Question

I have the following code, for an embedded platform where an int is 16 bits and a long int is 32 bits:

#define MULTIPLIER 0x1000

static void my_function(uint16_t i, void *p)
{
    uint32_t start = MULTIPLIER * i;
    ...
}

My compiler gives me the warning:

Warning 1 : lower precision in wider context: '*'

for this line.

What does this really mean? I can make the warning go away by changing the #define to

#define MULTIPLER 0x1000ul

(explicitly making it an unsigned long) but I would like to understand the warning.

Answer 1

It’s warning you that the multiplication will take place using 16-bit values, and the 16-bit result will then be converted to a 32-bit result. This might not be what you expect (the 16-bit multiplication might overflow), hence the warning.

This forum posting covers the issue