double d;
scanf("%f", &d);
printf("%f", d);
result:
input: 10.3
output: 0.00000
Why? i think output should be 10.3 visual studio 2008.
views:
80answers:
2
+5
A:
For scanf(), %f is for a float. For double, you need %lf. So,
#include <stdio.h>
main() {
double d;
scanf("%lf", &d);
printf("%f\n", d);
}
with input 10.3 produces 10.300000.
Ramashalanka
2010-03-07 10:05:33
+1 for %4.1lf format string Sergey could view this: http://www.cplusplus.com/reference/clibrary/cstdio/printf/
stacker
2010-03-07 10:13:06
`%lf` is needed for `scanf()`, but for `printf()`, `%f` means `double` (and works with `float` too, because `float` is promoted to `double` in the variable portion of the argument list). `%lf` is meaningless to `printf()`.
caf
2010-03-07 10:20:46
printf is a vararg function, so argument promotion need not apply - compiler does not know argument types beyond format string. Said that, %lf is needed.
el.pescado
2010-03-07 14:18:27
el.pescado: There are a particular set of argument promotions that are always applied to arguments in the variable portion of the argument list (they are the same as those applied to the arguments of a function declared without a prototype).
caf
2010-03-08 06:33:24