Thread: Trying to use strstr

Originally Posted by fxtdr79

Here is my new code based on all the suggestions. Obviously I'm too stupid to figure this out. Since it should take 2 minutes and its taken me way too long.

Code:

#include <stdio.h>
#include <string.h>

void checkString(char c);

int main()
{
   char c[50];
   fgets(c, 50, stdin);

   while (c != EOF)
   {
      checkValue(c);
      c = getchar();
   }
   return(0);
}

void checkString(char *c)
{
   char *ptr_to_string = malloc(sizeof(char) * 100);
   strcpy(ptr_to_string, *c);
   char *find = strstr(ptr_to_string, "lb");
   if (*find != null)
   {
      puts("Input contains lb");
   }
}

First of all you are not stupid! You are just frustrated, relax, you will feel better once this works, because it will be YOUR WORK NOT MINE. I meant it should take two minutes to make some changes and post it back, which is what you did in that exact time frame. Take a deep breath, let the anger out, relax!

Now back to work:

A few things to fix now so you have a clean framework:

1) Do you see the checkString function declaration above the main() function? It still has the old argument type char c instead of char *c. That is basically telling main: "Hey if you see checkString anywhere this is what it should look like and its argument should pe a pointer. So if you leave it as it is the compiler will give you the same warning you got before basically saying : "hey, you told me this is supposed to take a character, but you are passing the argument as a pointer! STOP DECEIVING ME!! "

2) Delete everything in checkString and do what you were trying to do before with strstr() (not exactly sure what you were checking but you CAN NOW USE strstr, because c is now a pointer to string).

3) Delete the while loop you have in main now. fgets actually returns NULL if it fails so what you want is to while loop like this:

Code:

/* read until there is nothing left */
while(fgets(c,50,stdin) != NULL){
   /* do stuff with array c*/
}

Originally Posted by fxtdr79

Obviously I'm stupid. I guess that's why I haven't figured it out yet. If I can't use char for an array to represent a string, what else can I use?

You've got the magic word there: array. Why not use an array to represent your string? EDIT: To be more specific, you have to change everything to be an array.

First if you want to have a string of 5 characters you should declare an array [6] to account for the '\0' as well.

My bad. It tells you that because there is not enough space in the array. Every iteration it reads another 5 characters and replaces the ones read in the previous iteration. So on the last iteration all there is left to read is "lb" which is why that's the only thing in the array at the end.

my god, the guy is just trying to get some string input tested and you're harassing him mercilessly and refusing to give any code like it's gold-plated or something.

now really, if you people are going to be such hardasses, then at least get your information correct: part of the problem that he's having -- and that you're allowing to perpetuate -- is that fgets() requires you to allocate your character array by at least TWO (2) chars larger than your largest possible input, because it collects the newline ('\n') character in addition to appending the null character ('\0').

if you're going to collect input from a user via the standard input (terminal) then you better really oversize the character buffer, unless you're prepared to develop some bulletproof error testing and input buffer overflow controls. but we're not going to go there, so for now just do it like this:

Code:

char c[500];                                  // yes, that's right.  memory is cheap.

while (fgets(c, sizeof c, stdin) != NULL)     // no magic numbers here
{
    checkstring(c);
}

Originally Posted by fxtdr79

What I'm saying is if I type 123456789lb shouldn't c be c = {1, 2, 3, 4, \n} so when I call checkString, it shouldn't be able to find the string "lb"?

not necessarily. when you overflow your array, the behavior is undefined. depending on your system/compiler it may "work" one way or "not work" in another. looks like in this case, your overflowed char array didnt crash into anything else, so it "works". try initializing another variable before or after that array and it probably won't.




BTW, have you considered using "strtol"? this will convert the numeric string to a (long) integer, and give back a pointer to the rest of the string. it might be a bit more useful for what you're trying to accomplish, overall. here's a lazy example that employs no error checking, just to give you an idea:

Code:

char str[20] = "12345 lbs";                        // notice it's "lbs" and not "lb"

char *ptr;
long value;

value = strtol(str, &ptr, 10)                      // 10 is the base to convert, i.e. decimal base 10

while(*ptr = ' ')                                  // is current single pointer character a space?
    ptr++;                                         // increment until it's not     

if (strncmp(ptr, "lb", 2))                         // using 'strncmp' just for the hell of it.  only looks at 2 characters
    printf ('%ld pounds.\n", value);

else if (strncmp(ptr, "kg", 2))
    printf ("%ld kilograms\n", value);

else 
    printf ("%ld [unknown units]\n", value);

Originally Posted by jephthah

now really, if you people are going to be such hardasses, then at least get your information correct: part of the problem that he's having -- and that you're allowing to perpetuate -- is that fgets() requires you to allocate your character array by at least TWO (2) chars larger than your largest possible input, because it collects the newline ('\n') character in addition to appending the null character ('\0').

No it doesn't. He's using it in a loop. It can be any size he wants. The only thing he risks is his substring being split across calls. But fgets doesn't require your input be 2 characters smaller. Fgets will read N bytes, or until it hits a newline, which ever comes first.

Code:

while( fgets( buf, 5, stdin ) != NULL )
{
    ...
}

If the input is: 1234567, then the following will happen:

1234\0 <-- first pass
567\n\0 <-- second pass

If you're going to correct people, make sure you're right.

Originally Posted by jephthah

not necessarily. when you overflow your array, the behavior is undefined. depending on your system/compiler it may "work" one way or "not work" in another. looks like in this case, your overflowed char array didnt crash into anything else, so it "works". try initializing another variable before or after that array and it probably won't.

He's not overflowing his buffer. He's using it in a loop. The whole point of fgets is so that you can't overflow it. He's not overflowing anything. He's not right in what he thinks is going to end up in the string, but he's not overflowing it either.


Quzah.

on the first point, that's what i was saying. if you're expecting a 5 character input then because fgets() collects the newline, your char buffer needs to be TWO characters larger than your expected input. the fact that he's using it in a loop doesnt matter... he will have it "split across two calls" which will screw up the operation of the program.


on the second point... yeah, you're right. he only thinks it's working because on teh second pass it finds the "lb". i must have been thinking of his previous post where he was using scanf, which would overflow. my bad.