# C Exercises: Compute the sum of the two specified integers. If one of the given integer value is in the range 10..20 inclusive return 18

## C-programming basic algorithm: Exercise-22 with Solution

Write a C program to compute the sum of the two given integers. Return 18 if one of the integer values given is in the range 10..20 inclusive.

**C Code:**

```
#include <stdio.h>
#include <stdlib.h>
int main(void){
printf("%d",test(3, 7));
printf("\n%d",test(10, 11));
printf("\n%d",test(10, 20));
printf("\n%d",test(21, 220));
}
int test(int x, int y)
{
return (x >= 10 && x <= 20) || (y >= 10 && y <= 20) ? 18 : x + y;
}
```

Sample Output:

10 18 18 241

**Explanation:**

int test(int x, int y) { return (x >= 10 && x <= 20) || (y >= 10 && y <= 20) ? 18 : x + y; }

The function takes two integer arguments x and y, and returns 18 if either x or y is between 10 and 20 (inclusive). Otherwise, it returns the sum of x and y.

**Time complexity and space complexity:**

The time complexity of this function is O(1) because it only performs a constant number of operations, regardless of the input size.

The space complexity of this function is O(1) because it only uses a constant amount of memory to store the input arguments and the return value.

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**Flowchart:**

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**Next:** Write a C program to check if it is possible to add two integers to get the third integer from three given integers.

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## C Programming: Tips of the Day

** What does (x ^ 0x1) != 0 mean?**

The XOR operation (x ^ 0x1) inverts bit 0. So the expression effectively means: if bit 0 of x is 0, or any other bit of x is 1, then the expression is true.

Conversely the expression is false if x == 1.

So the test is the same as:

if (x != 1)

and is therefore (arguably) unnecessarily obfuscated.

Ref :https://bit.ly/2NIisQM

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