string comparison

[junky_fellow@yahoo.co.in]

Consider the following piece of code:
char *str = "Hello";
if (str = "Hello")
printf("\nstrin g matches\n");

str is pointer to char and "Hello" is a string literal whose
type is "array of char".

How can we compare two different objects for equality ?
Is some conversion is being done before that comparison ?
If yes, which conversion rule from C standard is applied here ?

[Krishanu Debnath]

Re: string comparison


junky_fellow@ya hoo.co.in wrote:[color=blue]
> Consider the following piece of code:
> char *str = "Hello";
> if (str = "Hello")
> printf("\nstrin g matches\n");
>
> str is pointer to char and "Hello" is a string literal whose
> type is "array of char".
>
> How can we compare two different objects for equality ?
> Is some conversion is being done before that comparison ?
> If yes, which conversion rule from C standard is applied here ?[/color]

'=' is assignment operator.

Krishanu

[Sensei]

Re: string comparison

On 2005-08-09 12:38:58 +0200, junky_fellow@ya hoo.co.in said:
[color=blue]
> Consider the following piece of code:
> char *str = "Hello";
> if (str = "Hello")
> printf("\nstrin g matches\n");
>
> str is pointer to char and "Hello" is a string literal whose
> type is "array of char".
>
> How can we compare two different objects for equality ?
> Is some conversion is being done before that comparison ?
> If yes, which conversion rule from C standard is applied here ?[/color]


strcmp


Check the string.h header file for all the functions.

--
Sensei <senseiwa@tin.i t>

cd /pub
more beer

[junky_fellow@yahoo.co.in]

Re: string comparison


Sensei wrote:[color=blue]
> On 2005-08-09 12:38:58 +0200, junky_fellow@ya hoo.co.in said:
>[color=green]
> > Consider the following piece of code:
> > char *str = "Hello";
> > if (str = "Hello")
> > printf("\nstrin g matches\n");
> >
> > str is pointer to char and "Hello" is a string literal whose
> > type is "array of char".
> >
> > How can we compare two different objects for equality ?
> > Is some conversion is being done before that comparison ?
> > If yes, which conversion rule from C standard is applied here ?[/color]
>
>
> strcmp
>
>
> Check the string.h header file for all the functions.
>
> --
> Sensei <senseiwa@tin.i t>
>
> cd /pub
> more beer[/color]

I know that there's a C library function "strcmp" for string
comparison. What I wanted to ask is that the comparison
if (str == "Hello") doesn't give any compile time error.
Nor does it give any Warning. That means some conversion
should have been done because we cannot compare two objects
of different types.

[Stephane Zuckerman]

Re: string comparison

> I know that there's a C library function "strcmp" for string[color=blue]
> comparison. What I wanted to ask is that the comparison
> if (str == "Hello") doesn't give any compile time error.
> Nor does it give any Warning. That means some conversion
> should have been done because we cannot compare two objects
> of different types.[/color]

When you write
if (a_var == "a string") { /* ... */ }
you're really writing "is the value of a_var equals to the value of the
address that points to the constant string 'a string' ?" So, yes, there is
a conversion, but maybe not the one you thought...


--
"Je deteste les ordinateurs : ils font toujours ce que je dis, jamais ce
que je veux !"
"The obvious mathematical breakthrough would be development of an easy
way to factor large prime numbers." (Bill Gates, The Road Ahead)

[Krishanu Debnath]

Re: string comparison


junky_fellow@ya hoo.co.in wrote:[color=blue]
> Sensei wrote:[color=green]
> > On 2005-08-09 12:38:58 +0200, junky_fellow@ya hoo.co.in said:
> >[color=darkred]
> > > Consider the following piece of code:
> > > char *str = "Hello";
> > > if (str = "Hello")
> > > printf("\nstrin g matches\n");
> > >
> > > str is pointer to char and "Hello" is a string literal whose
> > > type is "array of char".
> > >
> > > How can we compare two different objects for equality ?
> > > Is some conversion is being done before that comparison ?
> > > If yes, which conversion rule from C standard is applied here ?[/color]
> >
> >
> > strcmp
> >
> >
> > Check the string.h header file for all the functions.
> >
> > --
> > Sensei <senseiwa@tin.i t>
> >
> > cd /pub
> > more beer[/color]
>
> I know that there's a C library function "strcmp" for string
> comparison. What I wanted to ask is that the comparison
> if (str == "Hello") doesn't give any compile time error.
> Nor does it give any Warning. That means some conversion
> should have been done because we cannot compare two objects
> of different types.[/color]

Okay, now you have realized that you are talking about equality
operator. Yes, "Hello" has type 'array 5 of char'. But in value
context it decays to pointer to the first element of the array
which has type 'pointer to char'.

Krishanu

[junky_fellow@yahoo.co.in]

Re: string comparison


Krishanu Debnath wrote:[color=blue]
> junky_fellow@ya hoo.co.in wrote:[color=green]
> > Sensei wrote:[color=darkred]
> > > On 2005-08-09 12:38:58 +0200, junky_fellow@ya hoo.co.in said:
> > >
> > > > Consider the following piece of code:
> > > > char *str = "Hello";
> > > > if (str = "Hello")
> > > > printf("\nstrin g matches\n");
> > > >
> > > > str is pointer to char and "Hello" is a string literal whose
> > > > type is "array of char".
> > > >
> > > > How can we compare two different objects for equality ?
> > > > Is some conversion is being done before that comparison ?
> > > > If yes, which conversion rule from C standard is applied here ?
> > >
> > >
> > > strcmp
> > >
> > >
> > > Check the string.h header file for all the functions.
> > >
> > > --
> > > Sensei <senseiwa@tin.i t>
> > >
> > > cd /pub
> > > more beer[/color]
> >
> > I know that there's a C library function "strcmp" for string
> > comparison. What I wanted to ask is that the comparison
> > if (str == "Hello") doesn't give any compile time error.
> > Nor does it give any Warning. That means some conversion
> > should have been done because we cannot compare two objects
> > of different types.[/color]
>
> Okay, now you have realized that you are talking about equality
> operator. Yes, "Hello" has type 'array 5 of char'. But in value
> context it decays to pointer to the first element of the array
> which has type 'pointer to char'.
>
> Krishanu[/color]

Can you please specify which section of C standard specify this
conversion ?

[junky_fellow@yahoo.co.in]

Re: string comparison


Krishanu Debnath wrote:[color=blue]
> junky_fellow@ya hoo.co.in wrote:[color=green]
> > Consider the following piece of code:
> > char *str = "Hello";
> > if (str = "Hello")
> > printf("\nstrin g matches\n");
> >
> > str is pointer to char and "Hello" is a string literal whose
> > type is "array of char".
> >
> > How can we compare two different objects for equality ?
> > Is some conversion is being done before that comparison ?
> > If yes, which conversion rule from C standard is applied here ?[/color]
>
> '=' is assignment operator.
>
> Krishanu[/color]
I am sorry, I meant (str == "Hello")

[Krishanu Debnath]

Re: string comparison


junky_fellow@ya hoo.co.in wrote:[color=blue]
> Krishanu Debnath wrote:[color=green]
> > junky_fellow@ya hoo.co.in wrote:[color=darkred]
> > > Sensei wrote:
> > > > On 2005-08-09 12:38:58 +0200, junky_fellow@ya hoo.co.in said:
> > > >[/color][/color][/color]

<snip>
[color=blue][color=green]
> >
> > Okay, now you have realized that you are talking about equality
> > operator. Yes, "Hello" has type 'array 5 of char'. But in value
> > context it decays to pointer to the first element of the array
> > which has type 'pointer to char'.
> >
> > Krishanu[/color]
>
> Can you please specify which section of C standard specify this
> conversion ?[/color]

c99 draft Section 6.3.2.1p3

--
krishanu@india. ti.com

[kernelxu@hotmail.com]

Re: string comparison

there may not be any warnnings or errors ,even though you had used
"(str == "Hello") ", they just campare their address value ,not the
strings.

if you use (str = "Hello") , the judge expression is always true .
"printf("\nstri ng matches\n");" will be excuted .

[Chen Ming]

Re: string comparison

u should use *lint* to check your program.

<junky_fellow@y ahoo.co.in>
??????:11235864 07.988940.29044 0@z14g2000cwz.g ooglegroups.com ...[color=blue]
>
> Sensei wrote:[color=green]
>> On 2005-08-09 12:38:58 +0200, junky_fellow@ya hoo.co.in said:
>>[color=darkred]
>> > Consider the following piece of code:
>> > char *str = "Hello";
>> > if (str = "Hello")
>> > printf("\nstrin g matches\n");
>> >
>> > str is pointer to char and "Hello" is a string literal whose
>> > type is "array of char".
>> >
>> > How can we compare two different objects for equality ?
>> > Is some conversion is being done before that comparison ?
>> > If yes, which conversion rule from C standard is applied here ?[/color]
>>
>>
>> strcmp
>>
>>
>> Check the string.h header file for all the functions.
>>
>> --
>> Sensei <senseiwa@tin.i t>
>>
>> cd /pub
>> more beer[/color]
>
> I know that there's a C library function "strcmp" for string
> comparison. What I wanted to ask is that the comparison
> if (str == "Hello") doesn't give any compile time error.
> Nor does it give any Warning. That means some conversion
> should have been done because we cannot compare two objects
> of different types.
>[/color]

[Default User]

Re: string comparison

junky_fellow@ya hoo.co.in wrote:
[color=blue]
> Consider the following piece of code:[/color]


Consider reading through the FAQ.




Brian

[pete]

Re: string comparison

Krishanu Debnath wrote:
[color=blue]
> "Hello" has type 'array 5 of char'.[/color]

(sizeof "Hello" == 6)

--
pete

[pete]

Re: string comparison

junky_fellow@ya hoo.co.in wrote:
[color=blue]
> What I wanted to ask is that the comparison
> if (str == "Hello") doesn't give any compile time error.[/color]

After you get that settled, you need to know that
("Hello" == "Hello")
isn't guaranteed true.
Identical string literals may either refer to the same object
or to different objects.

--
pete

[bwaichu@yahoo.com]

Re: string comparison

You are confused.

char *string; /* string is a pointer */

char *string = "hello"; /* string is a pointer that points to 'h' */

char *string; /* declare the pointer */
string = malloc(5); /* sets up memory to point to*/

string = "hello"; /* assigns "hello" to memory pointed to */

Here's a small C program:

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

int
main() {

char *string;
string = malloc(5);

string = "hello\n";

(void)printf("% s\n", string);

exit(0);
}

Here's it disassembled:

..section .rodata # read only data

..LC0:
.string "hello\n"

LC1:
.string "%s\n"

..section .text #program starts here

..global main

main:
pushq %rbp # set up the stack
movq %rsp, %rbp
subq $16, %rsp # allocate space on the stack for
# 16 bytes
movl $5, %edi # pass the size to malloc
call malloc
movq %rax, -8(%rbp) # %rax is the string variable
movq $.LC0, -8(%rbp) # this is the memory assign
movq -8(%rbp), %rsi # "hello\n"
movl $.LC1, %edi # "%s\n"
movl $0, %eax # NOP padding
call printf
movl $0, %edi # exit(0)
call exit

So now you want to compare two strings. If you look at libc strcmp:

int
strcmp(const char *s1, const char *s2)
{
while (*s1 == *s2++)
if (*s1++ == 0)
return (0);
return (*(unsigned char *)s1 - *(unsigned char *)--s2);
}

What is being looked at are the individual values of each element of
string. Since 'A' is really 0x41, when checking for comparison, you
are really comparing two values and doing simple subtraction.

C hides this from you. The assembly would use cmps or else you could
just manually do the compare.

Brian