For 循环未在 c 语言 cs50 ide 中运行

Question

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

string crypt(string msg, int key);

int main(int argc, string argv[])
{
    if (argc != 2)
    {
        printf("Sorry\n");
        return 1;
    }
    else
    {
        string h = get_string("ipt: ");
        h = crypt(h, atoi(argv[2]));
    }
}

string crypt(string msg, int key)
{
    string c = "";
    printf("%s\n", msg);
    int len = strlen(msg);
    for(int i = 0; i > len; i++)
    {
        char t = (msg[i] + key) % 26;
        printf("%c", t);
    }
    return c;
}

In this code it says Segmentation Fault

I Don't Understand the Problem

I'm Trying a lot But Don't get it I'm not trying to access Memory "That Doesn't belong to me" as the wiki says:-

Core Dump/Segmentation fault is a specific kind of error caused by accessing memory that “does not belong to you.”

I'm Using the cs50 ide on "Windows Microsoft Edge" My Computers Ram is 4 GB .

This code is supposed To Output the Coded version of the input: Text.

Plz help I Don't Understand The problem.

The Error Appears After The text Is inputted.

Answer 1

For starters if argc is equal to 2 then argv[2] is equal to NULL . That is according to the C Standard argv[argc] is equal to NULL .

So you need to change this statement

h = crypt(h, atoi(argv[2]));
                  ^^^^^^^

to

h = crypt(h, atoi(argv[1]));
                  ^^^^^^^

This is the reason of the segmentation fault.

From the C Standard (5.1.2.2.1 Program startup)

2 If they are declared, the parameters to the main function shall obey the following constraints:

— The value of argc shall be nonnegative.

— argv[argc] shall be a null pointer.

Also it seems you mean the following condition in the for loop

for(int i = 0; i < len; i++)
               ^^^^^^^

instead of

for(int i = 0; i > len; i++)
               ^^^^^^^

Also it is better to use the type size_t instead of the type int . It is the type of the value returned by the function strlen .

size_t len = strlen(msg);
for( size_t i = 0; i < len; i++)
//...

Pay attention to that within the function you are not changing the original string. You are just outputting its converted characters.

In this case the function does not make a great sense because it seems it should return the converted string to the caller.

And the returned by the function a pointer to an empty string

string c = "";
//...
return c;

does not make a sense.

The function should be defined at least the following way.

string crypt( string msg, int key )
{
    printf("%s\n", msg);
    
    size_t len = strlen( msg );
    for ( size_t i = 0; i < len; i++ )
    {
        msg[i] = (msg[i] + key) % 26;
        printf("%c", msg[i] );
    }

    return msg;
}

Answer 2

for(int i = 0; i > len; i++)

This says that the loop starts with i at 0 , and will continue only while i is bigger than len .

Think about that for a minute.
If i starts at 0 , it is never bigger than len .

The loop will not run.

You intend for the loop to run while i is less than len .