Help printing strings

**Frendin** · 12-09-2012

Hey guys!

I've been working on a program that will create a randomly generated password using the numbers 0-9 and the letters A-Z all in uppercase. I have my program created and have to come to my final impass before I complete this program.

I have been asked by my teacher that this has to be printed with the %s specifier. In the code below, it comes out weird. If I switch the array password[8] to a char from an int, it does a weird response as well. Since this is homework and I want to learn how to do this correctly, if you could possibly give me a hint as to what to do to solve this, that would be wicked. If you want to give the anwser, that is cool as well.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


void numbers12(int password[8]);
void characters34(int password[8]);
void numbers56(int password[8]);
void characters78(int password[8]);


int main ()
{
	srand(time(NULL));
	int password[8];


	numbers12(password);
	characters34(password);
	numbers56(password);
	characters78(password);


	printf("Your randomly generated password comprised of numbers and letters is %s.\n", password);


	return 0;
}


void numbers12(int password[8])
{
	int *ptr;


	ptr = &password[0];


	for(int i = 0; i < 2; i++)
	{
		ptr[i] = rand() % 10;
		if(ptr[1] == ptr[0])
		{
			while(ptr[1] == ptr[0])
			{
				ptr[1] = rand() % 10;
				break;
			}
		}
		printf("A part of your password is the number %d.\n", ptr[i]);
	}
}


void characters34(int password[8])
{
	int *ptr;


	ptr = &password[0];


	for(int i = 2; i < 4; i++)
	{
		ptr[i] = (rand() % 26) + 'A';
		if(ptr[3] == ptr[2])
		{
			while(ptr[3] == ptr[2])
			{
				ptr[3] = (rand() % 26) +'A';
				break;
			}
		}
		printf("A part of your password is the character %c.\n", ptr[i]);
	}
}


void numbers56(int password[8])
{
	int *ptr;


	ptr = &password[0];


	for(int i = 4; i < 6; i++)
	{
		ptr[i] = rand() % 10;
		if(ptr[4] == ptr[0] || ptr[1])
		{
			while(ptr[4] == ptr[0] || ptr[1])
			{
				ptr[4] = rand() % 10;
				break;
			}
		}
		if(ptr[5] == ptr[0] || ptr[1] || ptr[4])
		{
			while(ptr[5] == ptr[0] || ptr[1] || ptr[4])
			{
				ptr[5] = rand() % 10;
				break;
			}
		}
		printf("A part of your password is the number %d.\n", ptr[i]);
	}
}


void characters78(int password[8])
{
	int *ptr;


	ptr = &password[0];


	for(int i = 6; i < 8; i++)
	{
		ptr[i] = (rand() % 26) + 'A';
		if(ptr[6] == ptr[2] || ptr[3])
		{
			while(ptr[6] == ptr[2] || ptr[3])
			{
				ptr[6] = (rand() % 26) +'A';
				break;
			}
		}
		if(ptr[7] == ptr[2] || ptr[3] || ptr[6])
		{
			while(ptr[7] == ptr[2] || ptr[3] || ptr[6])
			{
				ptr[7] = (rand() % 26) + 'A';
				break;
			}
		}
		printf("A part of your password is the character %c.\n", ptr[i]);
	}
}

Thanks again =D
-Frendin

**Frendin** · 12-09-2012

I would also like to specify that those while loops are to prevent repeats of the numbers. There are supposed to be no repeated numbers or letters in the password. I am working on fixing that.

**c99tutorial** · 12-09-2012

Strings have to be char* (or arrays of char) and they have to be null-terminated. So, my suggestion:

1. switch back to char password[]
2. make it one more character long (e.g. char password[9])
3. make sure the last character is a '\0', e.g. set password[8] = '\0'

Then when you do printf("%s\n", password) it will make sure that password is printed correctly

**dwks** · 12-09-2012

This

Code:

while(ptr[7] == ptr[2] || ptr[3] || ptr[6])

is equivalent to

Code:

while((ptr[7] == ptr[2]) || (ptr[3] != 0) || (ptr[6] != 0))

instead of what you probably meant, i.e.

Code:

while((ptr[7] == ptr[2]) || (ptr[7] == ptr[3]) || (ptr[7] == ptr[6]))

You have a lot of repeated code, and a lot of code that references constants, like that bit I quoted -- why are you saying ptr[7]? You should rarely mention array indices like that directly. A lot of it is unnecessary too. Like this.

Code:

        if(ptr[3] == ptr[2])
        {
            while(ptr[3] == ptr[2])
            {
                ptr[3] = (rand() % 26) +'A';
                break;
            }
        }

I'll assume you didn't really mean to have the break in since it renders the while loop irrelevant. In that case, here's some equivalent code.

Code:

do {
    ptr[3] = rand() % 26 + 'A';
} while(ptr[3] == ptr[2]);

You should probably have a function for that. Maybe a function that takes in an array of letters you've already generated, so that it can iterate through the array and check for duplicates. That way you'd catch all repeats rather than just ptr[3] == ptr[2]. Example:

Code:

char random_letter_that_hasnt_been_generated_yet(char generated[], int generated_count) {
    for(;;) {
        char c = rand() % 26 + 'A';
        int i, already_generated = 0;
        for(i = 0; i < generated_count && !already_generated; i ++) {
            if(c == generated[i]) already_generated = 1;
        }
        if(!already_generated) {
            return c;
        }
    }
}

Of course it may be a good idea to not make that loop infinite, in case you try to generate 27 distinct letters (with only 26 different possibilities). The pigeonhole principle strikes again.

**Frendin** · 12-09-2012

Thanks for the help guys. My code is still coming up weird c99tutorial, so I will have to take a longer look at it. Thank you for the lesson dwks, I'll take a quick break from my code and then see if I can't change those while loops. I also don't know what those do functions do so I will look those up as well.

**Click_here** · 12-09-2012

Code:

printf("Your randomly generated password comprised of numbers and letters is %s.\n", password);


...


ptr[i] = rand() % 10;

Your numbers are not in ascii format.

You can do one of two things:
Add: rand() % 10 + '0'

OR

Change the printf at the end to print an number, number, character, character, ...

If I were you, I'd go through and change all the inputs to include + '0'

**Click_here** · 12-09-2012

Also, c99tutorial's advice needs to be taken with the string being of type char instead of int for the printf

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