A problem with my program.

This is a discussion on A problem with my program. within the C Programming forums, part of the General Programming Boards category; I tried to create a program that gets an integer and returns 1. Well, here it is: Code: int Option1(int ...

  1. #1
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    A problem with my program.

    I tried to create a program that gets an integer and returns 1.
    Well, here it is:
    Code:
    int Option1(int num)
    {		
    	if(num%2 == 0)
    	{
    		num = num/2;
    		printf("\nNext value: %d", num);
    		Option1(num);
    	}
    	if(num%2 != 0)
    	{
    		num = num*3+1;
    		printf("\nNext value: %d", num);
    		Option1(num);
    	}
    	
    	if(num == 1)
    	{
    		printf("\nFinal value: 1...DONE\n");
    	}
    	
    	return 0;
    }
    
    int main(void)
    {
    	int chosen_opt;
    	int num;
    	
    	printf("Loops and conditions:\n");
    	printf("----------------------------\n");
    	
    	printf("Choose an option:\n");
    	printf("#1 -Gets an integer and returns 1.\n");
    	printf("#2\n");
    	printf("#3\n");
    		
    	chosen_opt = scanf("%d", &chosen_opt);
    	if(chosen_opt == 1) 
    	{
    		 printf("\nEnter a number:\n");
    		 num = getchar();
    		 Option1(num);
    	}
    	
    	if(chosen_opt == 2) printf("\nIN CONSTRUCTION!!\n");
    	if(chosen_opt == 3) printf("\nIN CONSTRUCTION!!\n");
    	if(chosen_opt == 4) { printf("\nExiting in 5 seconds...\n"); sleep(5); }
    
    
    	getchar();
    	return 0;
    }
    I dunno why, but it runs an eternal loop (:O), why?
    Last edited by eXeCuTeR; 11-22-2007 at 12:05 PM.

  2. #2
    Aia
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    Code:
    num = getchar();
    will always obtain the '\n' left behind by the scanf() function called previously.

  3. #3
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    Quote Originally Posted by Aia View Post
    Code:
    num = getchar();
    will always obtain the '\n' left behind by the scanf() function called previously.
    Huh?

  4. #4
    Aia
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    Code:
    scanf("%d", &chosen_opt);
    When you enter an integer you are pressing ENTER and that's like '\n', scanf() reads the integer and leaves the new line in the buffer.
    Code:
    num = getchar();
    Comes along and read that '\n' character. That's the one you're processing in the Option1 function. Since is not a 0 it keeps looping by vitue of the the call to itself in the second if.

  5. #5
    Chinese pâté foxman's Avatar
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    I didn't test the program, but here's what i think it's doing... (by the way, when you have this kind of problem (infinite loop), you should try using your debugger in mode "step by step", can be really useful, in fact, it will give you the answer right over so...).

    Your fonction is looping infinitely because you are telling it to do so... you'll never converge with a condion such as
    Code:
    int Option1(int num)
    {		
    	if(num%2 == 0)
    	{
    		num = num/2;
    		printf("\nNext value: %d", num);
    		Option1(num);
    	}
    	if(num%2 != 0)
    	{
    		num = num*3+1;
    		printf("\nNext value: %d", num);
    		Option1(num);
    	}
     
    	// ...
    }
    Take num == 1 and look...

    Code:
        1 % 2 == 1
    =     <enter second if>
        4 = 3 + 1
    =     <recursive call with num == 4>
        4 % 2 == 0
    =     <enter the first if>
        2 = 4 / 2
    =     <recursive call with num == 2>
        2 % 2 == 0
    =     <enter first if>
        1 = 2 / 2
    =     <recursive call with num == 1>
        1 % 2 == 1
    =     <enter second if>
        4 = 3 + 1
    =     <and on and on and on....>
    Also, this will not resolve your problem but consider an else-if instead of two if.

  6. #6
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    Oh OK thanks
    How do I fix it now?
    I'm trying to think on many ways but can't find the right answer.
    I tried also:
    Code:
    if(num == 1) printf("w00t");
    else
    {
    if...
    if...
    }
    Last edited by eXeCuTeR; 11-22-2007 at 12:28 PM.

  7. #7
    Aia
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    Code:
    chosen_opt = scanf("%d", &chosen_opt);
    Do you know what scanf returns?
    It returns the total number of items successfully read, EOF when the input stream reached its end, or a negative number if an error occurs.

    No matter what integer you enter chosen_opt will always be 1 if its successful.

  8. #8
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    Oh, OK
    But how do I fix the eternal loop?

  9. #9
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    Please help!

  10. #10
    and the hat of wrongness Salem's Avatar
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    Don't bump your threads - see the rules.
    foxman all but gave you the answer already, test the exit condition before you decide how to recurse.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  11. #11
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    The exit condition? When you press 4?

    This is my code atm:
    Code:
    #include <stdio.h>
    #include <unistd.h>
    
    int Option1();
    
    int Option1()
    {
    	int num;
    	
    	printf("\nEnter a number:\n");
    	num = getchar() - 48;
    	if(&num==NULL) if(num < 0) printf("\nOnly native numbers are acceptable!\n"); // why &num == NULL and not num = NULL? I just tried it and it was compiled successfully.
    	
    	if(num != 1 && num > 0)
    	{
    		if(num&#37;2 == 0)
    		{
    			num = num/2;
    			printf("\nNext value: %d", num);
    		}
    		if(num%2 != 0)
    		{
    			num = num*3+1;
    			printf("\nNext value: %d", num);
    		}
    		if(num == 1) printf("\nFinal value: 1...\nDONE!\n");
    	}
    	return 0;
    }
    
    int main(void)
    {
    	int chosen_opt;
    	
    	printf("Loops and conditions:\n");
    	printf("----------------------------\n");
    	
    	printf("Choose an option:\n");
    	printf("#1 -Gets an integer and returns 1.\n");
    	printf("#2\n");
    	printf("#3\n");
    		
    	chosen_opt = getchar() - 48;
    	if(chosen_opt == 1) Option1();
    	if(chosen_opt == 2) printf("\nIN CONSTRUCTION!!\n");
    	if(chosen_opt == 3) printf("\nIN CONSTRUCTION!!\n");
    
    
    	getchar();
    	return 0;
    }
    And it's not working well =[
    I've been trying for hours to fix it, but failed..
    Last edited by eXeCuTeR; 11-22-2007 at 01:29 PM.

  12. #12
    Captain Crash brewbuck's Avatar
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    Please, for the love of your local deity print '\n' AFTER the string not before it.

    Don't do this:

    Code:
    printf("\nHello world!");
    Instead do this:

    Code:
    printf("Hello world!\n");
    Putting newlines BEFORE strings defies all logic and antagonizes the stdio buffer layer's natural buffering strategy.

  13. #13
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    I fixed it.
    But how do I fix my program now? =[

  14. #14
    and the hat of wrongness Salem's Avatar
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    You basically rewrote the code and now you have a completely different question.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  15. #15
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    Kinda, lol.
    I don't wanna open a lot of topics so I did this.
    Anyways,
    Could you help me with that?

    New code:
    IT WORKS NOW!
    2 questions, please answer these I have no idea:
    1. how to get more than 1 digit?
    2. how to make the program stop.
    for example,
    look at the if(num <= 0)...i typed exit(3); because I didn't know what to stop the program and don't let it execute the OddNumbers and EvenNumbers function.
    PLEASE HELP!!
    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int OddNumbers(int num);
    int EvenNumbers(int num);
    void Result1();
    
    int OddNumbers(int num)
    {
    	num *= 3;
    	num++;
    	if(num == 1) Result1();
    	else printf("Next value: &#37;d\n", num);
    	if(num%2 == 0) EvenNumbers(num);
    	if(num%2 != 0) OddNumbers(num);
    	
    	return 0;
    }
    int EvenNumbers(int num)
    {
    	num /= 2;
    	if(num == 1) Result1();
    	else printf("Next value: %d\n", num);
    	if(num%2 != 0) OddNumbers(num);
    	if(num%2 == 0) EvenNumbers(num);
    		
    	return 0;
    }
    
    void Result1()
    {
    	printf("Final value: 1...\nDONE!\n");	
    	exit(3);
    }
    
    int main(void)
    {	
    	printf("Loops and conditions:\n");
    	printf("----------------------------\n");
    	
    	int num;
    
    	printf("Enter a number:\n");
    	num = getchar() - 48;
    	if(num == 1) Result1();
    	if(num <= 0) { printf("ERROR::Only native numbers (doesn't include 0)!\n"); exit(3); }
    	if(num != 1 && num%2 == 0) EvenNumbers(num);
    	if(num != 1 && num%2 != 0) OddNumbers(num);
    
    	getchar();
    	return 0;
    }
    Last edited by eXeCuTeR; 11-22-2007 at 03:51 PM.

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