what's wrong with the following code?

This is a discussion on what's wrong with the following code? within the C Programming forums, part of the General Programming Boards category; Hello ppl, was wondering if you can help with me with some pointer code. Code: void Menu() { char* phone; ...

  1. #1
    Registered User
    Join Date
    Nov 2001
    Posts
    3

    Question what's wrong with the following code?

    Hello ppl, was wondering if you can help with me with some pointer code.

    Code:
    void Menu()
    {
    	char* phone;
    	char* name;
    
    	GetPhone(&phone);
    	GetName(&name);
    // use the following printf function to test my code. 
    // check bottom of msg for output and what my problem.
    	printf("%d - %s - %d - %s", &name, name, &phone, phone);
    
    }
    
    void GetName(char** nameRef)
    {
    	int ch;
    	char buffer[100];  /* buffer used to get input from the user.
    			      if the user enters in 105 characters for name
    			      the buffer will reset to 5
    			    */
    	printf("(Name must be between 3-20 characters in length)\nEnter a name to be added to the list: ");
    	fgets(buffer, sizeof(buffer), stdin);	// gets input from stdin and puts in buffer
    	while ((ch = getchar()) != '\n' && ch != EOF);
    	*nameRef = buffer;
    	//printf("%d - %s", &nameRef, *nameRef);
    	if((strlen(*nameRef)-1 < 3) || (strlen(*nameRef)-1 > 20))
    	{
    		printf("Name has to be between 3 and 20 characters long.\nPress [Enter] key to be prompted again.");
    		
    		GetName(nameRef);
    	}	
    }
    
    void GetPhone(char** phoneRef)
    {
    	int ch;
    	char buffer[50];
    	
    	printf("Enter the phone number to be added to the list: ");
    	fgets(buffer, sizeof(buffer), stdin);
    	//while((ch = getchar()) != '\n' && ch != EOF);
    	*phoneRef = buffer;
    	printf("%d - %s", &phoneRef, *phoneRef);
    	if(strlen(*phoneRef)-1 != 9)
    	{
    		printf("The phone number you entered has too many digits.\nMust be in either xxxx xxxx or xxxx-xxxx formats.\nReplace x with numbers.");
    		
    		GetPhone(phoneRef);
    	}
    	if(!((buffer[4] == ' ') || (buffer[4] == '-')))
    	{
    		printf("Must be in xxxx xxxx or xxxx-xxxx formats. Replace x with numbers.");
    		GetPhone(phoneRef);
    	}
    	if(validate_phone(*phoneRef) != 0)
    	{
    		printf("The phone number you have entered contains non-numeric data. Try again.");
    		GetPhone(phoneRef);
    	}
    }
    
    int validate_phone(char* phoneRef)
    {
    	unsigned x;
    	for (x = 0; x < strlen(phoneRef); x++)
    	{
    		if(phoneRef[x] == phoneRef[4])
    			break;
    		if(!isdigit(phoneRef[x])) return 1;
    	}
    	return 0;
    }
    output of test printf statement: & it does go to the new line like that
    601364 - paul
    - 601368 - 5074-0658

    problem is when i put a printf statement in the GetPhone and GetName functions and notice that the reference address of the name pointer is the same as the one for phone reference. problem is whenever i try and get the string entered in GetName i get weird ASCII characters.

    What i'm try to do is create a simple phone book using a link list structure to hold the data, as a exercise to try and understand pointers.

    Thx in advance

    catalyst

  2. #2
    ATH0 quzah's Avatar
    Join Date
    Oct 2001
    Posts
    14,826
    Code:
    void GetName(char** nameRef)
    {
    	int ch;
    	char buffer[100];
    Buffer is a local variable, which loses scope as soon as this function returns. Anything in this buffer is gone. Your pointer pointing to this buffer ends up pointing at something that no longer exists.

    Either make your buffer static, or use dynamic memory allocation (malloc or calloc). If you use *alloc functions, you will need to use 'free' on them before your program exits.

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

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