Little program

[mdh]

Can someone look at this little program and tell me why the last line
is not compiling.

#include <stdio.h>

int main (){
int i, j, k;
i=0;
j=9;
k=-8;
printf("Given i=0, j=9 and k= -8 \n");
printf( "Then for \" if ( i) , if (j) and if (k)\"\n");
printf("the Answers are %d %d %d ", if(i), if(j), if(k) ); /**parse
error before 'if'**/
return 0;
}

It's probably something really silly, or I am missing something, which
is more likely!! but all in the quest to learn C!!!!
Thanks in advance.

[jmcgill]

mdh wrote:

printf( "Then for \" if ( i) , if (j) and if (k)\"\n");

You cannot have a function named "if"

[mdh]

jmcgill wrote:

You cannot have a function named "if"

I thought you could (page 56, K&R II);

[Bart]

mdh wrote:

jmcgill wrote:

You cannot have a function named "if"

I thought you could (page 56, K&R II);

You are thoroughly confused about the use of the if-statement. Re read
that page and the previous one very carefully.

Regards,
Bart.

[jmcgill]

Bart wrote:

mdh wrote:

>jmcgill wrote:

>>You cannot have a function named "if"

I thought you could (page 56, K&R II);

You are thoroughly confused about the use of the if-statement. Re read
that page and the previous one very carefully.

I have a feeling he needs to back up a bit more than that. K&R2 is a
great book, but might not be the best tutorial for someone working from
scratch, on his own.

[mdh]

Bart wrote:

>
You are thoroughly confused about the use of the if-statement.

I could give a smart answer, but I will refrain. The reason some of us
ask, is that we wish to be enlightened. One of the ways of
enlightenment is to write little things that are causing confusion. So,
to be told that I am confused is a pointless exercise.
If you look at page 56 it says:

"Since an if simply tests the numeric value of an expression, certain
coding shortcuts are possible. The most obvious is writing

if(expression)

instead of

if (expression != 0).

I was simply trying to see if the !=0 includes negative values, hence
the program, which may not work for the reasons you say, but that does
not mean one should not try and get a better understanding.

Anyway, thanks for your help.

[Bart]

mdh wrote:

Bart wrote:

You are thoroughly confused about the use of the if-statement.

I could give a smart answer, but I will refrain. The reason some of us
ask, is that we wish to be enlightened. One of the ways of
enlightenment is to write little things that are causing confusion. So,
to be told that I am confused is a pointless exercise.

I was about to explain it at length, but I just realized that I'm
repeating what's written there in K&R. Since you have the book I
thought that it would be pointless, so I just suggested that you
re-read more carefully.

If you look at page 56 it says:

"Since an if simply tests the numeric value of an expression, certain
coding shortcuts are possible. The most obvious is writing

if(expression)

instead of

if (expression != 0).

I was simply trying to see if the !=0 includes negative values, hence
the program, which may not work for the reasons you say, but that does
not mean one should not try and get a better understanding.

Yes, but it also says that the general syntax is:

if (expression)
statement;
else
statement;

You can't just use the 'if' as though it were an expression. You can
only use it in a complete statement as above.

Regards,
Bart.

[Bart]

jmcgill wrote:

Bart wrote:

mdh wrote:

jmcgill wrote:
You cannot have a function named "if"
I thought you could (page 56, K&R II);

You are thoroughly confused about the use of the if-statement. Re read
that page and the previous one very carefully.

I have a feeling he needs to back up a bit more than that. K&R2 is a
great book, but might not be the best tutorial for someone working from
scratch, on his own.

Don't assume too much. There are languages where the if statement is
can be used as an expression, and it's quite concievable that he's
confused precicely for this reason.

Regards,
Bart.

[jmcgill]

/* consider this */
#include <stdio.h>
int
main (int argc, char **argv)
{
int i, j, k;
if( argc == 4 ){
i = atoi(argv[1]);
j = atoi(argv[2]);
k = atoi(argv[3]);
printf ("Given i=%d, j=%d and k=%d \n", i, j, k);
printf ("Then for \" if(i) , if(j) and if(k)\"\n");
printf ("the Answers are %d %d %d\n",
(i ? 1 : 0), (j ? 1 : 0), (k ? 1 : 0) );
} else {
printf("usage: %s i j k\n", argv[0]);
}
return 0;
}

[jmcgill]

mdh wrote:

>>Bart wrote:

You can't just use the 'if' as though it were an expression. You can
only use it in a complete statement as above.

Yes....I am beginning to see that!! :-)

If you do want to use a conditional as an expression, the ternary
operator is your friend. See my previous post for an example.