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

ok im kinda new to C , and i have the c programming language 2nd edition book but when i try to some of the programs from the book in it , it doesnt work!! I mean no errors show but the program isnt running the way is suppose to.

For example, in the 1st chapter where theres 3 parts to characters ,words , lines, and other stuff counting programs, it shows no error while compiling but it doesnt show the output from printf. I have the newest version of dev c++ (i think 4.9.9.2)
but it doesnt work; i tried to the settings and put options like
"support all ansi standard c programs " and "attempt to support some aspect of traditional c programming" and it doesnt change a thing.

please please please help! i am really stuck and dont know what to do.

You might want to post your problem over at the Dev-C++ forum.
The URL is http://sourceforge.net/forum/forum.php?forum_id=48211

Try placing

Code:

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system("PAUSE");

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just above return 0 in your source file.
Then in the line counting program for example .Type some lines pressing enter .Then press Ctrl + z then enter.

> it shows no error while compiling but it doesnt show the output from printf.
Post your code.
I've seen plenty of ways of messing this up which still compile, but which fail to run properly.

You're trying to compile the code as C, not C++, right?

Quote:

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it doesnt show the output from printf.

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

Code:

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#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.

Quote:

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You're trying to compile the code as C, not C++, right?

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Yes:

Quote:

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it shows no error while compiling

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I think Sephiroth's program compiled, but the console flashed and dissapeared before [s]he could read it.

I think getche() is non standard. Use getchar()

ok this is the code

Code:

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#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);
}

---

ok the problem is not that the program terminates so fast that i cant see anything. I did the getchar() thing at the end. problem is that i see no output like there should be (for example the printf("digits =") ) . I just dont get it! why doesnt it show?! oh yeah and i saved and compiled it as a C source not c++.

There's a FAQ on what to do when your program goes away before you like it to. In short, run it from a terminal, or put something at the end to prevent it from exiting until you hit enter or what not:

Code:

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while( (c = getchar()) != '\n' );

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Also, you can initialize arrays to zero (or anything really, provided you fill in all the elements manually, otherwise any you don't fill in default to zero) at declaration time:

Code:

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int array[ SOMESIZE ] = { 0 };

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This will set the first array element to zero. All uninitialized elements, provided you have manually initialized at least one, will be zero filled. Note, this does not fill them with the value of the first element, but with zero. Consider:

Code:

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int array[ SOMESIZE ] = { 1 };

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This doesn't fill them all with 1. It fills the first one with 1, and the rest with zero.

Code:

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int array[ SOMESIZE ] = { 0, 1, 2 };

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This fills the first with zero, the second with 1, the third with 2, and the remaining with zero.


Quzah.

no is not that the program window flashes and quits, the program is suppose to print an output counting the occurences of characters, words, lines, and other stuff but when i enter such characters, i receive no output.

The program is fine, Dev-C++ is fine. It is just that this program requires a little unusual input.

1. Type in your input.
2. Press ENTER.
3. Press F6. This is the Windows EOF input key. It must be on a line by itself.
4. Press ENTER.

You can follow these last three steps whenever a program requires an EOF. You can see that this program reads input until it receives an EOF by this line:

Code:

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      while ((c = getchar()) != EOF)

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O ya!! it works now . ok i gotta get this straight 1 more time.
so if the program or any program just require anything that has to do with EOF i have to do the F6 thing right? and if it doesnt call eof as an argument of any kind then i wont need to press F6 ever?

oh sorry i forgot also one last thing, is there a way to set dev c++ to do the F6 thing automatically whenever eof is called?

>> so if the program or any program just require anything that has to do with EOF i have to do the F6 thing right? and if it doesnt call eof as an argument of any kind then i wont need to press F6 ever? <<

Yes. Pretty much. You can also use a file as input. This is called redirection. In this case, the program will receive EOF (which stands for "end of file") at the end of the file. This example shows how to run a program using file.txt as the input.

Code:

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>yourprogram < file.txt

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>> oh sorry i forgot also one last thing, is there a way to set dev c++ to do the F6 thing automatically whenever eof is called? <<

No. How would it know when to supply the EOF? The EOF allows you to tell the program that you have finished supplying input. If your program only expects a single line of input, it might use the new line character to tell it when the input is complete.

> 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:

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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;
}

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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 :D

Salem, a

Code:

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#include <stdio.h>

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might help too. :)

Quote:

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Originally Posted by Salem

3. Fix the EOF problem inside your code.

Code:

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

---

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

Quote:

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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.

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

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:

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#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....

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:

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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.

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:

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#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......

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?

Quote:

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

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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:

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

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

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();
}

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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...

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.

Quote:

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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...

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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

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.

Quote:

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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.

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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 ).

Quote:

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Originally Posted by Sephiroth

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.

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c is a charater declaration. so anything that gets into c will be character.

if c = '1' , ascii value of character '1' is 49 so c = 49 too.

c = '1', in [c-'0'] c's ascii value will be subratted to the ascii value of charcter '0' which will yeild the diff of 1.

get the point?

ascii values:
'0' = 48
'1' = 49
'2' = 50
so on...

You know, I am doing the same as Sephiroth myself (working through K+R) and I am at the same place. I too was completely confused by the exact same thing.... but I think its just dawned on me whats happening with that little statement.

essentially its the asci code for the value in c - the acsi code for the value of 0 right? So if c were 2 then [c-'0'] would essentially be the same as:

50 - 48 = 2

Is that right?

Yes.

Code:

---

#include <stdio.h>

int main(void)
{
  char digit;
  for (digit = '0'; digit <= '9'; ++digit)
  {
      printf("'%c' = %d; ",    digit, digit);
      printf("'%c' - '%c' = ", digit, '0');
      printf("%d - %d = ",    digit, '0');
      printf("%d\n",          digit - '0');
  }
  return 0;
}

/* my output
'0' = 48; '0' - '0' = 48 - 48 = 0
'1' = 49; '1' - '0' = 49 - 48 = 1
'2' = 50; '2' - '0' = 50 - 48 = 2
'3' = 51; '3' - '0' = 51 - 48 = 3
'4' = 52; '4' - '0' = 52 - 48 = 4
'5' = 53; '5' - '0' = 53 - 48 = 5
'6' = 54; '6' - '0' = 54 - 48 = 6
'7' = 55; '7' - '0' = 55 - 48 = 7
'8' = 56; '8' - '0' = 56 - 48 = 8
'9' = 57; '9' - '0' = 57 - 48 = 9
*/

---

(Although it need not be ASCII.)

ok thanks now i understand what it means. But is that really necessary? I tried taking that out and it produces an error message. Why couldnt we just do [c-48]?? or that doesn't work?

o sorry 1 more thing, so '0' and just 0 r different. Cause in your program when u put things like this:

'0' = 0

thats actually different than 0 = 0 right?
So can you use '0' to stand for the value 0 too?

Quote:

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So can you use '0' to stand for the value 0 too?

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No. There was no "'0' = 0", only "digit = '0'", which in ASCII would be equivalent to "digit = 48".

> But is that really necessary? I tried taking that out and it produces an error message
So why didn't you post the code, the error message and a question?

ok let me try asking this a little clearer. for instance, [c-'0'] u said that is the same as 50-48 if c was 2. yeah but how do u know when c is either 2 or 50?? is it because 0 is written as '0'??

like how do u know that [c-'0'] is 50 - 48 or 2-0 if c was 2?
is the difference the single quotes outside the 0? I 'm confused over this part. I hope I made it clear this time.

Quote:

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Originally Posted by Sephiroth

is the difference the single quotes outside the 0?

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Yes that makes the difference. The single quotes are there for a purpose.
Kurt

Quote:

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for instance, [c-'0'] u said that is the same as 50-48 if c was 2. yeah but how do u know when c is either 2 or 50?? is it because 0 is written as '0'

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It sounds like c is an unfortunate choice of variable name.

'0' is the character zero. In ASCII it has the value of 48.
If a char ch = '2', then (ch-'0') is equivalent to ('2'-'0').
Now '2' in ASCII has the value of 50.
So ('2'-'0') is equivalent to (50-48), which is of course, 2 - but this is the integer 2, not the character '2'.

Quote:

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like how do u know that [c-'0'] is 50 - 48 or 2-0 if c was 2?

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In this case, we refer to an ASCII table, such as the one provided here.

yeah i thought it would be because of the single quotes beside the 0. I just wonder y people use ascii character value instead of the value itself ; it doesn't really look like it makes anything easier... :D

> I just wonder y people use ascii character value
But '0' isn't necessarily ASCII

Saying c - 48 limits your program to those implementations which use ASCII as their character set. In say EBCDIC your code is broken.

c - '0' always produces the correct answer no matter what your implementation character set is (C assumes that '0' to '9' are contiguous, but you can't assume the same for 'a' to 'z' for example).

What if the code said c - 77, would you instantly know what letter that was without reaching for your character tables, or would c - 'M' be more readable?