Converting an int array to a character array nicely printed

This is a discussion on Converting an int array to a character array nicely printed within the C Programming forums, part of the General Programming Boards category; Hello everyone, I am trying to build a function that takes a struct that contains an array of integers, which ...

  1. #1
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    Converting an int array to a character array nicely printed

    Hello everyone,

    I am trying to build a function that takes a struct that contains an array of integers, which I know the length (size) of, and convert it to a nicely formatted character array (string) output in C.

    I am trying to do a nicely printed output in my function like this:
    []
    [3]
    [3, 4]
    [3, 4, 6]

    My code is:
    Code:
    typedef struct {
    	/* A pointer to a dynamically allocated array to hold the elements. */
    	int *element;
    	/* The length of the array. */
    	int length;
    	/* The capacity as a list. */
    	int capacity;
    } IntArray;
    
    char *toString(IntArray *ar) {
        /*
         * Returns a string to be printed, from an IntArray.
         */
        char *str;
    
        // Check for empty array
        if (ar->length == 0) {
              str = (char *) malloc(3 * sizeof(char));
              if (str == NULL) {
                    printf("*** Memory request failed. ***\n");
                    exit(EXIT_FAILURE);
              }
    
              str[0] = '[';
              str[1] = ']';
              str[2] = '\0';
        }
    
        // Otherwise:
        else {
        	int i = 0, j = 0;
        	int size = ar->length * 3 + 3;
    	str = (char *) malloc(size * sizeof(char));
    
    	str[j] = '[';
    	while (i != ar->length) {
    		str[++j] = '0' + ar->element[i++];
    		str[++j] = ',';
    		str[++j] = ' ';
    	}
    	str[--j] = ']';
    	str[++j] = '\0';
        }
    
        return str;
    }
    I have a problem though because sometimes the int array has number bigger than 0-9. I tried using itoa() but it does not seem to solve my problem because it adds a null terminated character at the end. I also have problem finding how much memory I can to allocate as I don't know how big the integers will be.

    It works great for an int array of 0's, but when I have different numbers this function does not work anymore.

    I will really appreciate your help as I have been stuck on this function for quite a bit.
    Thank you very much.

  2. #2
    spurious conceit MK27's Avatar
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    Quote Originally Posted by Jary316 View Post
    I tried using itoa() but it does not seem to solve my problem because it adds a null terminated character at the end.
    Removing the null terminator at the end of a string is easy to do.
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    Thank you very much.

    Oh right, should I use itoa() and then strlen(str) to start at the null terminated charactetr?
    I think I got this working now:
    Code:
        else {
        	int i = 0, j = 0;
        	int size = ar->length * 3 + 3;
    		str = (char *) malloc(size * sizeof(char));
    
    		str[j] = '[';
    		while (i != ar->length) {
    			itoa(ar->element[i++], &str[++j], 10);
    			j = strlen(str);
    			str[j] = ',';
    			str[++j] = ' ';
    		}
    		str[--j] = ']';
    		str[++j] = '\0';
        }
    But I still have the problem of not knowing the right size for malloc(). I know the size of the integer array (how many elements) but each integer could be 1 element long or more. It could be 1, 3 or 6 for example. I do not know what size to allocate to malloc().

    Anyway I can find it so I don't use too many memory please? I could use realloc too, set a small initial malloc() but I do not know how to find out when the array is full (while itoa() is called or before it gets called). Thank you.
    Last edited by Jary316; 11-12-2008 at 08:53 AM.

  4. #4
    and the Hat of Guessing tabstop's Avatar
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    No int can have more then 10 digits plus maybe a negative sign. So that's your bound; that's what you should probably malloc.

  5. #5
    Registered User hk_mp5kpdw's Avatar
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    Code:
    str = (char *) malloc(size * sizeof(char));
    You shouldn't cast malloc in C, plus I believe the size of a character is guaranteed to be 1 so that whole thing should just boil down to:
    Code:
    str = malloc(size);
    ... likewise for the other malloc.
    "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."
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  6. #6
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    tabstop: That is correct but there is some extra space left though.

    I was thinking about something like that:
    Code:
        	// How many elements are in the array
        	int size = 3;
        	// Defines a string: size is 3 because we have 2 brackets and a null terminated character
    	str = (char *) malloc(size * sizeof(char));
    
    	str[j] = '[';
    	while (i != ar->length) {
    
    		// Define how many characters are in the next integer
    		int temp = ar->element[i];
    		if (temp < 0)
    			temp -= temp;
    		int count = 0;
    		while ((temp = temp / 10) > 0)
    			++count;
    
    		// The new size is the old size and the new element
    		size += count;
    		// Reallocate the array to be able to hold the next element
    		str = (char *) realloc(str, size * sizeof(char));
    
    		itoa(ar->element[i++], &str[++j], 10);
    		j = strlen(str);
    		str[j] = ',';
    		str[++j] = ' ';
    	}
    Basically I define that the array is 3 characters long: [] and the null character.
    Then I assign the next element to temp, check the number of characters by dividing it by 10 until it reaches 0, then calls realloc for the old value + the new value to hold the new element.

    Unfortunately, something is going wrong and it crashes.

    hk_mp5kpdw: You are right for the part that sizeof(char) is one but it's not a big deal. It is better, in my opinion, to typecast malloc because it returns a void pointer. Some compilers do not do the automatic conversion. But that is not a big problem

  7. #7
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Jary316
    It is better, in my opinion, to typecast malloc because it returns a void pointer. Some compilers do not do the automatic conversion.
    All standard conforming C compilers perform that conversion.
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  8. #8
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    Quote Originally Posted by laserlight View Post
    All standard conforming C compilers perform that conversion.
    And hopefully, VERY few people are still using compilers that are significantly over 15 years old. The ANSI standard came out in 1989, but compilers supported the "upcoming" ANSI standard before then - I worked with ANSI-like compilers in 1986-7, IIRC. Older compilers than that would be UNLIKELY to still be in use.

    --
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  9. #9
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    I think I did successfully solve it:
    Code:
        else {
        	int i = 0, j = 0;
    
        	// How many elements are in the array
        	int size = 3;
        	// Defines a string: size is 3 because we have 2 brackets and a null terminated character
    		str = malloc(size);
    
    		str[j] = '[';
    		while (i != ar->length) {
    
    			// Define how many characters are in the next integer
    			int temp = ar->element[i];
    			int count = 1;
    
    			if (temp != 0) {
    				if (temp < 0)
    					temp = -temp;
    
    				while ((temp = temp / 10) > 0)
    					++count;
    			}
    
    			// The new size is the old size and the new element, and the comma and space
    			size = size + (count * 3);
    			// Reallocate the array to be able to hold the next element
    			str = realloc(str, size);
    
    			itoa(ar->element[i++], &str[++j], 10);
    			j = strlen(str);
    			str[j] = ',';
    			str[++j] = ' ';
    		}
    		str[--j] = ']';
    		str[++j] = '\0';
        }
    This seems to be working.

    Thanks for all your help guys!

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