«

Hi

#include<stdio.h>
main()
{
char s[]="Taiwan University.";
float a;
scanf("%f",&a);
printf("%f",a);
printf("input string:");
gets(s);
puts("%s",s);
}

After compiled by Visual Studio, the above program can input a.
But after it shows a, then I cannot enter s string.
Why?

Mike

On Mar 12, 2:27 pm, Mike <Sulfate...@gmail.com> wrote:
> Hi
>
> #include<stdio.h>
> main()
> {
> char s[]="Taiwan University.";
> float a;
> scanf("%f",&a);
> printf("%f",a);
> printf("input string:");
> gets(s);
> puts("%s",s);
>
> }
>
> After compiled by Visual Studio, the above program can input a.
> But after it shows a, then I cannot enter s string.
> Why?
>
> Mike

The routine scanf will stop when met Enter key, then gets() get the
Enter key.

Mike <> writes:

> After compiled by Visual Studio, the above program can input a.
> But after it shows a, then I cannot enter s string.

Maybe yes, maybe no. The way you've written your code, you might never
be able to tell.

> #include<stdio.h>
> main()

int main(void)

would be significantly better.

> {
> char s[]="Taiwan University.";

Why fill s[] with a value just to overwrite it?

> float a;
> scanf("%f",&a);
> printf("%f",a);

If scanf() encounters a conversion error (say you enter "a" instead of
a float value), a will still be uninitialized when you print it, in
which case anything could happen.

> printf("input string:");

If stdout is line-bufered, you won't see this prompt before it starts
expecting input.

> gets(s);
> puts("%s",s);

Since gets() strips newlines, puts() won't print one, so you may not
see this output either.

Aside from that, gets() is unsafe.

Aside from reading further into your C book, you might take a peek at
the comp.lang.c FAQ http://c-faq.com/ for better understanding of
what's wrong with your code.

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...
http://micah.cowan.name/

Mike <> writes:
> #include<stdio.h>
> main()
> {
> char s[]="Taiwan University.";
> float a;
> scanf("%f",&a);
> printf("%f",a);
> printf("input string:");
> gets(s);
> puts("%s",s);
> }
>
> After compiled by Visual Studio, the above program can input a.
> But after it shows a, then I cannot enter s string.
> Why?

The comp.lang.c FAQ is at <http://www.c-faq.com/>. You've just asked
question 12.18a.

Some more points:

"main()" is more or less ok in some circumstances, but
"int main(void)" is better.

You should probably print a new-line after each item that you print.
For example,
printf("%f\n", a);
Otherwise your output will be illegibly joined together on a single
line.

Never use the gets() function. See question 12.23 in the
aforementioned FAQ.

The puts() function takes one argument, not two. Since you have the
required "#include <stdio.h>" (good for you), the compiler should have
caught this error. In other words, it's very likely that the code you
posted is not the same as your actual code. Always copy-and-paste the
*exact* code you're having trouble with; don't paraphrase it, and
don't try to re-type it. Recommended reading:
<http://www.catb.org/~esr/faqs/smart-questions.html>.

It's a good idea to add a "return 0;" at the end of the main function.

--
Keith Thompson (The_Other_Keith) <kst->
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On Mar 11, 11:27*pm, Mike <Sulfate...@gmail.com> wrote:
> Hi
>
> #include<stdio.h>
> main()
> {
> * char s[]="Taiwan University.";
> * float a;
> * scanf("%f",&a);
> * printf("%f",a);
> * printf("input string:");
> * gets(s);
> * puts("%s",s);
>
> }
>
> After compiled by Visual Studio, the above program can input a.
> But after it shows a, then I cannot enter s string.
> Why?

#include <stdio.h>
#include <string.h>

char *getsafe(char *buffer, int count)
{
char *result = buffer,
*np;
if ((buffer == NULL) || (count < 1))
result = NULL;
else if (count == 1)
*result = '\0';
else if ((result = fgets(buffer, count, stdin)) != NULL)
if (np = strchr(buffer, '\n'))
*np = '\0';
return result;
}

int main(void)
{
char s[256] = "Taiwan University.";
int converted;
float a;
puts("Enter a floating point number:");
if (getsafe(s, sizeof s)) {
converted = sscanf(s, "%f", &a);
if (converted == 1)
printf("%f\n", a);
else
puts("Failed to collect a float from standard input.");
puts("input string:");
if (getsafe(s, sizeof s))
puts(s);
else
puts("Failed to collect a string from standard input.");
} else {
puts("Failed to collect a floating point input from standard
input.");
}
return 0;
}

Mike wrote:
> Hi
>
> #include<stdio.h>
> main()
> {
> char s[]="Taiwan University.";
> float a;
> scanf("%f",&a);
> printf("%f",a);
> printf("input string:");
> gets(s);
> puts("%s",s);
> }
>
> After compiled by Visual Studio, the above program can input a.
> But after it shows a, then I cannot enter s string.
> Why?
>
> Mike

Just to add to the other posts, your string literal "Taiwan University."
is of type const char *. You should not assign anything to it, because
it might be stored in a place in memory to which your program cannot
write, at which point it will crash (this idea can be summarized easily:
never cast away the constness of string literals). Have a look at
malloc() and free() instead, or just declare an array of char, eg. char
s[81].

If you're feeling a bit overwhelmed with all these replies, being a
newbie, note that strings are much more complicated than other variables
in C.

--
--Falcon Kirtaran

Falcon Kirtaran <> writes:

> Mike wrote:
>>
>> #include<stdio.h>
>> main()
>> {
>> char s[]="Taiwan University.";
<snip>
>> }
<snip>
> Just to add to the other posts, your string literal "Taiwan
> University." is of type const char *.

Not in C. The elements have type char and are read-only but the
standard stops just short of saying that they have type "const char".

> You should not assign anything
> to it,

It is just as well, then, that the OP does not do that! 'char s[] =
"...";' is not the same as 'char *s = "...";' -- I am sure this is
covered in the FAQ.

To the OP: apart from using gets (never do that) it is perfectly OK
to change the contents of your array 's'. The string literal is used
only to initialise it, after that, you can change it as much as you
want.

Had you written 'char *s = "something";' then you can not change the
data pointed to by 's' -- the arrays created by string literals are
read-only and any attempt to change them is undefined (i.e. bad things
could happen).

> because it might be stored in a place in memory to which your
> program cannot write, at which point it will crash (this idea can be
> summarized easily: never cast away the constness of string
> literals).

I think a less confusing rule (for C) is that one should treat string
literals as if there were const.

--
Ben.

Falcon Kirtaran <> wrote:

> Just to add to the other posts, your string literal "Taiwan University."
> is of type const char *.

No, it's not. It's of type char * (not const), but the text is not
(reliably) writable. This is illogical, but it's for hysterical raisins.

Richard

Mike <> writes:
> #include<stdio.h>
> main()
> {
> char s[]="Taiwan University.";
[...]
> }
[...]

At least one person has remarked on the fact that the code set the
array s to the value "Taiwan University." and then overwrites it.

In fact, given the empty square brackets, the initializer does two
things: it determines the initial value of s and it determines its
size (19 bytes in this case). Copying the string "Taiwan University."
into s is a bit wasteful, but it's a fairly concise way to say that
you want s to be big enough to hold that particular string value.

An alternate way to do the same thing is:

char s[sizeof "Taiwan University." + 1];

but it's easy to forget the "+ 1" (needed to allow for the terminating
'\0').

I'm resisting the temptation to propose adding a new feature to the
language so you can write:
char s[] != "Taiwan University."; /* } */
to set the size without initializing the array.

(Of course, 19 bytes isn't big enough to make the program immune to
buffer overruns, but we've covered that.)

--
Keith Thompson (The_Other_Keith) <kst->
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

Keith Thompson wrote:

> Mike <> writes:

>> char s[]="Taiwan University.";

> it's a fairly concise way to say that
> you want s to be big enough to hold that particular string value.
>
> An alternate way to do the same thing is:
>
> char s[sizeof "Taiwan University." + 1];
>
> but it's easy to forget the "+ 1" (needed to allow for the
> terminating '\0').

Oops, it seems as easy to confuse sizeof with strlen, too.

Ralf