printf behaviour in C

How using the 'f' changes/influences the output?

The f at the end of a floating point constant determines the type and can affect the value.

4.567 is floating point constant of type and precision of double. A double can represent exactly typical about 2⁶⁴ different values. 4.567 is not one on them^*1. The closest alternative typically is exactly

4.56700000000000017053025658242404460906982421875 // best 4.56699999999999928235183688229881227016448974609375 // next best double 

4.567f is floating point constant of type and precision of float. A float can represent exactly typical about 2³² different values. 4.567 is not one on them. The closest alternative typically is exactly

4.566999912261962890625 // best 4.56700038909912109375 // next best float 

When passed to printf() as part of the ... augments, a float is converted to double with the same value.

So the question becomes what is the expected difference in printing?

printf("%f",4.56700000000000017053025658242404460906982421875); printf("%f",4.566999912261962890625); 

Since the default number of digits after the decimal point to print for "%f" is 6, the output for both rounds to:

4.567000 

To see a difference, print with more precision or try 4.567e10, 4.567e10f.

45670000000.000000 // double 45669998592.000000 // float 

Your output may slightly differ to to quality of implementation issues.

^*1 C supports many floating point encodings. A common one is binary64. Thus typical floating-point values are encoded as an sign * binary fraction * 2^exponent. Even simple decimal values like 0.1 can not be represented exactly as such.