Question

For the following code:

#include <stdio.h>
int main(){
    printf("%f", 5);
    printf("%d", 5.01);
}

The first statement will print 0.000000 and the second statement will print a large number.

I thought the printf see the format %f will pop a 4-byte argument from stack. Then I looked up some references that say the printf function will transform float into double, so 8-byte argument. So I thought it may print out a unpredictable value. But how could it print out a 0.000000.

The second one is also weired. The binary format for 5.01 should be 0 10000001 01000000101000111101100 (0x40A051EC), It should be 1084248556 in decimal format, but the result of the statement is 1889785610. Why is this happen?

Answer 1

This is confusing, but %d is the format for an integer, like %i (most people expect it to be for double since %f is for float, but it’s not). You can’t use it with a double, you must use %f. Same with %f, you can’t use an integer with it; you must use a float or double.

Also, printf doesn’t check the types you use it with, it just blindly interprets bytes as the types you tell it you gave it. Using the wrong types with the wrong type specifiers yields undefined behaviour which is why you’re seeing what you are.