CODEWITHSUNDEEP
c
JohnX
JohnX

Reputation: 29

C: Trouble with operators and results of function

I am a college student who is taking his first programming class, in the process of studying a past exam I found this question:

give the following values of the print function

#include <stdio.h>
int main(){
float x;
int a = 5, b = 0;
x = a&b ? 2/8.f*5+b++ : a|b
printf("%f\n", x);

}

The printf displays 5.000 after running. What is tripping me up is the ?: operator. As I understand it, it works almost like an if, else statement, where if (condition) is true, then x, if not true then y. I don't understand the flow of the function.

is it saying that because the a&b yields 0 that it is not true, therefore x = a|b which after the operator runs yields decimal value 5?

Upvotes: 2

Views: 65

Answers (2)

John Bode
John Bode

Reputation: 123578

is it saying that because the a&b yields 0 that it is not true, therefore x = a|b which after the operator runs yields decimal value 5?

Yup.

The expression parses out as follows:

(a & b) ? ((2 / 8.f) * 5 + (b++)) : (a | b)

& is the bitwise AND operator; it will perform an AND operation on each bit of the two operands. Since a is 5 (101) and b is 0 (000), the result of a&b is 101 & 000 == 000.

Since a&b evaluates to 0, the expression following the ? is not evaluated; instead, (a|b) is evaluated. | is the bitwise OR operator. 101 | 000 == 101, which is 5.

Hence your result.

Upvotes: 2

wolfPack88
wolfPack88

Reputation: 4203

Your understanding is exactly correct. The operator is called the ternary operator. In this case, the code evaluates a&b to be 0 or false, which causes it to use the value after the :, or a|b, which is 5. If, instead, b were to equal 1 for example, then a&b would evaluate to true, and x would equal the expression before the :, which evaluates to 2.25.

Upvotes: 2

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