C++ syntax error clarification and some stuff.

[A.D.A.M]

I got this mcq am wondering if i got the right idea:

3. Analyze the following code:

Code 1:

bool even;
int number;

if (number % 2 == 0)
even = true;
else
even = false;

Code 2:

bool even = (number % 2 == 0);
a. Code 2 has syntax errors.
b. Code 1 has syntax errors.
c. Both Code 1 and Code 2 have syntax errors.
d. Both Code 1 and Code 2 are correct, but Code 2 is better.

 

If i use VS2015 i can't compile without explicitly saying int number=0; (since without it num is undefined.) so answer would be "C" correct? Since 2 definitely can't run without initialisation of num. 

[AustinJ]

-SNIP-

 

int number = 0;bool even = (number % 2 == 0) ? true : false;

 

You could do something like that. You should always initialize a variable anyways before doing anything else.

[Timopomer]

yep. you cannot do an operation on number without defining it first.

 

its c

[A.D.A.M]

Ah ok thanks. Another thing if you don't mind.

say we have something like this 

int x;int y;x=3;switch (x+3){    case 6:  y = 0;       case 7:  y = 1;       default: y += 1;     }

this would cause the it to run each case yeah? Until it reaches default?

so first y would be assigned 0 , then 1 ,and then y= y+1 thus 1+1 so 2?

[Hunter259]

yep. you cannot do an operation on number without defining it first.

 

its c

You can do this, just matters on the compiler. Xcode gives no errors with this program.

What are you trying to do with this program though? If you were trying to make it random then this is not the way to do it.

[Hunter259]

Ah ok thanks. Another thing if you don't mind.

say we have something like this 

int x;int y;x=3;switch (x+3){    case 6:  y = 0;       case 7:  y = 1;       default: y += 1;     }

this would cause the it to run each case yeah? Until it reaches default?

so first y would be assigned 0 , then 1 ,and then y= y+1 thus 1+1 so 2?

Switch statements are not loops so no. If you were to put it within a do while, while, or for, this would be ok although it would never enter case 7 as x does not ever equal 7.

[A.D.A.M]

You can do this, just matters on the compiler. Xcode gives no errors with this program.

What are you trying to do with this program though? If you were trying to make it random then this is not the way to do it.

oh it's just a question i have to solve (spotting code errors and all that). Yeah my friends with dev C++ says it compiles fine...

[AustinJ]

Ah ok thanks. Another thing if you don't mind.

say we have something like this 

int x;int y;x=3;switch (x+3){    case 6:  y = 0;       case 7:  y = 1;       default: y += 1;     }

this would cause the it to run each case yeah? Until it reaches default?

so first y would be assigned 0 , then 1 ,and then y= y+1 thus 1+1 so 2?

 

x+3 = 6, so case 6 would be called. There is no looping in a switch.

[A.D.A.M]

Switch statements are not loops so no. If you were to put it within a do while, while, or for, this would be ok although it would never enter case 7 as x does not ever equal 7.

so how does it ever cout<<y; a 2?  because there isn't a break shouldn't it like well break...

[Timopomer]

Ah ok thanks. Another thing if you don't mind.

say we have something like this 

int x;int y;x=3;switch (x+3){    case 6:  y = 0;       case 7:  y = 1;       default: y += 1;     }

this would cause the it to run each case yeah? Until it reaches default?

so first y would be assigned 0 , then 1 ,and then y= y+1 thus 1+1 so 2?

switch just gives your 1 case. it doesnt do any more than that (loops or anything else you think it might do)

[Hunter259]

oh it's just a question i have to solve (spotting code errors and all that). Yeah my friends with dev C++ says it compiles fine...

Ah. On Xcode it just defaults ints to 32767. 

[Hunter259]

so how does it ever cout<<y; a 2?  because there isn't a break shouldn't it like well break...

I just ran it and yes Y comes out to 2 but I don't exactly know why. To my knowledge it should only return the value of Case 6 as it is the value that matches X.

[A.D.A.M]

I just ran it and yes Y comes out to 2 but I don't exactly know why. To my knowledge it should only return the value of Case 6 as it is the value that matches X.

the way i understand is its missing the break; so it runs till the end? Since only when i do that will it give the expected result

[Hunter259]

the way i understand is its missing the break; so it runs till the end? Since only when i do that will it give the expected result

After reading this http://stackoverflow.com/questions/252489/why-was-the-switch-statement-designed-to-need-a-break and retesting the code it is infact having a fall through occur allowing it to sort of loop through these expressions.

[A.D.A.M]

After reading this http://stackoverflow.com/questions/252489/why-was-the-switch-statement-designed-to-need-a-break and retesting the code it is infact having a fall through occur allowing it to sort of loop through these expressions.

so the previous assumption was correct?

[madknight3]

After reading this http://stackoverflow.com/questions/252489/why-was-the-switch-statement-designed-to-need-a-break and retesting the code it is infact having a fall through occur allowing it to sort of loop through these expressions.

 

Yes, if there's no break, a case will fall through to the next. You'll often see it used like this where more than one case runs the same code.

switch (x){    case 1:    case 2:        // Does something if x == 1 || x == 2        break;    case 3:        // Does something else if x == 3        break;    default:        // Does something else if the above cases aren't called}

Keep in mind that while fall through can be intentional, it can also be a bug because people can forget to add the break.

[A.D.A.M]

Yes, if there's no break, a case will fall through to the next. You'll often see it used like this where more than one case runs the same code.

Keep in mind that while fall through can be intentional, it can also be a bug because people can forget to add the break.

So its called a fall? Good thing to know. 

Also, off topic but you proficient in jQuery.If you are I'd appreciate if you would look at my other thread.

http://linustechtips.com/main/topic/456363-need-some-help-with-jquery-and-making-chrome-extensions/#entry6124428