Question

I’m getting MISRA type errors when I use “%f” specifier for snprintf with a parameter of type float.

According to my research, MISRA is correct because “%f” expectes a type of double.

Is there a floating point specifier or modifier that will use a float type parameter and not a double?

I’m working on an embedded system and don’t want to convert from 32-bit float to 64-bit double just to please the snprintf function. The code prints to the debug / console port and this is the only place where the conversion takes place.

For those of you needing a code example:

// This section is for those C++ purists and it fulfills the C++ tag.
#if __cplusplus
#include <cstdio>
#else
#include <stdio.h>
#endif

#define BUFFER_SIZE (128U)

int main(void)
{
    char buffer[BUFFER_SIZE];
    float my_float = 1.234F;

    // The compiler will promote the single precision "my_float"
    //   to double precision before passing to snprintf.
    (void)snprintf(buffer, BUFFER_SIZE, "%10.4f", my_float);
    puts(buffer);
    return 0;
}

All my research on SO and the Web is about printing a floating point value, not about which specifiers will require a float parameter so that no promotion to double takes place.

I am using IAR Embedded Workbench compiler for ARM7TDMI processor.

Answer 1

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.