isdigit

[mdh]

Could I get some help as to understanding why I am not getting a result
I expect.
#include <stdio.h>
#include <ctype.h>

int main (){
int i,j,k,c;
i=9;
c='A';
j=isdigit(9);
k=isdigit('A');
printf("\nThe value of \"isdigit(%d )\" is %d", i,j);
printf("\nAnd the value of \"isdigit('%c') \" is %d\n", c,k);

return 0;
}

Output:

The value of "isdigit(9) " is 0
And the value of "isdigit('A ')" is 0

Expected:

The value of "isdigit(9) " is 0
And the value of "isdigit('A ')" is 1

[Neil]

mdh wrote:

Could I get some help as to understanding why I am not getting a result
I expect.
#include <stdio.h>
#include <ctype.h>

int main (){
int i,j,k,c;
i=9;
c='A';
j=isdigit(9);
k=isdigit('A');
printf("\nThe value of \"isdigit(%d )\" is %d", i,j);
printf("\nAnd the value of \"isdigit('%c') \" is %d\n", c,k);

return 0;
}

Output:

The value of "isdigit(9) " is 0
And the value of "isdigit('A ')" is 0

Expected:

The value of "isdigit(9) " is 0
And the value of "isdigit('A ')" is 1

'A' is not a digit
9 is not a digit

'9' is a digit

[Cong Wang]

On Oct 6, 2:02 pm, "mdh" <m...@comcast.n etwrote:

Could I get some help as to understanding why I am not getting a result
I expect.

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

int main (){
int i,j,k,c;
i=9;
c='A';

j=isdigit(9);
k=isdigit('A');

printf("\nThe value of \"isdigit(%d )\" is %d", i,j);
printf("\nAnd the value of \"isdigit('%c') \" is %d\n", c,k);

return 0;

}Output:

The value of "isdigit(9) " is 0
And the value of "isdigit('A ')" is 0

Expected:

The value of "isdigit(9) " is 0
And the value of "isdigit('A ')" is 1

Do you consider 'A' as a hexadecimal digit? If so, use function
isxdigit() instead of isdigit().

[mdh]

'A' is not a digit
9 is not a digit

'9' is a digit

it's getting late...thank you

[Rnags]

Answer; Since c is a integer variable,when we try to store the
charecter in the integer variable Ascii value of that charecter is
stored in the variable,since Ascii value of the charecter is also digit
hence the isdigit()
function returns true

[shafi]

j=isdigit(9);

k=isdigit('A');

Hi,
The isdigit() function takes an argument of character type or EOF and
it return non zero value if
the passing argument is characters '0' , '1' . . . '9' else it returns
zero.
Now consider your actual out put.

1. The first part of the out put is true as you have supplied wrong
argument i.e. a integer
insite of supplying char.
2. Second part is also true. No need to explain.

Regards,
Shafi

"the difficult we do immediately, the impossible takes a little
longer", [US Army

[Keith Thompson]

"shafi" <sh*********@gm ail.comwrites:

>j=isdigit(9) ;
k=isdigit('A') ;

The isdigit() function takes an argument of character type or EOF and
it return non zero value if
the passing argument is characters '0' , '1' . . . '9' else it returns
zero.

More precisely,

... the argument is an int, the value of which shall be
representable as an unsigned char or shall equal the value of the
macro EOF. If the argument has any other value, the behavior is
undefined.

This is an important distinction. Something like this:

char c = some_value;
isdigit(c);

can invoke undefined behavior if (1) plain char is signed, and (2) the
value of c is negative (and not equal to EOF). For arguments other
than EOF, it's often a good idea to cast the argument:

char c = some_value;
isdigit((unsign ed char)c);

(This is one of the few cases where a cast is actually a good idea.)

Or, in this case, you could avoid the cast by changing the declaration
of c:

unsigned char c = some_value;
isdigit(c);

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Andrew Poelstra]

On Fri, 2006-10-13 at 22:22 -0700, shafi wrote:

j=isdigit(9);
k=isdigit('A');

Hi,
The isdigit() function takes an argument of character type or EOF and
it return non zero value if
the passing argument is characters '0' , '1' . . . '9' else it returns
zero.

isdigit(c) function takes an argument of type int that must be in the
range of unsigned char or EOF. It returns nonzero on '0'...'9',
otherwise returns 0.

Now consider your actual out put.

1. The first part of the out put is true as you have supplied wrong
argument i.e. a integer
insite of supplying char.

Wrong. 9 is in the range of unsigned char, and therefore is perfectly
valid. Whether the expression evaluates to true or not depends on the
charset.

2. Second part is also true. No need to explain.

Wrong. 'A' is not a digit.

--
Andrew Poelstra <http://www.wpsoftware. net/projects/>

[pete]

Andrew Poelstra wrote:

>
On Fri, 2006-10-13 at 22:22 -0700, shafi wrote:

j=isdigit(9);
k=isdigit('A');

Hi,
The isdigit() function takes an argument
of character type or EOF and
it return non zero value if
the passing argument is characters
'0' , '1' . . . '9' else it returns zero.

isdigit(c) function takes an argument of type int that must be in the
range of unsigned char or EOF. It returns nonzero on '0'...'9',
otherwise returns 0.

Now consider your actual out put.

1. The first part of the out put is true as you have supplied wrong
argument i.e. a integer
insite of supplying char.

Wrong. 9 is in the range of unsigned char, and therefore is perfectly
valid. Whether the expression evaluates to true or not depends on the
charset.

It doesn't depend on the char set.
isdigit is one of the ctype functions
that does not have locale specific behavior.

--
pete

[Keith Thompson]

pete <pf*****@mindsp ring.comwrites:

Andrew Poelstra wrote:

[...]

>Wrong. 9 is in the range of unsigned char, and therefore is perfectly
valid. Whether the expression evaluates to true or not depends on the
charset.

It doesn't depend on the char set.
isdigit is one of the ctype functions
that does not have locale specific behavior.

The value of isdigit(9) depends on the character encoding, which can
vary from one implementation implementation to another. (I don't know
whether "char set" is the right term for this.)

I believe it returns 0 on all existing implementations , but 9 could
theoretically be the encoding for some digit (but it can't be the
encoding for '9', since the digits are contiguous and '\0' can't be a
digit).

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.