Question

C Programming Help
What does this error message mean and how do I fix it?

Answer 1

@saifoo.khan @e.mccormick

Answer 2

One sec I'm posting it..

Answer 3

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

int main(int words, string letters[])
{
if(words < 2)
{
printf("Usage: ./caesar <phrases>\n");
return 0;
}
else
{
printf("What is the key?");
int key = GetInt();
for(int x = 1; x < words-2; ++x)
{
for (int y = 0; y < strlen(letters) - 1; ++y)
{
printf("%d", key);
}
}
}
}

Here is my code.

Answer 4

jharvard@appliance (~/pset2): make caesar

clang -ggdb3 -O0 -std=c99 -Wall -Werror caesar.c -lcs50 -lm -o caesar
caesar.c:20:40: error: incompatible pointer types passing 'string *' (aka 'char **') to parameter of type 'const char *'; dereference with *
[-Werror,-Wincompatible-pointer-types]
for (int y = 0; y < strlen(letters) - 1; ++y)
^~~~~~~
*
/usr/include/string.h:399:35: note: passing argument to parameter '__s' here
extern size_t strlen (const char *__s)
^
1 error generated.
make: *** [caesar] Error 1

And here is the error message. Should I take a screenshot or is this good? @saifoo.khan

Answer 5

@confluxepic Can you help?

Answer 6

@ikram002p @CausticSyndicalist @rvc @paki

Answer 7

@sleepyjess :)

Answer 8

@Jaynator495

Answer 9

HELLO!!!

Answer 10

Please help this poor fellow

Answer 11

>_<

Answer 12

One momment ill read it... *Gives evil glance over at sleepyjess*

Answer 13

Lol

Answer 14

Should I take a screenshot?

Answer 15

Before I read it, What programming language is it so i know if i can help or not XD

Answer 16

Read the question xD- It's in C

Answer 17

At the top lol

Answer 18

Theres a language called "C" ._. Learn somethign new every day XD

Answer 19

Ya, and C++

Answer 20

I know of c++ ._.

Answer 21

So you can'tz Helpz mez? meowww

Answer 22

Let me see if I can spot someone that can though xD

Answer 23

where's the ambassador help page

Answer 24

strlen = You want to determine the length of the array, right?

Answer 25

No it is to determine the length of the string that is in the parentheses- called letters

Answer 26

Length of string on which index of array? :P

Answer 27

Ummm.. Idk. By the way you know C programming right?

Answer 28

Here's a video that can help... http://cs50.tv/2012/fall/shorts/command_line_arguments/command_line_arguments-720p.mp4

Answer 29

I've never seen this error message before and I can't find anything wrong..

Answer 30

See, when you want to check something in an array (even if you want to check the length of the string inside the index[0]), you still need to give it the index of the array where you want to apply strlen().

Answer 31

Sorry i was watching the video.

Answer 32

I see what the problem is.
You're passing a string array.
Whereas you should be passing a char* agrv[].

Answer 33

As char* argv is a pointer type array.

Answer 34

Hmmm... I just started learning C I don't understand Sorry :C

Answer 35

Umm, don't know about pointers?
Don't worry, you'll learn them at a later stage.

For now, just change it to char* argv[].

Answer 36

umok

Answer 37

wait what should i change to char* argv[]

Answer 38

change strlen(letters) to strlen(*letters)

Answer 39

string is of type 'point to a pointer of a char'. strlen method expects type 'pointer to a char'. So you need to 'derefence' the letters variable. Hence the operator * in front of it.

Answer 40

@confluxepic why dont u accept messages?

Answer 41

I changed my settings now. @DavidUsa

Answer 42

@e.mccormick Please help, I still need it :)

Answer 43

What issue are you having now? Stillt he same one or something else?

Answer 44

Something else. I'll post now so I can give you medal :)

Answer 45

Calculate strlen once, store it, then use it as a condition. Don't use a linear \(O(n)\) solution to a constant \(O(1)\) problem.

Answer 46

The problem here is that `letters` is an array of `string`s. You only want to pass in a single `string`.
For example, if they ran the command
```
./caesar I am the law.
```
Then we'd have data similar to:
```
letters[0] = "I";
letters[1] = "am";
letters[2] = "the";
letters[3] = "law.";
words = 4;
```
First you want to iterate through your words.
Then you can iterate through each letter of each word.

By the way `*letters` is no different that `letters[0]`.

Answer 47

@wio why are you replying to a closed quesition? :P