", plaintext[i]);
    }
}
printf("\n");
}