3. What are the values of sum1 and sum2 if the operands in the expressions are e

Question

3. What are the values of sum1 and sum2 if the operands in the expressions are evaluated left to right ? Must include explanation
int fun(int *k){ *k += 10; return 5 * (*k); } void main(void){ int a = 6, b =12, sum1, sum2; sum1= (a/2) + fun(&a) sum2 = fun(&b) + (b/2); } 3. What are the values of sum1 and sum2 if the operands in the expressions are evaluated left to right ? Must include explanation
int fun(int *k){ *k += 10; return 5 * (*k); } void main(void){ int a = 6, b =12, sum1, sum2; sum1= (a/2) + fun(&a) sum2 = fun(&b) + (b/2); }
int fun(int *k){ *k += 10; return 5 * (*k); } void main(void){ int a = 6, b =12, sum1, sum2; sum1= (a/2) + fun(&a) sum2 = fun(&b) + (b/2); }

Explanation / Answer

Solution:

Value of sum1 = 83

Value of sum2 = 121

Explanation:

a) sum1 = (a/2) + fun (&a)

sum1 = (6/2) + 80 = 3 + 80 = 83

because when we pass the address of 'a' in function fun, then the changed value of 'a' will be reflected in the calling function. Now, we see the implementation of the function fun:

* is called value at address operator.

fun(int *k) => fun(*(&6)) => fun(6)

fun(6) here k = 6 => k = 10 + 6 = 16 => 5 * 16 = 80, thus it will return 80.

b)  sum2 = fun(&b) + (b/2);

similarly, sum2 = 110 + (22/2) = 110 + 11 = 121

because after calling fun(&b) the value of b will be 22. So fun(&b) would return 110 and b would 22.

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