Why does printf implicit float to int conversion not work?

Question

Please help me in understanding the following C Output:

#include
int main() {
    float x = 4.0;
    printf(\"%f\\n\",x);
    printf(\"%d\\n\",x);         


        

Answer 1

floats are automatically promoted to doubles when passed as ... parameters (similarly to how chars and short ints are promoted to ints). When printf() looks at a format specifier (%d or %f or whatever), it grabs and interprets the raw data it has received according to the format specifier (as int or double or whatever) and then prints it.

printf("%d\n",x) is wrong because you are passing a double to printf() but lie to it that it's going to be an int. printf() makes you pay for the lie. It takes 4 bytes (most likely, but not necessarily 4 bytes) from its parameters, which is the size of an int, it grabs those bytes from where you have previously put 8 bytes (again, most likely, but not necessarily 8 bytes) of the double 4.0, of which those 4 bytes happen to be all zeroes and then it interprets them as an integer and prints it. Indeed, powers of 2 in the IEEE-754 double precision format normally have 52 zero bits, or more than 6 8-bit bytes that are zero.

Those "most likely, but not necessarily" words mean that the C standard does not mandate a fixed size and range for types and they may vary from compiler to compiler, from OS to OS. These days 4-byte ints and 8-byte doubles are the most common (if we consider, e.g. the x86 platform).