Thread: How to do this

Halo all,

I've made a simple code in which i'm trying to read a string and compare it with another string. The problem is that i'm not able to get the string from the user neither from scanf nor from fgets. I get run time error. One more thing how can i assign the array size in the second argument of fgets when i'm using pointer format of a string.

Code:

#include<stdio.h>
#include<string.h>
#include<conio.h>
int main(void)
{
	char *p="hello",*q;
	int x;
	printf("Enter the string");
	//scanf("%s",*q); I did this one time
	fgets(q,10,stdin);// At the other time i did this
	x=strcmp(p,q);
	if(x==0)
	{
		printf("YES");
	}
	else
	{
		printf("NO");
	}
	getch();
}

Also what are the advantages of declaring a string as "char *s" rather than "char s[20]" with respect to memory? I mean if the string is of 10 chars then in the latter form 10 bytes will get wasted, does the same happens in the former also?

Originally Posted by BEN10

Also what are the advantages of declaring a string as "char *s" rather than "char s[20]" with respect to memory?

Char *s is a pointer with no memory.

Char s[20] can hold 20 bytes.

So there is not an advantage, and using using *q as you do in your program will probably cause an overflow/overwrite (= segmentation fault, your runtime error).

> Also what are the advantages of declaring a string as "char *s" rather than "char s[20]" with respect to memory?
The latter points to something you can use :-)

"q" doesn't point to anything meaningful (most certainly an area of memory you're not allowed to access).

Don't be worried about "wasting" 10 bytes. Unless you're on some super-memory-restricted-machine, which I doubt.

Perhaps a more compliant example:

Code:

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

int main(void)
{
   const char * p = "hello";
   char q[32];
   int x = 0;

   printf("Enter the string");
   fflush(stdout);

   fgets(q, sizeof q, stdin);

   x = strcmp(p, q);

   if(x == 0)
      printf("Yes");
   else
      printf("No");
   
   fflush(stdout);

   getchar();
   return 0;
}

Originally Posted by zacs7

> Also what are the advantages of declaring a string as "char *s" rather than "char s[20]" with respect to memory?
The latter points to something you can use :-)

"q" doesn't point to anything meaningful (most certainly an area of memory you're not allowed to access).

Don't be worried about "wasting" 10 bytes. Unless you're on some super-memory-restricted-machine, which I doubt.

Perhaps a more compliant example:

Code:

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

int main(void)
{
   const char * p = "hello";
   char q[32];
   int x = 0;

   printf("Enter the string");
   fflush(stdout);

   fgets(q, sizeof q, stdin);

   x = strcmp(p, q);

   if(x == 0)
      printf("Yes");
   else
      printf("No");
   
   fflush(stdout);

   getchar();
   return 0;
}

My question is how can I input a string using char *q format? Or is it that char *q is helpful only while initializing the string, in case of inputting I should use char q[32]? And why did you use fflush(stdout)?

So it means it'll be better to use the array form i.e char a[10] when I want to input the string from the user and when I want to initialize the string then i may use char *a form.
Why you have done this

Code:

char *buffer[10];//shouldn't it be char buffer[10]
     char *q = buffer;   /*  or equivalently  char *q = &buffer[0]; */

I think grumpy was just trying to provide an example. *buffer[10] would be an array of pointers -- in your program, you don't want that.

You can create a variable to fit the exact size of the input if you want, but you cannot do it in the actual fgets statement; you still have to use a buffer there of a predetermined size. Afterword, you can call strlen() on the buffer, and then malloc() a separate variable of that size and strcpy() into it. Of course, there is no point in doing that in main to save memory, since "buffer" will never die. But if you wrote it into a separate function, you could save memory, because once the function is finished it's local variables will be freed/recycled. Just to let you know.

But in the case of user input, where we are talking about < a single kb, there is not much of a point to that.

I modified my code and still not getting the correct output.

Code:

#include<stdio.h>
#include<conio.h>
#include<string.h>
int main(void)
{
	char *p="hello";
	char q[10];
	int x;
	printf("Enter the string");
	fgets(q,10,stdin);
	x=strcmp(p,q);
	if(x==0)
	{
		printf("YES");
	}
	else
	{
		printf("NO");
	}
	getch();
}

I'm inputting hello but still getting printed NO. I printed x and it shows me a value of 1, why x not getting equal to 0?
And finally what is the need of fflush(stdout) in zac7's code?

An easy way to test this would be with a printf:

Code:

printf("->%s<-",q); fflush(stdout);

Notice I enclosed the ->string<-. If you add this into your program, you will probably see:

->hello
<-

Because fgets includes the newline, so q is "hello\n" (which does not match).

The purpose of fflush(stdout) is to force the output to be "flushed" immediately, since output is actually held until it reaches a certain size or some other circumstance (like, the program is over). Generally, fflush is not necessary. However, it is important when using printf to debug (like above) because is you are trying to (eg) locate the source of a segmentation fault, the output buffer may not be flushed, so you will be misled as to where the fault is.

Originally Posted by c.user

BEN10,

Code:

    char *p="hello";

you know, all strings write to read-only memory, all strings like printf("one two three\n"); - "one two three\n" literal constant it goes to read-only memory and have address, that address then pass to function printf, it reads it and print then, so your "hello" is string literal too, it goes to read-only memory and every it's charachter has an adress (six characters), address of first character assigns to p pointer; that is why you can't change it, when you try to change it, you try to change read-only memory
in K&R 5.5

ZOMG, lolz , if it is a "read only memory" in the first place, how on earth would you put your string there?.

Originally Posted by stevesmithx

ZOMG, lolz , if it is a "read only memory" in the first place, how on earth would you put your string there?.

Because it is set to read-only by the OS _AFTER_ the OS has stored it there in the first place.

The technical details of this is that the OS has control of the page-table which determines what process has what access-rights to which bits of memory. The OS itself can change the attributes, and it can also bypass the attributes by using a secondary mapping of the same memory, with a different set of attributes.

--
Mats

Thanks for the clarification, guys.
"write to read-only memory" sounded a bit like an oxymoron to me.
A term like "protected memory" instead of "read-only memory" which might refer to the ROM of the machine would be less ambiguous i guess.

But read-only memory is a common term for "non-writable" memory in modern systems. It is in RAM, but the OS has marked it to be "non-writable". And it can not be written BY THE APPLICATION, but it CAN be written to by the OS (if the OS decides that it wants to).

The code itself is almost always read-only, but if you use a debugger, the OS will write breakpoint instruction(s) into the code itself [and store away the previous instruction so that it can be put back in again when the breakpoint is no longer in use].

Like most things, you have to get used to the nomenclature, whether it actually makes sense or not.

--
Mats