What is the correct format specifier for double
in printf? Is it %f
or is it %lf
? I believe it's %f
, but I am not sure.
#include <stdio.h> int main() { double d = 1.4; printf("%lf", d); // Is this wrong? }
We can print the double value using both %f and %lf format specifier because printf treats both float and double are same. So, we can use both %f and %lf to print a double value.
%d stands for decimal and it expects an argument of type int (or some smaller signed integer type that then gets promoted). Floating-point types float and double both get passed the same way (promoted to double ) and both of them use %f .
"%f"
is the (or at least one) correct format for a double. There is no format for a float
, because if you attempt to pass a float
to printf
, it'll be promoted to double
before printf
receives it1. "%lf"
is also acceptable under the current standard -- the l
is specified as having no effect if followed by the f
conversion specifier (among others).
Note that this is one place that printf
format strings differ substantially from scanf
(and fscanf
, etc.) format strings. For output, you're passing a value, which will be promoted from float
to double
when passed as a variadic parameter. For input you're passing a pointer, which is not promoted, so you have to tell scanf
whether you want to read a float
or a double
, so for scanf
, %f
means you want to read a float
and %lf
means you want to read a double
(and, for what it's worth, for a long double
, you use %Lf
for either printf
or scanf
).
1. C99, §6.5.2.2/6: "If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions." In C++ the wording is somewhat different (e.g., it doesn't use the word "prototype") but the effect is the same: all the variadic parameters undergo default promotions before they're received by the function.
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