cprintfscanfformat-string

What is the difference between %f and %lf in C?

I have seen these two parameters in a C example in a C book, but the author didn't elaborate what the difference between the two is. I know that %f specifies that a float should take its place. I have tried looking this up but have had a hard time finding an explanation. What about %lf?

Solution

  • The short answer is that it has no impact on printf, and denotes use of float or double in scanf.

    For printf, arguments of type float are promoted to double so both %f and %lf are used for double. For scanf, you should use %f for float and %lf for double.

    More detail for the language lawyers among us below:

    There is no difference between %f and %lf in the printf family. The ISO C standard (all references within are from C11), section 7.21.6.1 The fprintf function, paragraph /7 states, for the l modifier (my emphasis):

    Specifies that a following d, i, o, u, x, or X conversion specifier applies to a long int or unsigned long int argument; that a following n conversion specifier applies to a pointer to a long int argument; that a following c conversion specifier applies to a wint_t argument; that a following s conversion specifier applies to a pointer to a wchar_t argument; or has no effect on a following a, A, e, E, f, F, g, or G conversion specifier.

    The reason it doesn't need to modify the f specifier is because that specifier already denotes a double, from paragraph /8 of that same section where it lists the type for the %f specifier:

    A double argument representing a floating-point number is converted to decimal notation

    That has to do with the fact that arguments following the ellipse in the function prototype are subject to default argument promotions as per section 6.5.2.2 Function calls, paragraph /7:

    The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.

    Since printf (and the entire family of printf-like functions) is declared as int printf(const char * restrict format, ...); with the ellipsis notation, that rule applies here. The default argument promotions are covered in section 6.5.2.2 Function calls, paragraph /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.

    For the scanf family, it mandates the use of a double rather than a float. Section 7.21.6.2 The fscanf function /11:

    Specifies that a following d, i, o, u, x, X, or n conversion specifier applies to an argument with type pointer to long int or unsigned long int; that a following a, A, e, E, f, F, g, or G conversion specifier applies to an argument with type pointer to double; or that a following c, s, or [ conversion specifier applies to an argument with type pointer to wchar_t.

    This modifies the /12 paragraph of that section that states, for %f:

    Matches an optionally signed floating-point number, infinity, or NaN, whose format is the same as expected for the subject sequence of the strtod function. The corresponding argument shall be a pointer to floating.

    The reason why the default argument promotions don't come into play for scanf is because you're not actually passing a float that would be converted (or you shouldn't be). Instead, you're meant to pass a pointer to a float which is not affected by those promotions.

    The printf family takes values because it just needs to use them. The scanf family needs pointers to the values because its intent is to change those values, hence needs to know where to write the new values (C is strictly pass-by-value: pass-by-reference is emulated by using pointers and dereferencing).