Unexpected double value output when using printf

Question

#include <stdio.h>
int main()
{
    double  a=7.1;
    printf("%d",a);
    return 0;
}

Output:

1717986918

When I am writing this program and using signed format specifiers to the method printf() . Why I am getting this value?

Answer 1

Its a format specifier problem.
Using "%d" , the compiler is being told that the type of data being printed is of type int , but the variable is of type float , causing undefined behavior , resulting in this case, the value 1717986918 .

The format specifier for type float is "%f" , not "%d" .

Change this:

printf("%d",a);

To this:

printf("%f",a);

If you compile with -Wall or equivalent in your compiler settings, a warning for this will notify you.

Answer 2

Your code has undefined behavior. It's not defined what should be the output of your code and as far as any programmer should be concerned, your code spawns nasal demons .

why I am getting this output in my Cprogram?

I am assuming lot about your computer and compiler implementation, such as that your compiler uses IEEE754 standard to represent floating point numbers and the size of an int is 4 bytes and that your computer uses little endian. According to this site the double number 7.1 when represented as a IEEE754 double precision number is stored as 8 bytes as a 0x401C666666666666 number. Then those bytes are passed to printf as variadic arguments. Because of how on your architecture arguments are passed, the least significant 4 bytes from that number are extracted for printf %d format specifier, which results in printing the base 10 representation of 0x66666666 number, which is 1717986918 .