scanf "%n" format specifier is not supported

Sun May 31 17:25:00 GMT 1998

Hi Alexander,
Alexander Chernov wrote:
> ...For example, the following piece of code
> left variable n value as 0 instead of 2. ...
>> #include <stdio.h>
> #include <string.h>
>> int
> main()
> {
> int v = 0, n = 0, r = 0;
>> r = sscanf("32", "%d %n", &v, &n);
> printf("v = %d\nn = %d\nr = %d\n", v, n, d);
> return 0;
> }

The problem is that the scanner gives up as it doesn't find the space
that your format string specifies. The %n is not processed, and so n is
left unchanged. Try this:
#include <stdio.h>
#include <string.h>
int
main()
{
 int v = 0, n = 0, r = 0;
 r = sscanf("32", "%d %n", &v, &n);
 printf("v = %d\nn = %d\nr = %d\n", v, n, r);
 r = sscanf("32", "%d%n", &v, &n);
 printf("v = %d\nn = %d\nr = %d\n", v, n, r);
 return 0;
}
Should output
v = 32
n = 0
r = 1
v = 32
n = 2
r = 1
so long, benny
======================================
Benjamin Riefenstahl (benny@crocodial.de)
Crocodial Communications EntwicklungsGmbH
RuhrstraÃƒÂŸe 61, D-22761 Hamburg, Germany
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