I don't want to write a code I don't understand ..!!

Hi,
I'm new in the world of programming, I just begun to program in C langauage.. I started to write my first program using Borland C++ and I can use different styles :

1.

Code:

---

#include <stdio.h>
  main() {
    printf("Help meee");
  }

---

2.

Code:

---

#include <stdio.h>
  int main() {
    printf("Help meee");
  return 0;
  }

---

3.

Code:

---

#include <stdio.h>
  int main(void) {
    printf("Help meee");
  return 0;
  }

---

I suggest choosing one of the latter two since the first lacks a return type for main(). The last one comes directly from the C Standard. You should also keep the indentation consistent:

Code:

---

#include <stdio.h>
int main(void) {
    printf("Help meee");
    return 0;
}

---

Now, what do you not understand?

I don't understand the differences, I installed Div C++ and I got another code for hello world app:

Code:

---

int main(int argc, char *argv[])
{
  printf("Hello World!\n");

  printf("Press ENTER to continue...\n");
  getchar();
  return 0;
}

---

In windows the console window will disapear as soon as the program finishes. Printing a message than waiting for input keeps the window open until the user responds.

Quote:

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I don't understand the differences, I installed Div C++ and I got another code for hello world app:

---

You probably mean Dev-C++. Anyway, what you are looking for is explained in an FAQ on main and another FAQ on command line arguments.

There are basically two prototypes for main - both as you see. In C, you must specify void in the parameter list if it does not take any arguments. int main() is technically wrong as it should be int main(void), so there's two to choose from!

Quote:

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Originally Posted by laserlight

You probably mean Dev-C++. Anyway, what you are looking for is explained in an FAQ on main and another FAQ on command line arguments.

---

Yes Dev-C++ :) .. I'm gonne read it but now I have another Q:

Code:

---

#include <stdio.h>

int main(int argc, char *argv[])
{
  int a[10];
  int i, j, temp;
  printf("Give me 10 integers:\n");
 
  for (i = 0; i<10; i++){
      printf("Give number &#37;d: ",i+1);
      scanf("%d",&a[i]);
  }
 
  printf("\nAfter sorting ..\n");
  getchar(); // why do I have to add this ??
 
  for (i = 0; i<10; i++){
      for (j = i; j<10; j++) {
          if (a[j] > a[i]) {
                  temp = a[i];
                  a[i] = a[j];
                  a[j] = temp;
          }
      }     
  } 
 
  for (i = 0; i<10 ; i++) {     
      printf("Number %d : %d\n",i+1,a[i]);
  }
 
  printf("Press any key to exit ...\n");
  getchar();
  return 0;
}

---

Firstable, I couldn't run it after compiling but I added getchar(); I just did that .. :D

Getchar() basically waits for a keypress. It halts the program execution until you press a key - I'm assuming you need to pause to see that message before proceeding?

Quote:

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Firstable, I couldn't run it after compiling but I added getchar(); I just did that ..

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Perspective and Elysia both answered that, but we actually have an FAQ on it. You might want to read the FAQs.

Quote:

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Originally Posted by Elysia

Getchar() basically waits for a keypress. It halts the program execution until you press a key - I'm assuming you need to pause to see that message before proceeding?

---

Mmm so Getchar() waits for keypress, O.K .. I get it but why do I have to add it ?? why do my program stop if I don't add Getchar() ???


Normally it has to process untill if arrives at }

What? No? Your program does not stop if you don't add it - that's the reason you do add it.
The second statement is also false; execution does not stop when it gets to a }.
In machine language, there are no }. There are just instructions. Loads of them and nothing more.

Right, I suppose what you mean is that the "window just disappears" after you enter the 10 numbers. This is because the scanf() function leaves a newline dangling in the input buffer, so your first getchar() actually removes the newline, then the second getchar() waits for you pressing enter.

In standard C library, all input is "cooked", meaning that the input is not given directly to the application, but rather stored temporarily until a newline is hit, so any "getchar()" will wait for newline to enter. Try this:

Code:

---

int main()
{
  char ch;
  do {
      ch = getchar();
      printf("char = %c\n", ch);
  } while(ch != '\n');
  printf("Hit enter to continue\n");
  getchar();
  return 0;
}

---

If you use that, you will see that all the "char = ..." output comes out in one hit just after you hit enter, rather than one char at a time like you may expect. This helps the application in that it doesn't have to deal with detecting backspace, arrows, delete keys and other complications and deal with these - you get what the user has confirmed, not all the typos and changes of mind.

--
Mats

:eek: :mad: :rolleyes: :devil: :confused:
I don't understandddd ...

You are talking about the function scanf () which leaves according to you a newline dangling in the input buffer !!!

BUT

Code:

---

#include <stdio.h>
int main(void) {
        printf("Help meee");
        return 0;
}

---

If I write this program in Dev-C++ and compile+run it, I don't see anything because the black window disapears after runnig but I tried the same program in Borland C++ and I could see my output !!!!!!

:eek: :mad: :rolleyes: :confused: :devil:

That's IDE-specific. Windows automatically closes a dos prompt after the program finishes.

Yes, well this is covered in the FAQ. http://faq.cprogramming.com/cgi-bin/...&id=1043284385

matsp also just explained it quite well in post #12.

Nevertheless, I'll say something too . . . the console window -- that black window that your program uses to get input and print output -- this window doesn't always act the same way. Some editors will automatically keep it open for you (like MSVC and, apparently, Borland C++), but you can't be certain of this. Some editors don't keep it open for you (like Dev-C++), and when you don't run it from an editor it's not kept open either. (Unless you run it from the command prompt.)

So, keeping that in mind, you can force the window to be kept open in case it isn't done automatically. Something like

Code:

---

getchar()

---

works most of the time, although if you're using scanf() for input you might have to use something more like

Code:

---

while(getchar() != '\n') {}
getchar();

---

to eat up any characters already existing in the standard input.