Can't run, dunno what is the problem

This is a discussion on Can't run, dunno what is the problem within the C Programming forums, part of the General Programming Boards category; Hi I am just a beginner in programming. Please help as I can't run the program. Code: #include<stdio.h> void displayMenu(); ...

  1. #1
    Registered User
    Join Date
    Oct 2010
    Posts
    16

    Can't run, dunno what is the problem

    Hi I am just a beginner in programming. Please help as I can't run the program.

    Code:
    #include<stdio.h>
    
    void displayMenu();
    void addition();
    void looping();
    
    void main()
    
    {
    
    	displayMenu();
    	looping();
    }
    
    void displayMenu()
    
    {
    	int choice;
    
    	printf("(1)Addition problem\n");
    	printf("(2)Subtraction problem\n");
    	printf("(3)Quit the Program\n");
    	printf("Plese press a number to make your selection: ");
    	scanf("%d",&choice);
    }
    
    void addition()
    {
    	int a,b,c,d;
    
    	printf("2+3= \n");
    	scanf("%d",&a);
    
    	if(a==5)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    
    	printf("3+4= \n");
    	scanf("%d",&b);
    
    	if(b==7)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    
    	printf("6+2= \n");
    	scanf("%d",&c);
    
    	if(c==8)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    
    	printf("3+1= \n");
    	scanf("%d",&d);
    
    	if(d==4)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    }
    
    void subtraction()
    {
    	int a,b,c,d;
    
    	printf("3-3= \n");
    	scanf("%d",&a);
    
    	if(a==0)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    
    	printf("4-3= \n");
    	scanf("%d",&b);
    
    	if(b==1)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    
    	printf("6-2= \n");
    	scanf("%d",&c);
    
    	if(c==4)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    
    	printf("3-1= \n");
    	scanf("%d",&d);
    
    	if(d==2)
    
    		printf("Right\n");
    
    	else
    
    		printf("Wrong\n");
    }
    
    void looping(int choice)
    
    {
    	if(choice==1)
    	{
    		addition();
    	}
    
    	else if(choice==2)
    	{
    		subtraction();
    	}
    
    	else
    	{
    		printf("You must select 1 or 2\n");
    	}
    }

  2. #2
    and the hat of int overfl Salem's Avatar
    Join Date
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    Location
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    Posts
    32,848
    displayMenu() does not return a value (your choice)

    looping() is declared and called with zero parameters, yet the definition expects a parameter.
    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.

  3. #3
    Registered User
    Join Date
    Oct 2010
    Posts
    16
    Thank you, I have just change my codes to:

    Code:
    #include<stdio.h>
    
    int displayMenu();
    void addition();
    void subtraction();
    void looping(int);
    
    void main()
    
    {
    	int c;
    	c=displayMenu();
    	looping(c);
    }
    
    int displayMenu()
    
    {
    	int choice;
    
    	printf("(1)Addition problem\n");
    	printf("(2)Subtraction problem\n");
    	printf("(3)Quit the Program\n");
    	printf("Plese press a number to make your selection: ");
    	scanf("%d",&choice);
    	return choice;
    }
    
    void looping(int c)
    
    {
    	if(c==1)
    	{
    		addition();
    	}
    
    	else if(c==2)
    	{
    		subtraction();
    	}
    
    	else
    	{
    		printf("You must select 1 or 2\n");
    	}
    }
    I always have a hard time remembering the formats/syntax.

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