Thread: int to char casting/itoa

  1. #1
    eat my shorts!
    Join Date
    Apr 2002

    int to char casting/itoa

    char binary(int number)
    	int remainder, i=0;
    	char c[1];
    	char str[22];
    		remainder = number%2;
    		number = number >> 1;
    		itoa(remainder, c, 32);
    		str[i] = c[0];
    		cout << "\nremainder: " << remainder;
    	} while (number > 1);
    	if(number <= 1) {
    		cout << "Number: " << number;
    		itoa(number, c, 2);
    		str[i] = c[0];
    	str[i] = '\0';
    	cout << "STRING: " << str << endl;
    	return *str;	
    it converts the number into binary but crashes after displaying the last cout statement. Could some1 tell me why is that?
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  2. #2
    Lurking whiteflags's Avatar
    Join Date
    Apr 2006
    United States
    You are returning a pointer to a character. Change your return type to reflect this.

  3. #3
    carry on JaWiB's Avatar
    Join Date
    Feb 2003
    Seattle, WA
    You have to ensure that the buffer you pass to itoa can store the string. In this case, you pass a char array of one element (c) which probably won't be large enough. I don't think you really understand itoa; perhaps you should read this page carefully.

    >You are returning a pointer to a character. Change your return type to reflect this.

    No, he is returning a character.
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  4. #4
    Confused Magos's Avatar
    Join Date
    Sep 2001
    He's prolly trying to return a string. In that case use std::string's instead as returning local non-static character arrays is gonna cause trouble.

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