How is conversion of float/double to int handled in printf?

Question

Consider this program

int main()
{
        float f = 11.22;
        double d = 44.55;
        int i,j;

        i = f;         //cast float to int
        j = d;         //cast double to int

        printf("i = %d, j = %d, f = %d, d = %d", i,j,f,d);
        //This prints the following:
        // i = 11, j = 44, f = -536870912, d = 1076261027

        return 0;
}

Can someone explain why the casting from double/float to int works correctly in the first case, and does not work when done in printf?
This program was compiled on gcc-4.1.2 on 32-bit linux machine.

---

EDIT: Zach's answer seems logical, i.e. use of format specifiers to figure out what to pop off the stack. However then consider this follow up question:

int main()
{

    char c = 'd';    // sizeof c is 1, however sizeof character literal
                     // 'd' is equal to sizeof(int) in ANSI C

    printf("lit = %c, lit = %d , c = %c, c = %d", 'd', 'd', c, c);
    //this prints: lit = d, lit = 100 , c = d, c = 100
    //how does printf here pop off the right number of bytes even when
    //the size represented by format specifiers doesn't actually match 
    //the size of the passed arguments(char(1 byte) & char_literal(4 bytes))    

 return 0;
}

How does this work?

Answer 1

There's no such thing as "casting to int in printf". printf does not do and cannot do any casting. Inconsistent format specifier leads to undefined behavior.

In practice printf simply receives the raw data and reinterprets it as the type implied by the format specifier. If you pass it a double value and specify an int format specifier (like %d), printf will take that double value and blindly reinterpret it an an int. The results will be completely unpredictable (which is why doing this formally causes undefined behavior in C).