come in(I have a puzzled)

[Kevin]

Source:
#include <stdio.h>
#include <ctype.h>
void main()
{
char a[]="this is the beautiful world!";
char b[20];
strcpy(b,a);
printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
if(strcmp(a,b)==0)
printf("a equal to b!\n");
}

Now , question as follows:

1. Why "b" size unequal "a" size , but "b" can output "a" content?
2. Why "if(strcmp(a,b)==0)" is true?

 

[Jens Thoms Toerring]

Source:
#include <stdio.h>
#include <ctype.h>

That's not needed but you're missing

void main()

main() always returns an it, so make that

int main( void )

{
char a[]="this is the beautiful world!";
char b[20];
strcpy(b,a);

Pfff. You just wrote past the end of array 'b'. All bets are off.

printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
if(strcmp(a,b)==0)
printf("a equal to b!\n");

You're missing a

return 0;

}

Now , question as follows:

1. Why "b" size unequal "a" size , but "b" can output "a" content?

Because you invoked undefined behaviour by writing past the end of b.
Everything can happen from now on, the program may even look as if it
would be working correctly.

2. Why "if(strcmp(a,b)==0)" is true?

Because you invoked undefined behavior and the way memory is
set aside for both the arrays by your compiler happened to
make that result possible.
Regards, Jens

 

[Bart]

Source:
#include <stdio.h>
#include <ctype.h>
void main()
{
    char a[]="this is the beautiful world!";
    char b[20];
    strcpy(b,a);
    printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
    if(strcmp(a,b)==0)
       printf("a equal to b!\n");

}

Now , question as follows:

1. Why "b" size unequal "a" size , but "b" can output "a" content?
2. Why "if(strcmp(a,b)==0)" is true?

a and b are different types:

a is a pointer to a string
b /is/ a string.

But a can obviously point to a string identical to what's in b.

The size of a will be 4 or whatever, the size of b will be 20
(although it should be more in this case otherwise your "this is..."
string will overflow it).

strcmp() compares two pointers to strings; but b is automatically
converted to such a pointer, thanks to the way C handles arrays.

 

[ReplyDude]

replied!

Source:
#include <stdio.h>
#include <ctype.h>
void main()
{
    char a[]="this is the beautiful world!";
    char b[20];
    strcpy(b,a);
    printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
    if(strcmp(a,b)==0)
       printf("a equal to b!\n");

}

Now , question as follows:

1. Why "b" size unequal "a" size , but "b" can output "a" content?
2. Why "if(strcmp(a,b)==0)" is true?

a and b are different types:

a is a pointer to a string
b /is/ a string.

But a can obviously point to a string identical to what's in b.

The size of a will be 4 or whatever, the size of b will be 20
(although it should be more in this case otherwise your "this is..."
string will overflow it).

strcmp() compares two pointers to strings; but b is automatically
converted to such a pointer, thanks to the way C handles arrays.

 

[Keith Thompson]

Source:
#include <stdio.h>
#include <ctype.h>
void main()
{
    char a[]="this is the beautiful world!";
    char b[20];
    strcpy(b,a);
    printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
    if(strcmp(a,b)==0)
       printf("a equal to b!\n");

}

Now , question as follows:

1. Why "b" size unequal "a" size , but "b" can output "a" content?
2. Why "if(strcmp(a,b)==0)" is true?

a and b are different types:
Yes.

a is a pointer to a string

No, a is an array of type char[29] (the length of the literal used to
initialize it plus 1 for the trailing '\0').

b /is/ a string.

No, b is an array of type char[20]. An array can *contain* a string.

A "string" in C is a a data format, not a data type.

But a can obviously point to a string identical to what's in b.

The object named ``a'' can't point to anything, since it's an array,
not a pointer. However, the expression ``a'', in most but not all
contexts, has the value of a pointer to (the address of) the first
element of the object named ``a''.

The size of a will be 4 or whatever, the size of b will be 20
(although it should be more in this case otherwise your "this is..."
string will overflow it).

No, sizeof a == 20.

strcmp() compares two pointers to strings; but b is automatically
converted to such a pointer, thanks to the way C handles arrays.

I think the other errors in the program have already been pointed out,
but ...

Drop the ``#include <ctype.h>''; it's not needed.
Add ``#include <string.h>''; it *is* needed.
Change ``void main()'' to ``int main(void)''.
Change ``char b[20];'' to, for example, ``char b[30];''.
In the printf call, change ``sizeof(b)'' to ``(int)sizeof b'';
the "%d" format requires an int, not a size_t.
Before the closing brace, add ``return 0;''.
Enable warnings in your compiler, and buy some whitespace (it's really
cheap).

 

[Bart]

Source:
#include <stdio.h>
#include <ctype.h>
void main()
{
    char a[]="this is the beautiful world!";
    char b[20];
    strcpy(b,a);
    printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
    if(strcmp(a,b)==0)
       printf("a equal to b!\n");
}
Now , question as follows:
1. Why "b" size unequal "a" size , but "b" can output "a" content?
2. Why "if(strcmp(a,b)==0)" is true?

a and b are different types:
Yes.

a is a pointer to a string

No, a is an array of type char[29] (the length of the literal used to
initialize it plus 1 for the trailing '\0').

Yes, of course, I read it as char *a (I think char a[] is pointer in
some other context).

b /is/ a string.

No, b is an array of type char[20].  An array can *contain* a string.

So some informality.. Here the intention is clearly a string.

A "string" in C is a a data format, not a data type.


The object named ``a'' can't point to anything, since it's an array,
not a pointer.  However, the expression ``a'', in most but not all
contexts, has the value of a pointer to (the address of) the first
element of the object named ``a''.


No, sizeof a == 20.

Don't know about that, it seems a bit more. But different from sizeof
b anyway, which was the original confusion.

Well if a and b are both arrays, then the reason why strcmp matches
them but sizeof is different is a little different!

(strcmp() only matches characters up to the nul terminator, ignoring
any trailing characters in the array that still contribute to the
sizeof property.)

 

[Keith Thompson]

Source:
#include <stdio.h>
#include <ctype.h>
void main()
{
    char a[]="this is the beautiful world!";
    char b[20];
    strcpy(b,a);
    printf("char array b size is(%d),The content of b is:%s\n",sizeof(b),b);
    if(strcmp(a,b)==0)
       printf("a equal to b!\n");

}

Now , question as follows:

1. Why "b" size unequal "a" size , but "b" can output "a" content?
2. Why "if(strcmp(a,b)==0)" is true?

a and b are different types:
Yes.

a is a pointer to a string

No, a is an array of type char[29] (the length of the literal used to
initialize it plus 1 for the trailing '\0').

Yes, of course, I read it as char *a (I think char a[] is pointer in
some other context).

``char a[]'' as a function parameter declaration declares ``a'' as a
pointer (unfortunately, IMHO).

b /is/ a string.

No, b is an array of type char[20].  An array can *contain* a string.

So some informality.. Here the intention is clearly a string.

The intention is that b is an array that contains a string (or rather,
it would contain a string after the strcpy() call *if* b were big
enough). Before the strcpy() call, the contents of b are
indeterminate; it may or may not contain a string.

Don't know about that, it seems a bit more. But different from sizeof
b anyway, which was the original confusion.

Sorry, my mistake. sizeof a == 29; sizeof b == 20.

Well if a and b are both arrays, then the reason why strcmp matches
them but sizeof is different is a little different!

(strcmp() only matches characters up to the nul terminator, ignoring
any trailing characters in the array that still contribute to the
sizeof property.)

Right. But again, the program as originally posted invokes undefined
behavior. One likely outcome is that, after the call to strcpy(), b
will contain the 20 characters "this is the beautifu" *without* a
trailing '\0'. (There might happen to be a trailing '\0' after the
end of b -- or the strcpy might have clobbered something vital before
anything is printed.)