Why the code only works when I use the %lf or %f place holder (second placeholder), but its printing 0 when I use %d?

#include <stdio.h>

void main()
{

    long id;
    id = 123456789;
    double Hourly;
    Hourly = 30;
    int HoursAday, daysAweek, Fired, Hired;
    HoursAday = 8; daysAweek = 5; Fired = 2021; Hired = 2019;
    printf("bob, id: \"%d\" should get %lf", id, (Hourly * HoursAday * daysAweek) * (Fired - Hired));

CodePudding user response：

The type of (Hourly * HoursAday * daysAweek) * (Fired - Hired) is double so the correct specifier is %lf. Because of variadic arguments default conversions the code will also work with %f (*). But %d is undefined behavior.

Also, the correct specifier for id (which is of type long) is %ld. %d is undefined behavior.

See the documentation for printf: https://en.cppreference.com/w/c/io/fprintf

(*) A float variadic argument always gets converted to double. So the printf family of functions will never receive a float, they will always receive a double. That's why %f and %lf are for practical reasons equivalent.

CodePudding user response：

This expression

(Hourly * HoursAday * daysAweek) * (Fired - Hired)

has the type double due to the usual arithmetic conversions because the variable Hourly has the type double. That is if one operand has the type double and other operand has an integer type in a binary operation then the operand with the integer type is converted to the type double.

Using the wrong conversion specifier %d designed to output objects of the type int with an object of the type double invokes undefined behavior.

Pay attention to that in this conversion specifier %lf the length modifier l has no effect. So it is enough to use %f.