Why does scanf() need "%lf" for doubles, when printf() is okay with just "%f"?


Why is it that scanf() needs the l in "%lf" when reading a double, when printf() can use "%f" regardless of whether its argument is a double or a float?

Example code:

double d;
scanf("%lf", &d);
printf("%f", d);
10/7/2017 2:00:21 PM

Accepted Answer

Because C will promote floats to doubles for functions that take variable arguments. Pointers aren't promoted to anything, so you should be using %lf, %lg or %le (or %la in C99) to read in doubles.

5/22/2016 3:55:41 AM

Since ะก99 the matching between format specifiers and floating-point argument types in C is consistent between printf and scanf. It is

  • %f for float
  • %lf for double
  • %Lf for long double

It just so happens that when arguments of type float are passed as variadic parameters, such arguments are implicitly converted to type double. This is the reason why in printf format specifiers %f and %lf are equivalent and interchangeable. In printf you can "cross-use" %lf with float or %f with double.

But there's no reason to actually do it in practice. Don't use %f to printf arguments of type double. It is a widespread habit born back in C89/90 times, but it is a bad habit. Use %lf in printf for double and keep %f reserved for float arguments.


scanf needs to know the size of the data being pointed at by &d to fill it properly, whereas variadic functions promote floats to doubles (not entirely sure why), so printf is always getting a double.


Because otherwise scanf will think you are passing a pointer to a float which is a smaller size than a double, and it will return an incorrect value.


Using either a float or a double value in a C expression will result in a value that is a double anyway, so printf can't tell the difference. Whereas a pointer to a double has to be explicitly signalled to scanf as distinct from a pointer to float, because what the pointer points to is what matters.


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