Did you try long long or unsigned long long? You could always just read a string and check each digit to see if it's a number until you run out of string.Code:char buf[ BUFSIZ ]; scanf( "%[^0-9]\n", buf ); printf( "You entered: %s\n", buf );
Quzah.
Last edited by quzah; 07-17-2011 at 05:25 AM.
Hope is the first step on the road to disappointment.
Unless you are using some brain dead compiler from then 1980s...
Will accept positive numbers up to 2^64 -1 or .... 18,446,744,073,709,551,615Code:#include <stdint.h> #include <stdio.h> int main (void) { uint64_t x; // also try.... unsigned long long int x; printf("Enter your number : "); scanf("%lld",&x); printf("%lld",x); return 0; }
Bear in mind that if you're using visual studio, or one of the many MinGW based ports (which rely on the Microsoft runtime), that you have to use a non-standard format conversion for reading in long long's
Format Specification Fields: scanf and wscanf Functions (CRT)
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
If you're going to post example code, it should be code that cannot segfault. Here a buffer overrun is possible if a user enters a really long string of numbers.Originally Posted by quzah
There's also a bug: ^ shouldn't be there.
here's a better version, although I'd suggest using CommonTater's code if possible:
Code:char buf[ 1024]={'\0'}; scanf( "%1023[0-9]\n", buf ); printf( "You entered: %s\n", buf );
It is too clear and so it is hard to see.
A dunce once searched for fire with a lighted lantern.
Had he known what fire was,
He could have cooked his rice much sooner.
There you go. Nice catch. As to overflowing BUFSIZ, good luck with that. I'd be surprised if your shell allowed you to type that much. Oh, sure, I suppose you could pipe a file in, but you have to go out of your way to break it.
Quzah.
Hope is the first step on the road to disappointment.
