Trying to randomly generate a string in order to get a conditional to execute

Question

Seems that strings in C are my weak point, need your guys' help again. I should have included this in my first one, so I apologize about that.

So I have this function that I got working correctly, it fills a string with random chars from the set {r, e, s, e, t} and returns it:

char *inputString()
{
    static char string[6];
    const char *digits = "reset";
    int i;

    for (i = 0; i < 5; i++) 
    {
        string[i] = digits[ rand() % 4 + 0 ];
    }

    return string;
}

So the char at string[5] is '\0', correct? And string[1] through string[4] are random combinations of the other characters (r, e, s, e, t).

Now, I have another function that calls this one. These purpose of it is trying to get the printf statement to execute:

other_fxn() 
{
  char *s;

  while (1) 
  { 
    s = inputString();
    if (s[0] == 'r'&& s[1] == 'e' && s[2] == 's' && s[3] == 'e' && s[4] == 's'
        && s[5] == '\0') 
    {
      printf('worked!\n');
      break;
    }
  }
}

main:

{
   srand(time(NULL));
   other_fxn();
}

It compiles fine but the loop runs forever, with the if statement never executing. Can somebody help me out why it won't execute. Thanks once again.

yes, sorry I printed it wrong here the first time. Its inside the loop. — John Doe, Jul 18 '15 at 00:28
You never actually assign string[5] = '\0'. So the c-string you return in from input_string may not be null terminated. This may cause your infinite loop behavior as well. — Timothy Murphy, Jul 18 '15 at 00:30
`random chars from the set {r, e, s, e, t}` : It does not include `t` because `rand() % 4 + 0` — BLUEPIXY, Jul 18 '15 at 00:33
@TimothyMurphy `string` is `static`. So It is initialized to `0`. — BLUEPIXY, Jul 18 '15 at 00:36
If you fix your broken printf statement and add your missing headers, the code [works as expected](https://ideone.com/B5EWRq). — Kerrek SB, Jul 18 '15 at 00:42
@KerrekSB I think you should post that as the answer since it's the only one that really answers the question. — tangrs, Jul 18 '15 at 01:54

Pavel Oganesyan · Accepted Answer · 2015-07-18T00:48:28.053

1

  while (1) 
  {    
  s = inputString();
    if (s[0] == 'r'&& s[1] == 'e' && s[2] == 's' && s[3] == 'e' && s[4] == 't'
        && s[5] == '\0') 
    {
      printf('worked!\n');
      break;
    }

  }

First of all, in your example you are looking for word "reses". And you just don't change a string inside the loop. So you will get out of the loop only if random string is reset after the initial generation. I don't know why do you need this, but code above will get out of the loop. Maybe not so soon, but it will.

Also, what is this one:

string[i] = digits[ rand() % 4 + 0 ];

Why + 0? And why % 4, not % 5?

Better go with this:

char *inputString()
{
    static char string[6];
    const char *digits = "reset";
    for (int i = 0; i < 5; i++) 
    {
        string[i] = digits[ rand() % 5];
    }
    return string;
}

And try using debugger, it really helps.

edited Jul 18 '15 at 00:48

answered Jul 18 '15 at 00:28

Pavel Oganesyan

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sorry the function was supposed to be inside the loop, posted it wrong the first time – John Doe Jul 18 '15 at 00:30
Isn't string [5] already '\0' anyways? when I declared it: string[6] I thought that meant null byte was at string[5] with indexes 0 -4 for characters – John Doe Jul 18 '15 at 00:34
1

Correct me if I am wrong, but aren't `static` variables always initialised to zero due to their being in the `data` section? If so, `string[5] = '\0';` is redundant as it is already 0 – Levi Jul 18 '15 at 00:37
@Levi http://stackoverflow.com/questions/11229477/are-all-char-arrays-automatically-null-terminated – Pavel Oganesyan Jul 18 '15 at 00:43
1

@Levi and also this one - http://stackoverflow.com/questions/1414215/initial-value-of-int-array-in-c And to make it clear - yes, you are right, static makes it initialized. – Pavel Oganesyan Jul 18 '15 at 00:45

Levi · Answer 2 · 2015-07-18T00:43:18.123

You are only running the inputString function once, so you will only ever receive one string. This can be fixed by moving the function call to the while (1) loop:

while (1) 
{ 
  s = inputString();
  if (s[0] == 'r'&& s[1] == 'e' && s[2] == 's' && s[3] == 'e' && s[4] == 's'
    && s[5] == '\0') 
    {
      printf('worked!\n');
      break;
    }
}

Also, to make your code slightly cleaner, you can use C's strcmp function to compare the string, rather than checking each individual character:

#include <string.h>

while (1) 
{ 
  s = inputString();
  if (strcmp(s, "reset") == 0) 
    {
      printf('worked!\n');
      break;
    }
}

Adding on to what @Pavel said about your random character selection, to make it a bit more reusable, you can use sizeof instead of a hard-coded 5:

string[i] = digits[ rand() % (sizeof(string) - 1)]; // Also removed extraneous 0

sorry the function was supposed to be inside the loop, posted it wrong the first time — John Doe, Jul 18 '15 at 00:30

Timothy Murphy · Answer 3 · 2015-07-18T00:45:44.520

I have two changes for you: You can make sure string[5] is 0. (It should already be because it's static)

char *inputString()
{
    static char string[6];
    const char *digits = "reset";
    int i;

    for (i = 0; i < 5; i++) 
    {
        string[i] = digits[ rand() % 4 + 0 ];
    }
    string[5] = '\0'; // null terminate your string.

    return string;
}

you can avoid doing while(1) by just putting the condition inside the while conditional check. and then perform the print after the while loop.

other_fxn() 
{
   char *s = inputString();

   while (s[0] != 'r' || s[1] != 'e' || s[2] != 's' || s[3] != 'e' || s[4] != 's' || s[5] != '\0') 
   { 
    s = inputString();
   }
   printf('worked!\n');
}

Trying to randomly generate a string in order to get a conditional to execute

3 Answers3