Question of printf, printf

Question of printf, printf

I used to implement printf when I learned the book "write an operating system by myself" by yuan, but it is a relatively simple version. I recently read the programmer interview book to answer this question:
# Include <stdio. h>
Int main {
Printf ("% f", 5 );
Printf ("% d", 5.01 );
}

What is the output?
The question is not difficult but very detailed: The first output is 0.000000, because the printf function encounters % f and considers the parameter as a double type (float in the printf function is automatically converted to a double type ), 5 is an int type with only four digits, so the output takes 8 digits out of bounds and outputs 0.000000. The second output takes only four digits and outputs a large number.

So I thought: What Will printf ("% f % d", 5, 5.01 );

In fact, both outputs are large numbers, but there are many zeros behind the first output floating point number, and the second output is the same as the second output.

Reasons for personal opinion analysis:

First, the stack grows down, while the printf function (in fact, all functions) parameters go to the stack from the right to the left. Then 5.01 first goes to the stack, and 5 re-enters, that is, 5.01 is at the high address, 5 at the low address, when % f is output, take all the bits of 5 and the four low byte of 5.01, output a large number, output % d to take the high byte, also output a large number. Http://hovertree.com/menu/c/
To intuitively see this analysis:
We use the following statement for testing:

Printf ("% f % d", 5,120.5 );

120.5 of the storage formats are as follows:

0 100 0000 0101 1110 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
4-byte values are all 0, and 4-byte values are large numbers,
Therefore, the output should be 0.000000 and a large number.
The actual output is as follows:

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