check my c code.

This is a discussion on check my c code. within the C Programming forums, part of the General Programming Boards category; I am trying to format a code to print hello fred, so far it only prints hello. the code cannot ...

  1. #1
    rfm
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    check my c code.

    I am trying to format a code to print hello fred, so far it only prints hello. the code cannot be replaced with a simple printf statement. Can anybody help me figure out how to print fred too. the code is
    Code:
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    
    //
    // A short program to print the string "Hello, Fred"
    //
    
    int main (int argc, const char *argv[])
    {
    	char name[5];
    	char ch = name[5];
    
    	char *last_char = &name[5];
    	ch = *last_char;
    
    	char *src = "Fred";
    	char *dst;
    	dst = name;
    
    	// copy the name
    	while (*src)
    	{	
                    src != 0;
                    ++src;
    		*dst = *src;
    	}
    
    
    	char *msg = (char *) malloc (5);
    	strcpy (msg, "Hello, ");
    	strcat (msg, name);
    	puts(msg);
    
    	//free (last_char);
    	free (msg);
    	*last_char = 'A';
    
           //free (msg);
    	msg = (char *) malloc (5);
    	msg = src;
    
    	puts (msg);
    	return 0;
    }
    Last edited by rfm; 09-23-2009 at 05:10 PM.

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    I am mystified, as I see no way for your while loop to end -- src never changes inside the loop so once you get in you can never get out.

  3. #3
    rfm
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    oh i am sorry
    it should be
    src != 0;
    ++src;
    *dst = *src;

  4. #4
    and the Hat of Guessing tabstop's Avatar
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    You would certainly need to increment after updating, otherwise you risk greeting "red" instead of "Fred".

    Also that declaration of the variable ch is invalid -- name[5] doesn't exist.

  5. #5
    Registered User hk_mp5kpdw's Avatar
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    Code:
    char name[5];
    char ch = name[5];
    
    char *last_char = &name[5];
    ch = *last_char;
    
    ...
    
    *last_char = 'A';
    name[5] is an area of memory that you shouldn't access. The valid indexes for the name array are 0 through 4. The last character of the array is name[4].

    Code:
    char *msg = (char *) malloc (5);
    You should avoid casting the return value of malloc in C.

    Code:
    char *msg = (char *) malloc (5);
    strcpy (msg, "Hello, ");
    strcat (msg, name);
    The string "Hello, " requires a buffer of 8 characters (7 + 1 for the terminating null) to store properly. Concatenating "Fred" requires an additional 4 spaces so your buffer must be at least 12 characters but you've only allocated 5.

    Code:
    msg = (char *) malloc (5);
    msg = src;
    This is a memory leak, you've allocated space but then repoint the pointer elsewhere immediately after the allocation. You've now lost the pointer to that memory location and have no way to free it.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

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