Re: C and NULL character

[jameskuyper@verizon.net]

mkeles84 wrote:

I have a problem about NULL's.
for example, a variable is name :
--------------
char name[6]="MEHMET"; /* 6 characters */
if (strcmp(name,"M EHMET") == 0){
printf("true");
}else{
printf("false") ;
}
I think result must bt "true" but I saw "false" on screen.
how can I compare it. I don't want to use NULL characker after variable.

The other responses you've received have explained why 'name' needs to
contain a Null-Terminated Character String (NTCS). However, it
occurred to me that you might not be aware of why creating a NTCS is
normally not very difficult.

For instance, the second "MEHMET" in your program causes a piece of
memory to be allocated for an array of 7 chars, with the contents
{'M', 'E', 'H', 'M', 'E', 'T', '\0'}. "MEHMET" itself is replaced in
your code with the address of the first character of that array. As a
result, it is automatically null-terminated, without you having to say
so.

The same thing did NOT happen with the initialization of the 'name'
array, because you explicitly declared it as having a length of 6. As
a result, the definition is equivalent to:

char name[6] = {'M', 'E', 'H', 'M', 'E', 'T'};

Note that there is no terminating null character in that array. If,
instead, you had either explicity made it big enough to hold the
terminating null character:

char name[7] = "MEHMET";

or, more simply, just left the array length unspecified:

char name[] = "MEHMET";

your definition would have been equivalent to:

char name[7] = {'M', 'E', 'H', 'M', 'E', 'T', '\0'};

Cross-posted to comp.lang.c, followups redirected there, since
comp.lang.c is more appropriate for questions about how to use C.

[--]

Re: C and NULL character

jameskuyper@ver izon.net wrote:

mkeles84 wrote:

>I have a problem about NULL's.
>for example, a variable is name :
>--------------
>char name[6]="MEHMET"; /* 6 characters */
>if (strcmp(name,"M EHMET") == 0){
> printf("true");
>}else{
> printf("false") ;
>}
>I think result must bt "true" but I saw "false" on screen.
>how can I compare it. I don't want to use NULL characker after variable.

>
The other responses you've received have explained why 'name' needs to
contain a Null-Terminated Character String (NTCS). However, it
occurred to me that you might not be aware of why creating a NTCS is
normally not very difficult.
>
For instance, the second "MEHMET" in your program causes a piece of
memory to be allocated for an array of 7 chars, with the contents
{'M', 'E', 'H', 'M', 'E', 'T', '\0'}. "MEHMET" itself is replaced in
your code with the address of the first character of that array. As a
result, it is automatically null-terminated, without you having to say
so.
>
The same thing did NOT happen with the initialization of the 'name'
array, because you explicitly declared it as having a length of 6. As
a result, the definition is equivalent to:
>
char name[6] = {'M', 'E', 'H', 'M', 'E', 'T'};
>
Note that there is no terminating null character in that array. If,
instead, you had either explicity made it big enough to hold the
terminating null character:
>
char name[7] = "MEHMET";
>
or, more simply, just left the array length unspecified:
>
char name[] = "MEHMET";
>
your definition would have been equivalent to:
>
char name[7] = {'M', 'E', 'H', 'M', 'E', 'T', '\0'};
>
Cross-posted to comp.lang.c, followups redirected there, since
comp.lang.c is more appropriate for questions about how to use C.

I have a question, don't C arrays start with the first index in 0 as foo[2]
in fact having 3 "positions" foo[0], foo[1], foo[2] ?
So I think MEHMET could be placed inside of name[6];
name[0] = M, name[1] = E, name[2] = H,name[3] = M,name[4] = E,name[5] = T,
name[6]='\0'.
Like:
#include <stdio.h>

int main(){
char test[] = "ba";
printf("%c",tes t[0]);

return 0;
}

Prints b correctly...
So IMHO this is not the problem. It is 3 AM here and there are ages I don
touch C but I think the previous explanation is not *at least totally*
correct.

[Keith Thompson]

Re: C and NULL character

-- <vfbsilva@gmail .comwrites:
[...]

I have a question, don't C arrays start with the first index in 0 as foo[2]
in fact having 3 "positions" foo[0], foo[1], foo[2] ?
So I think MEHMET could be placed inside of name[6];
name[0] = M, name[1] = E, name[2] = H,name[3] = M,name[4] = E,name[5] = T,
name[6]='\0'.
Like:
#include <stdio.h>
>
int main(){
char test[] = "ba";
printf("%c",tes t[0]);
>
return 0;
}
>
Prints b correctly...
So IMHO this is not the problem. It is 3 AM here and there are ages I don
touch C but I think the previous explanation is not *at least totally*
correct.

Yes, C arrays are indexed starting at 0, but given

char name[6];

the array "name" still has only 6 elements, numbered 0 through 5. The
previous explanation was quite correct.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Nick Keighley]

Re: C and NULL character

On 22 Jul, 06:36, -- <vfbsi...@gmail .comwrote:

jameskuy...@ver izon.net wrote:

mkeles84 wrote:

I have a problem about NULL's.
for example,  a variable is name :
--------------
char name[6]="MEHMET"; /* 6 characters */
if (strcmp(name,"M EHMET") == 0){
    printf("true");
}else{
    printf("false") ;
}
I think result must bt "true" but I saw "false" on screen.
how can I compare it. I don't want to use NULL characker after variable.

>

The other responses you've received have explained why 'name' needs to
contain a Null-Terminated Character String (NTCS). However, it
occurred to me that you might not be aware of why creating a NTCS is
normally not very difficult.

>

For instance, the second "MEHMET" in your program causes a piece of
memory to be allocated for an array of 7 chars, with the contents
{'M', 'E', 'H', 'M', 'E', 'T', '\0'}.  "MEHMET" itself is replaced in
your code with the address of the first character of that array. As a
result, it is automatically null-terminated, without you having to say
so.

>

The same thing did NOT happen with the initialization of the 'name'
array, because you explicitly declared it as having a length of 6. As
a result, the definition is equivalent to:

>

        char name[6] = {'M', 'E', 'H', 'M', 'E', 'T'};

>

Note that there is no terminating null character in that array. If,
instead, you had either explicity made it big enough to hold the
terminating null character:

>

        char name[7] = "MEHMET";

>

or, more simply, just left the array length unspecified:

>

        char name[] = "MEHMET";

>

your definition would have been equivalent to:

>

        char name[7] = {'M', 'E', 'H', 'M', 'E', 'T', '\0'};

>

Cross-posted to comp.lang.c, followups redirected there, since
comp.lang.c is more appropriate for questions about how to use C.

>
I have a question, don't C arrays start with the first index in 0

yes

as foo[2]
in fact having 3 "positions" foo[0], foo[1], foo[2] ?

no. foo has two cells foo[0] and foo[1]

So I think MEHMET could be placed inside of name[6];
name[0] = M, name[1] = E, name[2] = H,name[3] = M,name[4] = E,name[5] = T,
name[6]='\0'.

no

Like:
#include <stdio.h>
>
int main(){
        char test[] = "ba";

this is equivalent to
char test[3] = {'b', 'a', '\0'};
(assuming I made no typos!)
that is the compiler automatically leaves space for the terminating
null
character

        printf("%c",tes t[0]);

you need to either put a '\n' in the string
or call fflush(stdout) to ensure the output appears
(actually stdout is probably flushed when the program
terminates so strictly^2 it *doesn't* need an explicit
fflush()...)

>
        return 0;
>
}
>
Prints b correctly...

as it should

So IMHO this is not the problem. It is 3 AM here and there are ages I don
touch C but I think the previous explanation is not *at least totally*
correct.

the previous explanation was totally correct

--
Nick Keighley

"High Integrity Software: The SPARK Approach to Safety and Security"
Customers interested in this title may also be interested in:
"Windows XP Home"
(Amazon)

[--]

Re: C and NULL character

Thanks a lot for the explanations I just understood it now.

Keith Thompson wrote:

-- <vfbsilva@gmail .comwrites:
[...]

>I have a question, don't C arrays start with the first index in 0 as
>foo[2] in fact having 3 "positions" foo[0], foo[1], foo[2] ?
>So I think MEHMET could be placed inside of name[6];
>name[0] = M, name[1] = E, name[2] = H,name[3] = M,name[4] = E,name[5] =
>T, name[6]='\0'.
>Like:
>#include <stdio.h>
>>
>int main(){
> char test[] = "ba";
> printf("%c",tes t[0]);
>>
> return 0;
>}
>>
>Prints b correctly...
>So IMHO this is not the problem. It is 3 AM here and there are ages I don
>touch C but I think the previous explanation is not *at least totally*
>correct.

>
Yes, C arrays are indexed starting at 0, but given
>
char name[6];
>
the array "name" still has only 6 elements, numbered 0 through 5. The
previous explanation was quite correct.
>