Will not run.

This is a discussion on Will not run. within the C Programming forums, part of the General Programming Boards category; Hi, I am fairly new to the c language and I am working on a project for my class. When ...

  1. #1
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    Will not run.

    Hi, I am fairly new to the c language and I am working on a project for my class. When I run this program in cmd prompt, and enter my menu selection, the program errors out on me.

    Here is my code:
    Code:
    #include <stdio.h>
    
    
    void DisplayMenu();
    void Count10();
    void Countby2();
    void CountDown();
    int main()
    {
    	void DisplayMenu();
    	int menu;
    	do
    	{
    		DisplayMenu();
    printf ("Enter the number of the operatrion you would like to perform: ");
    		scanf ("%d", menu);
    		
    		switch (menu)
    		{
    		case 1: Count10(); break;
    		case 2:	Countby2(); break;
    		case 3:	CountDown(); break;
    		case 4:	printf ("Quitting the Program"); break;
    		default : printf ("Invalid Input\n\n"); break;
    		}
    		
    	}while (menu != 4);
    
    	
    	return 0;
    }
    //**************************************
    //**************************************
    void DisplayMenu()
    {
    	printf ("1. Count to 10\n");
    	printf ("2. Count to 100 by 2\n");
    	printf ("3. Countdown from 10\n");
    	printf ("4. Exit\n");
    }
    
    //**************************************
    //**************************************
    void Count10()
    {
    int i;
    	for(i=1;i<=10;i++)
    	{
    		printf ("%d", i);
    	}
    }
    
    //*************************************
    //*************************************
    void Countby2()
    {
    	int i=0;
    	while (i<=100)
    	{
    		printf ("%d\n", i);
    		i+=2;
    	}
    }
    //***********************************
    //***********************************
    void CountDown()
    {
    int f;
    	for(f=10;f>0;f--)
    	{
    		printf ("%d\n", f);
    	}
    
    }

  2. #2
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    Code:
    int main()
    {
    	void DisplayMenu();
    Why are you declaring a function inside a function, given that you've already declared it before? I don't think that's needed.
    Code:
    		scanf ("%d", menu);
    WE HAVE A WINNNNNAAAARRR!!!!!
    scanf requires the address of variables to stick stuff into, so you want &menu, not menu. Plus, for bonus points, using scanf can be hazardous to your health, as if you were trying to read a string into menu it may be too long (scanf wouldn't know how long your string buffer is). So, in future, may I suggest:-
    Code:
    char temp[BUFSIZ];
    
    fgets(temp, BUFSIZ, stdin);
    sscanf(temp, "%d", &menu);
    That will save you many tears in the future, when you want more than just a number.

    EDIT: fgets does solve one or two other problems as well, such as flushing the input buffer, but that may be a bit beyond at the mo.
    Last edited by SMurf; 03-16-2006 at 03:29 PM.

  3. #3
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    well the thing is

    we havent learned most of that stuff u just said, we have been told to use scanf. Also, i stated that i was a noob so u dont need to be dick. if your going to be a dick, dont reply.

  4. #4
    ATH0 quzah's Avatar
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    Quote Originally Posted by Vectorl33t
    if your going to be a dick, dont reply.
    You've got that slightly off. It's more like: If you're going to be a dick, at least make sure you're not wrong.

    Quote Originally Posted by SMurf
    Code:
    int main()
    {
    	void DisplayMenu();
    Why are you declaring a function inside a function, given that you've already declared it before? I don't think that's needed.
    It's not declaring a function within a function. (That would be a nested function.) It's a function prototype. You're allowed to prototype functions nearly anywhere you please, and it's still both legal, and allowed.


    Quzah.
    Last edited by quzah; 03-16-2006 at 04:16 PM.
    Hope is the first step on the road to disappointment.

  5. #5
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    Quzah: The same thing also appears four lines above that snippet.

    Vectorl33t: I'd love to know in what way I've offended. I neither poked fun nor swore at you, and I wrote the words "in future" for a reason, i.e. that you don't need/know fgets at the moment, but you will.

  6. #6
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    Quote Originally Posted by SMurf
    WE HAVE A WINNNNNAAAARRR!!!!!
    Thats what he is being a dick about. If he was not, then sorry for me saying that as i must have read it wrong.
    Last edited by Vectorl33t; 03-16-2006 at 04:36 PM.

  7. #7
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    Allow me to explain: the need for the address of menu to be passed as an argument to scanf was the correct answer to the implied question "Why won't this run?". Hence, as many a game show host would say, "We have a winner", although they usually shout it in the manner I tried to convey.

    I guess light-heartedness doesn't work here.

  8. #8
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    lol, tis hard to detect sarcasm over typing

  9. #9
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    Anywho...

    ya, i got it running, now... i wanna run case 1 2 and 3 but this time with and inputting number from the user. So, what i did was add 3 new case statements. I cannot figure out how to run a loop with user inputted data, being i suck at programming. Here is what i added for one of my loops.
    Code:
     
    //*************************************
    //*************************************
    void Inputcountup()
    {
    int p;
    int g;
    	Printf ("Enter a number to count up to: ");
    	scanf ("%d", &p);
    		for(g=1;g<=p;g++)
    {
    			printf ("%d\n", g);
    }
    But when i run the program, it errors out again and says syntax error at the end of input. And i do not no what it means by this. and yes, before u ask i did include it at the top such as void inputcountup(); and it is also in my case statement and included in my menu as an option.

  10. #10
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    Here is all my code so far

    Here is all my code so far. And it tells me syntax error before ')' token on line 87.
    Code:
    #include <stdio.h>
    //***************************
    //***************************
    void DisplayMenu();
    void Count10();
    void Countby2();
    void CountDown();
    void Inputcountup();
    int main()
    {
    	void DisplayMenu();
    	int menu;
    	do
    	{
    		DisplayMenu();
    		printf ("Enter the number of the operatrion you would like to perform: ");
    		scanf ("%d", &menu);		
    		switch (menu)
    		{
    		case 1 : Count10(); break;
    		case 2 : Countby2(); break;
    		case 3 : CountDown(); break;
    		case 4 : Inputcountup(); break;	
    		case 5 : printf ("Quitting the Program"); break;
    		default : printf ("Invalid Input\n\n"); break;
    		}
    		
    	}
    		while (menu != 4);	
    	return 0;
    }
    //**************************************
    //**************************************
    void DisplayMenu()
    {
    	printf ("1. Count to 10\n");
    	printf ("2. Count to 100 by 2\n");
    	printf ("3. Countdown from 10\n");
    	printf ("4. Input Count up\n");
    	printf ("5. Exit\n");
    }
    
    //**************************************
    //**************************************
    void Count10()
    {
    int i;
    	for(i=1;i<=10;i++)
    	{
    		printf ("%d\n", i);
    	}
    }
    //*************************************
    //*************************************
    void Countby2()
    {
    	int i=0;
    	while (i<=100)
    	{
    		printf ("%d\n", i);
    		i+=2;
    	}
    }
    //***********************************
    //***********************************
    void CountDown()
    {
    int f;
    	for(f=10;f>0;f--)
    	{
    		printf ("%d\n", f);
    	}
    	}
    //***********************************
    //***********************************
    void Inputcountup()
    {
    int p;
    int g;
    	Printf ("Enter a number to count up to: ");
    	scanf ("%d", &p);
    		for(g=1;g<=p;g++)
    {
    			printf ("%d\n", g);
    }
    }
    }

  11. #11
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    Remove a bracket at the end. If you had formatted correctly that would be readily apparent.

    You will also need to decapitalize Printf in function Inputcountup

  12. #12
    ATH0 quzah's Avatar
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    Quote Originally Posted by SMurf
    Quzah: The same thing also appears four lines above that snippet.
    What's your point? You can prototype a function as many times as you like, provided all prototypes are the same. It doesn't matter that it was four lines up. It's still valid to prototype a function within another funciton. Even if you prototype it outside that function.


    Quzah.
    Hope is the first step on the road to disappointment.

  13. #13
    Frequently Quite Prolix dwks's Avatar
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    What's your point?
    I think the point was that while it's valid to prototype a function multiple times, it's unnessesary.
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
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  14. #14
    ATH0 quzah's Avatar
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    But that wasn't the original point. The original point was that they were nesting functions. They weren't. They were prototyping, which as used, wasn't incorrect. Like I said, if you're pointing out "errors", at least make sure you're correct.


    Quzah.
    Hope is the first step on the road to disappointment.

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