C: Data Types & Casting with printf()

Question

I'm fairly new to programming, and am starting to get used to C. I apologize if this is a repeat question (I don't know the underlying process i.e. what to search for).

I'm working with a simple program to get used to the nuances of data types:

main(){
    int i;
    float a,b;


    i = 2;
    a = 2.0;
    b = 4.0;

    printf("%d %1.1f", i/b,a/b);
}

I expected the program to print 0 0.5 (since a and b are both floats and I am printing their ratio as a float), yet the program printed 0 0.0 (I'm using gcc -o). However, when I reverse the printf order (without switching the order of corresponding variables), that is:

printf("%1.1f %d", i/b,a/b);

The print result is 0.5 0. I'm not exactly sure what's happening here. It appears that in the first program b is being converted to int in i/b and fails to be converted to float in a/b. However, in the second variant, b doesn't have any trouble printing out as two different types. Can ints not be coerced into floats? Can someone explain this or point me in the right direction?

Thanks!

Answer 1

You are invoking undefined behaviour.

printf is a vararg function, there is no link between the format string and how the other arguments are passed to the function. The usual rules for / apply so i/b stays a double and is passed as it is to printf, this causes undefined behaviour when the function tries to read it as an int.

If you compile with -Wall you will see these warnings

a.c: In function ‘main’:
a.c:12:12: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘double’ [-Wformat=]
     printf("%d %1.1f", i/b,a/b);
             ^

Answer 2

I get a different result.

#include <stdio.h>

int main(){
    int i;
    float a,b;


    i = 2;
    a = 2.0;
    b = 4.0;

    printf("%d %1.1f", i/b,a/b);
}

1432933728 0.5

gcc -Wall provides a warning. You should always be running with -Wall to get helpful warnings.

$ gcc -Wall try.c 
try.c:12:24: warning: format specifies type 'int' but the argument has type 'float' [-Wformat]
    printf("%d %1.1f", i/b,a/b);
            ~~         ^~~
            %f
1 warning generated.

printf is doing something bizarre when it interprets the float as an integer. C is not intelligent about data and will simply apply the raw binary floating point number as an integer. It will usually get garbage.

When I fix that by changing it to printf("%1.1f %1.1f", i/b,a/b); I get the expect result of 0.5 0.5.

This is Apple's "gcc", which is actually clang.

$ gcc --version
Configured with: --prefix=/Applications/Xcode.app/Contents/Developer/usr --with-gxx-include-dir=/usr/include/c++/4.2.1
Apple LLVM version 6.1.0 (clang-602.0.53) (based on LLVM 3.6.0svn)
Target: x86_64-apple-darwin14.4.0
Thread model: posix

While I got a different result, the lesson is you can't cast a float to a decimal with %d. Generally, function argument types will not be cast for you, even printf. You have to do it with explicit type casting. You could have written this...

printf("%1.1f %d", i/b, (int)(a/b));

Note the extra parens are necessary because type casting has a higher precedence than division.

Rather than relying on type casting, you get better control by explicitly rounding a floating point result using ceil(), floor(), round() and related functions.