Thread: Some C test questions: challenge

29 d. both | and || are binary operators.....
bool operator | (arg1,arg2)
bool operator || (arg1,arg2)

both take 2 arguments.

21 c. you can play with the bits in a floating point number with bitwise operators with some creative casting but not directly.

Why exactly was the answer 'd' in that previous question?? I got 'c'. Can someone please explain this?

//radius = (radius++) + (z++)

radius = radius + z;
radius = radius + 1;
z = z++;

OK, so that's how you cook the books to get 48

But that's only one possible interpretation of that expression.

And this maybe, is how moi got 47 with
> lcc for windoze v2.5 output radius=47

Code:

//radius = (radius++) + (z++)
int temp1 = radius;      // save old value before incrementing it
int temp2 = z;
radius = radius + 1;     // increment radius (ie radius++)
z = z+1;                 // increment z (ie z++)   z=z++ is broken 
radius = temp1 + temp2;  // calculate the sum

Oh no!, the increment of radius is lost because it got written to store before the sum did.

Here's another thread where the poster managed to dig a big hole around expressions with multiple side effects and undefined behaviour.

Originally posted by Sebastiani
Why exactly was the answer 'd' in that previous question?? I got 'c'. Can someone please explain this?

Code:

// general form of the switch statement

switch (expression)
{
    case template_1: statement(s); break;
    case template_2: statement(s); break;
    ...
    case template_n: statement(s); break;
    default: statement(s);
}

I guess the result is '19' because 'break' is missing at the end of each statement. If a match is found between the expression and one of the templates, the statement that follows the case label is executed. After that execution is transferred to the first statement following the switch statement's closing brackets. But as a result of the missing 'break', the statements of the following case labels are executed too!

Code:

int num;
int alpha = 10;

scanf(“%d\n”, &num);
switch (num)
{
    case 3  : alpha++;
    case 4  : alpha = alpha + 2;
    case 8  : alpha = alpha + 3;
    default : alpha = alpha + 4;
}
printf(“%d\n”, alpha);

Originally posted by Mister C

Code:

21. The bitwise operators can be used to manipulate the bits of variables of type __________.
a) float
b) double
c) long
d) long double

Mr. C

i agree that it is poor form, but one can manipulate the bits of a float (even though in this particular example the bitwise operators are not used, they could be): http://cboard.cprogramming.com/showt...threadid=24350

I guess the result is '19' because 'break' is missing

Ok, nevermind, I overlooked the missing break's...

You could do a question about the binary operator ^^, what it is or what would happen if this was executed:

Code:

bool a = TRUE, b = TRUE;

if (a ^^ b){
    printf("TRUE");
} else {
    printf("FALSE");
}

There is no ^^ operator in C, there also isn't a bool data type or TRUE and FALSE macros unless you define them yourself. I have heard of people wondering about a logical exclusive OR which would probably be ^^, but I also remember them saying that the reason it wasn't created is because there's no point. I think you meant the binary XOR operator, which is just one ^.

Code:

#include <stdio.h>

#define TRUE (1==1)
#define FALSE !TRUE

typedef int bool;

int main() {
	bool a = TRUE, b = TRUE;

	if (a ^ b) {
		printf("TRUE\n");
	} else {
		printf("FALSE\n");
	}

	return 0;
}

Bebop

There's no point in using trick questions unless you want your students to miss them. The last time I checked people didn't go to school to pass or fail, they go to learn. There's no practical application for knowing that ^^ doesn't exist, so it's a stupid question, you don't get anything from having it.

Bebop

I think that since this is programming it might be a little more important than in other subjects, because of the precision required. Never try to argue semantics with a computer.