How to evaluate a C expression with literal suffixes U L F?

Question

I need help understanding how the expression below should be interepted, because I have no clue. help please.

Expression: 11U/22L*(3.75F-2)+3./6+.25/1.F;
Choices:
A) 0.5       B) 0.25
C) 0.0       D) 0.75

What value do you get if you use the expression in code, then run that code and print the answer? — Adrian Mole, Jul 06 '21 at 21:58
i got 0.75 .but i want to know how to get this result without running the code. — user16394548, Jul 06 '21 at 22:11
Do you see that `11U/22L` will be `0` due to integer division? So D) 0.75 by inspection: 0 + 0.5 + 0.25. The `1.F` is a `float` with value 1. — Weather Vane, Jul 06 '21 at 22:12
Correct. In addition you should know that integers without the suffix are `int` and fractionals are `double` — Eugene Sh., Jul 06 '21 at 23:05
`11U/22L` means "I don't really understand integer promotion so I'll just toss in some suffixes and hope for the best". — Lundin, Jul 07 '21 at 09:57

Erdal Küçük · Accepted Answer · 2021-07-07T06:06:38.170

U: unsigned int
L: long int
F: float

integral literal: [signed] int
FP literal (without suffix f/F): double

11U / 22L * (3.75F - 2) + 3. / 6 + .25 / 1.F
-> unsigned / long * (float - int) + double / int + double / float

Then the expression evaluates to:

unsigned / long -> promoted to long
*
(float - int) -> promoted to float
+
double / int -> promoted to double
+
double / float -> promoted to double

Put the values in:

11U / 22L -> 0L
*
(3.75F - 2) -> 1.75F
+
3. / 6 -> 0.5
+
.25 / 1.F -> 0.25

=> 0L * 1.75F + 0.5 + 0.25

=> 0F + 0.5 + 0.25

=> 0.5 + 0.25

=> (double) 0.75

1 Answers1