if and switch

[Bill Cunningham]

If won't work in evaluating integers will it? That's my interpretation
of what I've been reading. You would have to use switch in those interger
cases. I had some code.

if (ans=='y') printf("ok"); exit(0); /* ans is a char */

But what I'm asking for here is the ascii value for y. An integer. So my
compiler complained. Now if I did this,

if (ans=="y") printf("ok"); exit(0);

Then it would work right? Otherwise,

switch (ans=='y')
case1: printf ("ok"); exit (0);
break;
case 2: if (ans!='y') printf("error"); exit(1);
break;

Have I proofreaded myself pretty good on this one? I don't really know
switch well though. I should go with the if (ans=="y") since a string is
what I want.

Bill

[Default User]

Bill Cunningham wrote:

If won't work in evaluating integers will it?

You haven't finised the previous problem. One of things keeping you
from learning is that you are hopping from question to question with
ever completing anything.

Go back and answer those questions I gave you previously in that
thread. Keep working on your chunk() routine until it is complete and
correct.

I'm not going to give you any further input until you do that.

Brian

[santosh]

Bill Cunningham wrote:

If won't work in evaluating integers will it?

It can evaluate any legal expression that produces a non-void value.

That's my interpretation of what I've been reading. You would have to
use switch in those interger cases.

No.

I had some code.

if (ans=='y') printf("ok"); exit(0); /* ans is a char */

This is probably not what you meant. What you perhaps meant to write
was:

if (ans == 'y') {
printf("ok\n");
exit(0);
}

Notice the compound statement? And the newline after "ok"? You probably
also want the test condition to be:

if (ans == 'y' || ans == 'Y') { ... }

But what I'm asking for here is the ascii value for y.

C supports other character encoding than ASCII. Using hardcoded ASCII
values in your program ensures unnecessary loss of portability. Just
use character constants.

An integer.

A character constant is an integer.

So my compiler complained. Now if I did this,

if (ans=="y") printf("ok"); exit(0);

Then it would work right?

This broken. You cannot directly compare strings in C. You need to use
something like strcmp() or strncmp().

Otherwise,

switch (ans=='y')
case1: printf ("ok"); exit (0);
break;
case 2: if (ans!='y') printf("error"); exit(1);
break;

Have I proofreaded myself pretty good on this one?

No.

I don't really know
switch well though. I should go with the if (ans=="y") since a string
is what I want.

switch is overkill for this purpose.

[Bill Cunningham]

Thanks Santosh

Bill

[Army1987]

santosh wrote:

Bill Cunningham wrote:

> If won't work in evaluating integers will it?

It can evaluate any legal expression that produces a non-void value.

struct foo { int bar; };
struct foo baz(void) { struct foo ret = { 42 }; return ret; }
/* ... */
if (baz()) {
/* What about that? */
}

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Army1987 (Replace "NOSPAM" with "email")