Manipulating strings in C

Slougi · Mar 22, 2003

Hi all,
I am on my yearly try to learn how to code in C once again

Got further than usually this time =)

Anyway, I was wondering how to modify individual charactes in a string.

Say I have the following code:

Code:

#include <stdio.h>

int main(int argc, char *argv[])
{
	char  abc[20];
	strcpy(abc, "Wahoo!\0");
	printf("%s", abc);
	return 0;
}

This works fine. But this:

Code:

#include <stdio.h>

int main(int argc, char *argv[])
{
	char  abc[20];
	strcpy(abc, "Wahoo!\0");
	printf("%s", abc);
	printf("%s", abc[1]);
	return 0;
}

Gives me a segmentation fault. Why is that? Also, since apparently direct access to the chars does not work, how can I modify individual chars in the string?

If this is completely ridiculous please enlighten me why so

icepir8 · Mar 22, 2003

Slougi said:
Hi all,
I am on my yearly try to learn how to code in C once again Got further than usually this time =)

Anyway, I was wondering how to modify individual charactes in a string.

Say I have the following code:

Code:

#include <stdio.h> int main(int argc, char *argv[]) { char abc[20]; strcpy(abc, "Wahoo!\0"); printf("%s", abc); return 0; }

This works fine. But this:

Code:

#include <stdio.h> int main(int argc, char *argv[]) { char abc[20]; strcpy(abc, "Wahoo!\0"); printf("%s", abc); [COLOR=red] printf("%s", abc[1]);[/COLOR] return 0; }

Gives me a segmentation fault. Why is that? Also, since apparently direct access to the chars does not work, how can I modify individual chars in the string?

If this is completely ridiculous please enlighten me why so

The line in red is wrong in that you are passing a char to it and telling it that you are passing a string.
I belive that this will work.

printf("%c", abc[1]);

Slougi · Mar 22, 2003

Re: Re: Manipulating strings in C

icepir8 said:
The line in red is wrong in that you are passing a char to it and telling it that you are passing a string.
I belive that this will work.

printf("%c", abc[1]);

/me slaps forehead with a large hammer

Doomulation · Mar 22, 2003

Heh. And of course, you can still modify a string:

PHP:

str[5] = 'g';

As an example.
Otherwise it's always better to use char* strings unless using buffers. They can be modified at will.

Slougi · Mar 22, 2003

Doomulation said:
Heh. And of course, you can still modify a string:

PHP:

str[5] = 'g';

As an example.
Otherwise it's always better to use char* strings unless using buffers. They can be modified at will.

I thought that was only in c++?
/me goes read some more

Doomulation · Mar 22, 2003

Nope. Pointers and char are c.
Classes and references are c++.
Ie,

PHP:

class whatever { };
int& whatever;

Pointers are a very useful thing to learn.
The power of c++ lies within classes.

Slougi · Mar 22, 2003

Doomulation said:
Nope. Pointers and char are c.
Classes and references are c++.
Ie,

PHP:

class whatever { }; int& whatever;

Pointers are a very useful thing to learn.

I meant the str[5]= 'g'; bit.

Doomulation · Mar 22, 2003

I know you did

But to answer your question, it's possible in c

References and classes are mostly the only that isn't possible.

Slougi · Mar 22, 2003

Bear with me, but I don't see how str[5]='g'; has anything to do with pointers

Doomulation · Mar 22, 2003

Heh. You're right. It has nothing to do with pointers.
Char* is pointer. That's what i tried to point out

PHP:

char* test = "hello";
char test = new char[20];
test = "hello";

That's pointers.

PHP:

char test = new char[20]; <--- pointer
test[5] = 'g'; <--- not pointer
test = "hello"; <--- pointer

Slougi · Mar 22, 2003

Doomulation said:
Heh. You're right. It has nothing to do with pointers.
Char* is pointer. That's what i tried to point out

PHP:

char* test = "hello"; char test = new char[20]; test = "hello";

That's pointers.

PHP:

char test = new char[20]; <--- pointer test[5] = 'g'; <--- not pointer test = "hello"; <--- pointer

Doom, I know that

It is just that for some reason while posting this my brain refused to cooperate with me and notice the %s. Nothing more about this thread

Still, thanks for the help

Cyberman · Mar 22, 2003

well Ok let me give you
ye olde' Cybies tips on using the array operator(tm)

Tip 1
To prevent access violations and exceptions always do a bounds check before accessing a location in any array.

Code:

   char Str[20];
   int Index;

   Str[2]= ' '; // this is BAD programing practice!
   Index = 0;
   while (Index < sizeof(Str)) // good programing practice
   {  Str[Index] = ' ';
      Index++;
   }

Now you are probably wondering a few things why is this any better? Notice the array has 20 characters if you access outside of the array you are writting in a location you have NO BUSINESS to write to and you will get an Illegal Access exception error.

Tip 2
All pointers are arrays except type void.

This means if you assign a pointer allocate memory or typecast one.. be sure the SIZE is correct or an exception WILL ocure.

Code:

#define BUFFSIZE 128
   char *Src;
   int Pos;

   //Src = new char[128]; // this is BAD don't do this 
   Src = new char[BUFFSIZE]; // safer here is why
   Pos = 0;
   while (Pos < BUFFSIZE) // if you just change BUFFSIZE
   {  Src[Pos] = 0;    // you will automatically take care of bounds
   }
   return (void *)Src; // good but you should always retain an 
//original pointer copy instead of using Src could be disasterous 
//later ;)

Type void has no size, hence you can't have an array of type void. You can have an array of void pointers so be very careful.

Cyb

Slougi · Mar 22, 2003

Well, in case you all wonder what I was doing, here is my 1337 1337 filter

The commented out printf thingies were for debugging, i had i <= argc... Thanks Jabo for pointing that out

Code:

#include < stdio.h >
 
int main(int argc, char *argv[])
{
   int i,j,k;
   char abc[20],regular[8],leet[8];
   regular[0]='a';
   regular[1]='c';
   regular[2]='e';
   regular[3]='i';
   regular[4]='l';
   regular[5]='o';
   regular[6]='s';
   regular[7]='t';
   leet[0]='4';
   leet[1]='(';
   leet[2]='3';
   leet[3]='l';
   leet[4]='1';
   leet[5]='0';
   leet[6]='5';
   leet[7]='7';
   //printf("for starts now!\n");
   for (i=1; i < argc; i++)
     {
        strcpy(abc, argv[i]);
        //printf("for 1\n");
        for (j=0; j < strlen(abc); j++)
          {
             //printf("for 2\n");
             for (k=0; k <= 7; k++)
               {
                  //printf("for 3\n");
                  //printf("i:%d,j:%d,k:%d\n", i, j, k);
                  if (abc[j]==regular[k])
                   {
                      abc[j]=leet[k];
                   }
               }
          }
        printf("%s ", abc);
     }
   printf("\n");
   return 0;
}

Now, if someone thinks this code is bad, please tell me why, and how to fix it

EDIT: DAMN HTML!!!! :angry: Please ignore extra spaces

Slougi · Mar 22, 2003

Hmm, ok cyberman's post had some good points in it

Had clicked reply before the post was there

Cyberman · Mar 22, 2003

regular and leet appear to be constants.
you would be better served doing this

Code:

#include < stdio.h >

#define MAX_PARSE 128
int main(int argc, char *argv[])
{
   int i,j,k;
   char abc[20];//,regular[8],leet[8];
   const char regular[] = { "aceilost" }; // these are constants
   const char leet[] = " { "(3l1057" }; // ditto :)
/*   regular[0]='a';
   regular[1]='c';
   regular[2]='e';
   regular[3]='i';
   regular[4]='l';
   regular[5]='o';
   regular[6]='s';
   regular[7]='t';*/
/*   leet[0]='4';
   leet[1]='(';
   leet[2]='3';
   leet[3]='l';
   leet[4]='1';
   leet[5]='0';
   leet[6]='5';
   leet[7]='7';*/
   //printf("for starts now!\n");
   for (i=1; i < argc; i++)
     {
        strcpy(abc, argv[i]);
        //printf("for 1\n");
        for (j=0; j < (MAX_PARSE > strlen(abc))?MAX_PARSE;strlen(abc); j++) // so you don't exceed DOS's built in buffer of 128 chars or whatever
          {
             //printf("for 2\n");
             for (k=0; k <= 7; k++)
               {
                  //printf("for 3\n");
                  //printf("i:%d,j:%d,k:%d\n", i, j, k);
                  if (abc[j]==regular[k])
                   {
                      abc[j]=leet[k];
                   }
               }
          }
        printf("%s ", abc);
     }
   printf("\n");
   return 0;
}

This gives you errors if you acidentally access one of the check arrays and write to them.

Slougi · Mar 22, 2003

Thanks for the constant suggestion

I was looking for a better way to do this.

As to DOS limitations, no worry, I use Linux

Code:

louai@Gondolin louai $ uname -a
Linux Gondolin 2.4.20-gentoo-r2 #1 Sat Mar 22 11:40:02 EET 2003 i686 AMD Duron(tm) 
Processor AuthenticAMD GNU/Linux

Perfect for someone who wants to learn C. It really makes it much easier than Windows/DOS

Doomulation · Mar 22, 2003

As I said, pointers is flexible and good.
Instead of filling each array element, use a pointer.

PHP:

char test[20];
test[0] = 'h';
test[1] = 'e';
...
test[19] = '!';

= BAD!

PHP:

char* test;
test = "he...!"; // Good!

Pointers are flexible and with them you can basically do whatever you want and an error won't occour. Remember, though, that you can't use this as buffer.

PHP:

char* buffer;
buffer = ReadAFile(filename_here);

As it's writing data directly into the varaible, you'll get a memory error later. Especially when trying to delete it. However, you can do this:

PHP:

char* test;
test = "hello";
test = "helooooooooooooooooooo";
test = "t";

It'll work. Just remember that pointers need be released with delete.

To further demonstrate the incredible flexinbleness of pointers, you can use them to get arrays with any numbers of elements you don't know.
Just pass for example an array of classes to a function. Use a class pointer of that type and you can do a loop to go through all the elements like a dynamic array. Be careful not to go past the boundary, though.

Slougi · Mar 22, 2003

I don't quite understand why reading a file would not work, but assigning a value to it would?

Doomulation · Mar 22, 2003

Yup.
It's like when you assign a value to it, c++ does something with it. Heh, I don't know quite well actually.
But reading from a stream as a buffer from these functions simply won't work.
Try and you'll see.

aprentice · Mar 23, 2003

Stop confusing slougi with jibberish examples of your "l33t skills"

First of all, pointers point at the address of the array holding the data. There, was that too hard to explain?

Manipulating strings in C

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