Problems with array of strings

Hello,
I would like to as how shoul i input strings into array from keyboard?

Thanks in advance.

my code is:

Code:

---

#include <stdio.h>

void main () {

 char *string[3];
 int i;

 for(i=0;i<3;i++)
    scanf("%s",&string[i]);

 for(i=0;i<3;i++)
      printf("%s",string[i]);

}

---

:confused:

Firstly, don't use void main, use int main.
Secondly, you must have some sort of space to store the strings. Simply defining a pointer will not give you space. You have to manually call malloc and later free to do that. Or you can simple use an array.

char string[3][100];

Or the like.
This is very important. Be very careful when using scanf to read strings. Read this for more information:
http://cboard.cprogramming.com/showp...37&postcount=9

probably char string[3][100] to have 3 strings of 100 bytes each

Right. Good correction. Not used to 2D arrays, since I hardly need them in C++...

ok, i tired to correct mistakes.. but..there are some left..

Code:

---

#include <stdio.h>

int main () {

        char string[3][100];
        int i, j;

        j=0;
                while (j<4)
                {fgets(*string,100,stdin);
            for(i=0;i<100;i++)
                        {if(string[j][i]=='\n')
                                {string[j][i]='\0';
                            j+=1; break;}}
        return 0;

---

Erm. This will fail to compile in every sort of way.
You need to learn indenting. You can use an auto-indenting IDE such as Visual Studio or Code::Blocks if on Windows.
And as for the fgets... do you know how arrays work?

1. indentation
2. you should update different strings
3. use for loop

Code:

---

for(j=0;j<3;j++)
{
  fgets(string[j], sizeof string[j], stdin);

---

Quote:

---

Originally Posted by Elysia

Erm.
And as for the fgets... do you know how arrays work?

---

no..i'm new to C.. that's why i'm having problems with such a simple task :(

http://www.cprogramming.com/tutorial/c/lesson8.html

Thanks very much, i did it (atctually you did it) ;) I had some practice programing with Delphi, but it's much easier to work with strings there :)

And the other thing i wanted to ask what should i do, if i need other chars, which are not stored in ANSI C, for example Russian.

If you want to attempt non-english languages, I suggest you use unicode or wide char.
Usually denoted by wchar_t instead of char.
Wide strings also uses an L before the string, so "My string" would be come L"My string".
You also need to use appropriate library functions. But they should all be available in documentation if you need to find them.

But if I need to save all Russian chars, isn't it possible to set whole language alphabet somehow, using UTF-8 locale?

I imagine it might be, but code locales isn't directly my area.
If you use wide chars, you don't really have much about code locales to worry about. Or, at the very least, less to worry about.

So if I'll try to solve this problem using wide chars, I should define every char, which isn't included to ANSI C?

Quote:

---

Originally Posted by CactusC

But if I need to save all Russian chars, isn't it possible to set whole language alphabet somehow, using UTF-8 locale?

---

You do have russian encoding in ANSI - like Windows(Cyrillic), KOI-8 or ISO(Cyrillic) or DOS(Cyrillic)

You will have encoded russian chars in the codes from 128 to 255 (using different tables for each of the above) - and be able still work with the standard string functions

If you go for the wide char encoding - you will need to change the type od the string to wchar
and the function used to the w-version

Problem with ANSI is that the extended set isn't standard. It may look like russian on your computer, but it's not going to do so on everybody else's computer.
Which is, at least partially, why unicode exists today.

I need to set char set Latin 4, it's also called ISO-8859-4, but i still can't find anything, explaning how to deal with that. :rolleyes:

Quote:

---

Originally Posted by Elysia

Problem with ANSI is that the extended set isn't standard. It may look like russian on your computer, but it's not going to do so on everybody else's computer.
Which is, at least partially, why unicode exists today.

---

Yes, you'r right. But it would be enough for me to display it correctly on my computer :)

Quote:

---

Originally Posted by CactusC

I need to set char set Latin 4, it's also called ISO-8859-4, but i still can't find anything, explaning how to deal with that. :rolleyes:

---

First maybe try this http://www.russianlocalization.com/Texts/Encoding.htm
To see if you can see text displayed in the specified encoding?

PS. Also note they are metntioning ISO 8859-5 not 4

Quote:

---

Originally Posted by vart

First maybe try this http://www.russianlocalization.com/Texts/Encoding.htm
To see if you can see text displayed in the specified encoding?

PS. Also note they are metntioning ISO 8859-5 not 4

---

Yes, I've already tested that, I can't see russian symbols correctly, that's why I have chosen the othe char set ISO-8859-4, which is displayed correctly in my computer.

Also I wanted to ask how to print these strings, because every try computer writes me 'runtime error'

Code:

---

#include <stdio.h>

int main () {

        char string[100][3];
        int i, j;
            j=0;
        while (j<4)
                {fgets(string[j], sizeof string[j], stdin);
                  for(i=0;i<3;i++)
                {if(string[j][i]=='\n')
                    {string[j][i]='\0';
                                j+=1; break;}}
                        }
        return(0);

---

Code:

---

#include <stdio.h>
#include <string.h>

int main ()
{
        char string[3][100];
        int j;
        for(j=0;j<3;j++)
        {
                char* p;
                fgets(string[j], sizeof string[j], stdin);
                p = strchr(string[j],'\n');
                if(p) *p = '\0';
        }

        for(j=0;j<3;j++)
        {
                printf("String number &#37;d: \"%s\"\n", j, string[j]);
        }
        return 0;
}

---

Thanks a lot :cool: The problem was that I haven't included string.h Thank's very much :)

Hello again. Could anyone help me to find mistakes why this programs prints out only the lats string.

Code:

---


#include <stdio.h>
#include <string.h>
#define SIZE 124

int main ()
{
        char *string[3];
        char buf[SIZE], input[SIZE];
        int j, num;
        for(j=0;j<3;j++)
            {
              printf("\nPlease input string number &#37;d: ",j+1);
              fgets(input, SIZE, stdin);
              num = sscanf(input, "%s", &buf);
              if (num==1)
                string[j] = buf;
              else printf("error");
            }

        for(j=0;j<3;j++)
            printf("%s\n",string[j]);
        return 0;
        }

---

Code:

---

        for(j=0;j<3;j++)
            printf("&#37;s\n",string[0]);

---

Erm, maybe your problem is here.

Quote:

---

Originally Posted by mike_g

Code:

---

        for(j=0;j<3;j++)
            printf("&#37;s\n",string[j]);

---

Erm, maybe your problem is here.

---

Oh, sorry, I've forrgotten to change, i was trying to print the first string, but it is like the third one.

I have founded the mistake :)

Could anyone explain me why this program prints out
"Enter string 0"
"Enter string 1"
at one time and doesn't read the string[0][100] ?

Code:

---

#include <stdio.h>
#include <string.h>

int main ()
{
        char string[3][100];
        int j;
        for(j=0;j<3;j++)
        {
                char* p;
                                printf("Enter string &#37;d",j);
                fgets(string[j], sizeof string[j], stdin);
                p = strchr(string[j],'\n');
                if(p) *p = '\0';
        }

        for(j=0;j<3;j++)
        {
                printf("String number %d: \"%s\"\n", j, string[j]);
        }
        return 0;
}

---

I have no explanation to that - is that the entire program, or is that "reduced", and if it is reduced, does it actually show the problem in reduced form. My instinct tells me that you have a scanf() before the fgets() that leaves a newline in the input buffer.

--
Mats

you also need to add fflush(stdout) between
printf and fgets

Thanks :cool:

Hello,

I would like to ask how could I check if the input is right:
the program asks user to input a float from a keybord, but user inputs a char. Is there any funcion which can check that?

Thanks in advance.

The most common way is to read a string with fgets, then convert to float using strtod, and check the end pointer argument to see where the function stopped parsing. If it stopped parsing right away (the start is the same as the end), then it means it's an invalid input.
You can see documentation for strtod.

Thanks for an idea, but therefore I've faced another problem. I decided to use isdigit to check if the inputed number is not a char and if it isn't-then convert it with atof or atoi.

But then i started to use these funcions, the for loops began working incorrectly: they are being repeated more times than the num value is, for example if num=3, each loop is repeated10 times and so on.

Do you have an idea how could I correct this to work properly?

Code:

---


  char t;

              for(i=0;i<num;i++)
                    {
                char* p;
                printf("Enter string nr.&#37;d\n",i+1);
                fgets(string[i], sizeof string[i], stdin);
                p = strchr(string[i],'\n');
                if(p) *p = '\0';
            }

                for(i=0;i<num;i++)
        {
                      printf("Enter float nr.%d \n",i+1);
                      fgets(&t, sizeof &t, stdin);
                      if (isdigit(t)) floatt[i]=atof(&t);
                      else {printf("Error");return 0;}
                }

 
              for(i=0;i<num;i++)
                {       
                    printf("Enter integer nr.%d \n",i+1);
                    fgets(&t, sizeof &t, stdin);
                    if (isdigit(t)) integer[i]=atoi(&t);
                    else {printf("Error");return 0;}
        }

---

You are aware that there are valid floating point numbers that do not start with a digit:

Code:

---

+1.2
-1.2
.01

---

Above are three examples of such numbers.

By using strtod(), you can make a better check if the number is valid (and completely accepted - you then can decide if "123.4A" is a valid number or not).

--
Mats

I don't know how string is defined, but this is absolutely wrong:

Code:

---

fgets(&t, sizeof &t, stdin);
if (isdigit(t)) floatt[i]=atof(&t);

---

When working with C-strings, you need arrays, not a single char. Never. All C-strings contains one byte for each character + one character for NULL. So when reading that char, fgets write the char + NULL, thus buffer overrun. The same for atoi - it also expects a NULL char, and does a buffer overrun looking for it.
http://cpwiki.sf.net/Buffer_overrun

But you seriously need some indenting lessons too. Here are some reasources:
http://cpwiki.sf.net/Indentation
http://cpwiki.sf.net/User:Elysia/Indentation

Than you. I've changed

Code:

---

char t;

---

to

Code:

---

char t[100];

---

and nows everything is working in a right way ;)

I would like to make a program, which does smth when one of the arrows (of keyboard) is pressed. So I would like to ask which funcion should I use (scanf, getch...?) to get the pressed arrow. :rolleyes:

I did it.

How? What compiler are you using? If you are using Dev-C++, did you use getche()? (Just out of curiosity.)

By the way, that smiley that you keep using ( :rolleyes: ) is not used when asking for help. Actually, it indicates sarcasm, which usually means you're annoyed or something -- which I don't think is what you want to convey.

I'm using lcc-win32.

http://support.microsoft.com/kb/57888

P.S. Sorry for the smiley, it has a different meaning in our country :)