Why Does Printf() Promote a Float to a Double

Why does printf() promote a float to a double?

Yes, float arguments to variadic function are promoted to double.

The draft C99 standard section 6.5.2.2 Function calls says:

[...]and arguments that
have type float are promoted to double. These are called the default argument
promotions.[...]

from the draft C++ standard section 5.2.2 Function call:

[...]a floating point type that is subject to the floating point
promotion (4.6), the value of the argument is converted to the
promoted type before the call. [...]

and section 4.6:

A prvalue of type float can be converted to a prvalue of type double. The value is unchanged

cppreference covers the default conversions for variadic function in C++ well:

• std::nullptr_t is converted to void*
• float arguments are converted to double as in floating-point promotion
• bool, char, short, and unscoped enumerations are converted to int or wider integer types as in integer promotion

We can see in C and presumably in C++ this conversion was kept around for compatibility with K&R C, from Rationale for International Standard—Programming Languages—C (emphasis mine):

For compatibility with past practice, all argument promotions occur as
described in K&R in the absence of a prototype declaration, including
the not always desirable promotion of float to double.

Why when using printf float and double show the same number of digits?

sizeof returns size_t. To print size_t you need %zu instead of %d

If you want to see the real difference between float and double you need to print more digits using %.NUMBERf

Like:

#include <stdio.h>

int main(void)
{
    float x=1.2222222222222222f;
    printf("%.70f %zu\n", x,sizeof(x)); 
    double y=1.2222222222222222;  
    printf("%.70f %zu\n", y,sizeof(y)); 
    return 0;
}

Output:

1.2222222089767456054687500000000000000000000000000000000000000000000000 4
1.2222222222222220988641083749826066195964813232421875000000000000000000 8

Is there a printf specifier that requires float not double?

No, because printf and its friends are variadic functions, so a float parameter undergoes automatic conversion to double as part of the default argument promotions (see section 6.5.2.2 of the C99 standard).

I'm not sure why this triggers a MISRA warning though, I can't think of any way in which this might be dangerous.

How does printf and co differentiate between float and double

It doesn't differentiate. It's not possible to receive a float as a vararg: any float argument that you provide is first promoted to double.

6.5.2.2/6 defines "default argument promotions", and /7 states that default argument promotions are applied to "trailing arguments", that is varargs denoted by ....

how does it work for scanfs/sscanf?

The %f format for scanf requires a pointer to float. %lf requires a pointer to double, %Lf requires a pointer to long double.

copying the value to a temp and casting(is this right?)

If you provide a float argument, then the implementation creates a temporary of type double, initializes it with the float value, and passes this as the vararg. Casting by definition is explicit conversion by use of the cast operator -- you can cast if you like in order to make it exactly clear to the reader what's going on, but float f = 3; printf("%f", f); is exactly the same as float f = 3; printf("%f", (double)f);. The default argument promotion has the same meaning as the cast.

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

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.

Correct format specifier for double in printf

"%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).

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_{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.}

Does float always auto-convert to double when multiplying mixed data types?

It must refer to the result of binary arithmetic operations of float * float format. In the pre-standard versions of C operands of such expressions were promoted to double and the result had double type.

For example, here's a quote from "C Reference Manual"

If both operands are int or char, the result is int. If both are
float or double, the result is double.

In C89/90 already this behavior was changed and float * float expressions produce float result.

• If either operand has type long double, the other operand is converted to long double
• Otherwise, if either operand is double, the other operand is converted to double.
• Otherwise, if either operand is float, the other operand is converted to float.

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