using scanf in C/C++

问题:

To read an int using scanf we use:

scanf("%d",&i);

What if i is long not int??

Note: when using %d with long it gives me an irritating warning..

Thanks!

回答1:

Just use

long l;

scanf("%ld", &l);

it gives me an irritating warning..

That warning is quite right. This is begging for stack corruption.

回答2:

For gods sake:

long n;
scanf( "%ld", & n );

回答3:

scanf("%ld", &i);

You can also use "%Ld" for a long long (and depending on your compiler, sometimes also "%lld").

Take a look at the Conversions section of the scanf man page for more. (Just Google it if your system doesn't have manpages).

回答4:

Each conversion specifier expects its corresponding argument to be of a specific type; if the argument's type does not match the expected type, then the behavior is undefined. If you want to read into a long with scanf(), you need to use the %ld conversion specifier:

long i;
scanf("%ld", &i);

Check the online draft C standard (.pdf file), section 7.19.6.2, paragraph 11 for a complete listing of size modifiers and expected types.

回答5:

Check this, here is the answer: "%I64d"