How come Dev C++ cant run some programs??

**Salem** · 09-11-2005

> while ((c = getchar()) != EOF)
Unfortunately, EOF is a sticky state, so attempting to do
while( (c = getchar()) != '\n' );
simply doesn't do what you want it to when you've just pressed ctrl-z to exit your first while loop.

1. The easy thing to try is to open a console prompt and run the program from there, using say
cd c:\code
myprog.exe
All of the traditional examples (including those in K&R) assume you're working at a command prompt interface.

2. Use the external pause program to do the wait

Code:

int main ( ) {
    // your code here
    system("PAUSE");
    return 0;
}

It's not ideal, but gets past the EOF problem.

3. Fix the EOF problem inside your code.

Code:

int main ( ) {
      int c, i, nwhite, nother;
      int ndigit[10];
      
      nwhite = nother = 0;
      for (i = 0; i < 10; ++i)
           ndigit[i] = 0;
           
      while ((c = getchar()) != EOF)
          if (c >= '0' && c <= '9')
              ++ndigit[c-'0'];
          else if (c == ' ' || c == 'n' || c == '\t')
                   ++nwhite;
          else
              ++nother;
       
      printf("digits =");
      for (i = 0; i < 10; ++i)
           printf(" %d", ndigit[i]);
      printf(", white space = %d, other = %d\n",
            nwhite, nother);

    printf ("Press [Enter] to continue");
    clearerr( stdin );
    while ((c = getchar()) != '\n' && c != EOF);
    return 0;
}

**Sephiroth** · 09-11-2005

lol now that i thought about the 2nd part of my question, it was kinda dumb. but yeah thanks alot now i can continue on

**dwks** · 09-12-2005

Salem, a

Code:

#include <stdio.h>

might help too.

**Antigloss** · 09-12-2005

Originally Posted by Salem

3. Fix the EOF problem inside your code.

Code:

int main ( ) {
    /* ... ... */
    printf ("Press [Enter] to continue");
    clearerr( stdin );
    while ((c = getchar()) != '\n' && c != EOF);
    return 0;
}

What does this clearerr(stdin) for?
I found that even without calling clearerr, this program still worked fine

**Antigloss** · 09-12-2005

Originally Posted by dwks

Yeah, put a system("pause") or getche() before the return 0:

Code:

#include <stdio.h>

int main(void) {
    printf("Hello, World!\n");
    getche();
    return 0;
}

With getche(), you can press any key, not just <enter>, to close the console.

Post your program.

getche is non-standard function.
to use it, we should include conoi.h(windows) or ncurse.h(unix)

**Sephiroth** · 09-13-2005

ok i have a new problem....

the following program is suppose to count blanks tabs and newlines but it only counts newlines on mine

Code:

#include <stdio.h>

main()
{
      int c, nl, nt, nb; /* declaration of char counting variables */
      
      nl = 0;
      nt = 0;
      nb = 0;
      
      while((c = getchar()) != EOF) /* if c isn't the end of file then */
          if (c == '\n') 
             ++nl;
          if (c == '\t')
             ++nt;
          else if (c == ' ')
             ++nb;
      printf(" %d %d %d\n", nl, nt, nb);

      getchar();
}

argh i dont get it!! i dont think theres any errors but it just doesnt run the way is suppose to....

**cwr** · 09-13-2005

C is not Python

You don't have a { } block after your while, so only the first if (c == '\n') ++nl; is being executed within the loop. The other if (c == '\t') ++nt; else if (c == ' ') ++nb; are getting executed once when the while loop finishes, and c is now EOF, not '\t' or ' '.

You want:

Code:

while((c = getchar()) != EOF) /* if c isn't the end of file then */
{
    if (c == '\n') 
        ++nl;
    if (c == '\t')
        ++nt;
    if (c == ' ')
        ++nb;
}

Also, you are writing C89 code, because you put main() instead of int main(), so you should put the required return 0; before main finishes.

If you were compiling it as C99 code, you wouldn't need to put return 0, but then C99 would make using main() without the int illegal.

**Sephiroth** · 09-14-2005

ok now that u told me that whenever theres eof i need to the f6 thing but i dunno if this is an exception but it prints output without f6:

Code:

#include <stdio.h>

main()
{
      int c;
      
      c = 0;
      
      while((c = getchar()) != EOF) {
           putchar(c);
           if(c == ' ' || c == '\t' || c == '\n') /* newlines useless */
              printf("\n");     
              }
      getchar();
}

it shows the output of c without f6 ... weird......

**./fsck** · 09-14-2005

thats because your printf statement is within your while loop this time so it will print each time it goes through the while loop instead of waiting for EOF. In this case when you supply EOF the program simply calls getchar() and then exits since there is no other code after the while loop to execute. Make sense?

**dwks** · 09-14-2005

getche is non-standard function.
to use it, we should include conoi.h(windows) or ncurse.h(unix)

Usually I would add that, too, but with his compiler ("dev c++ (i think 4.9.9.2)"), just <stdio.h> works perfectly (at least I think it does; it works with 4.0).

Code:

if(c == ' ' || c == '\t' || c == '\n')

Try the isspace() function in <ctype.h>.

**Sephiroth** · 09-15-2005

alright this books confusion to me just keeps on goin.....

In this array program:

Code:

#include <stdio.h>

main()
{
      int c, i, nwhite, nother;
      int ndigit[10];
      
      nwhite = nother = 0;
      for (i = 0; i < 10; ++i)
           ndigit[i] = 0;
           
      while ((c = getchar()) != EOF)
          if (c >= '0' && c <= '9')
              ++ndigit[c-'0'];
          else if (c == ' ' || c == 'n' || c == '\t')
                   ++nwhite;
          else
              ++nother;
       
      printf("digits =");
      for (i = 0; i < 10; ++i)
           printf(" %d", ndigit[i]);
      printf(", white space = %d, other = %d\n",
            nwhite, nother);
 
      getchar();
}

ok the book says that the part that contains " ++ndigit[c-'0'];"
that that integer expression tells that is a digit of 0 - 9. okkk....
but how does it show that c is a numeric value of 0 -9 in c-'0' ???
looks like a subtraction sign to me or something...

**cwr** · 09-15-2005

It is a subtraction sign.

When you enter '0', '0'-'0' becomes the value 0.
When you enter '1', '1'-'0' becomes the value 1.
et cetera.

The author is using the fact that the characters '0' through '9' are contiguous in the character set.

Edit:

If you didn't already know, character constants like '0' are just integer values. On the ASCII character set, '0' is 48, '1' is 49, '2' is 50, '3' is 51, etc. 'A' is 65, 'a' is 97.

**ZuK** · 09-15-2005

Originally Posted by Sephiroth

ok the book says that the part that contains " ++ndigit[c-'0'];"
that that integer expression tells that is a digit of 0 - 9. okkk....
but how does it show that c is a numeric value of 0 -9 in c-'0' ???
looks like a subtraction sign to me or something...

The ascii-value of '0' is 48 and the acsii-value of '1' is 49.
So if c ='1' the expression c - '0' actually is 49-48 => 1.
Kurt
edit: too late

**Sephiroth** · 09-16-2005

ok wait let me think about it for a sec. Ok so u mean that "[c-'0']"
means that whatever c is , is going to turn into the value 0????
but then how come cwr says is a subtraction sign?

Does the value of c (whatever u inputted) just gets subtracted by '0' which is 48??? I am still kind of confused.

**ZuK** · 09-16-2005

Originally Posted by Sephiroth

Does the value of c (whatever u inputted) just gets subtracted by '0' which is 48??? I am still kind of confused.

I would phrase it this way.
The value of '0' witch is 48 is subtracted from the value of c.
Kurt
Edit: guess that's what you said. ( english somtimes confuses me ).