Help w/C++ syntax

[smdawson]

A question in my C++ class is asking me what will be printed by the following code


x = 15;
x += 3;
y = ++x % 4;
z = y--;
cout << “x = “ << x << “ y = “ << y << “ z = “ << z;

more than anything I need to know what the operations are.
does x += 3 mean x = x + 3?

does y = ++x % 4 mean y = 1+x % 4?

does z = y-- mean z = y?

[akvino]

It is easy to test that!

Just put
cout<<"\nx is "<<x<<endl;

[WildeBeest]

x += 3; -> x = x +3
y = ++x % 4; -> y = (x + 1) % 4
z = y--; -> z = (y - 1)

[dwhitney67]

It is easy to test that!

Just put
cout<<"\nx is "<<x<<endl;

The OP is probably scratching his rear, and saying "Write code? Are you crazy!"

[Zugzwang]

@OP: Look up the definitions of the operators, and you'll get to know the answer. Note that this is likely to be the purpose of this homework question. Just one Hint: WildeBeest's solution is incorrect, I'm afraid.

[smdawson]

@dwhitney67. well, that is the last time I ask for help. :???:

[akvino]

well, that is the last time I ask for help. :???:

Hahaha - what we gave you is the way to troubleshoot any possible homework assignment in the future without asking us to do it for you. In essence we did not give you a meal, but thought you how to fish.

[WildeBeest]

@OP: Look up the definitions of the operators, and you'll get to know the answer. Note that this is likely to be the purpose of this homework question. Just one Hint: WildeBeest's solution is incorrect, I'm afraid.


Youare correct I missed one.

z = y, y = y - 1

[dwhitney67]

Youare correct I missed one.

z = y, y = y - 1

And this one too...


y = ++x % 4;

[dwhitney67]

@dwhitney67. well, that is the last time I ask for help. :???:

Why not write yourself 10+ lines of code that duplicates what you asked? At every stage, you could print the value of the variable and compare with your assumptions.

[smdawson]

Yeah, I guess it would be easy enough to output the variables and backtrack through the equations real quick. I just didn't want to make a mistake and continue writing correctly. Time to fish.

[akvino]

And this one too...


y = ++x % 4;


Not to poke to much but this is the same as

y = (1+x) % 4;

[dwhitney67]

Not to poke to much but this is the same as

y = (1+x) % 4;

Yes, 'y' will have the same value regardless of which of these two operations are used:


y = ++x % 4

// or

y = (1 + x) % 4;

However, the assignment indicated that it wanted to know about all of the values: x, y, and finally z.

Thus, the "y = (1 + x) % 4" does NOT take into account that the value of 'x' would have been incremented if "y = ++x % 4" were taken into consideration. In conclusion, you are wrong.

[akvino]

Yes, 'y' will have the same value regardless of which of these two operations are used:


y = ++x % 4

// or

y = (1 + x) % 4;

However, the assignment indicated that it wanted to know about all of the values: x, y, and finally z.

Thus, the "y = (1 + x) % 4" does NOT take into account that the value of 'x' would have been incremented if "y = ++x % 4" were taken into consideration. In conclusion, you are wrong.

No I am not wrong. Y value is the same and he wanted to know if that expression (++x) is equal to (1 + x) which is correct.

:-)

[DZ*]

I always felt that ++, -- operators ain't being used creatively enough


#include <iostream>
using namespace std;

struct X {.
int x;.
X () : x (0) {}.
X& operator ++ (int) { x++; return *this; }.
X& operator ++ () { ++x; return *this; }.
X& operator -- (int) { x--; return *this; }.
X& operator -- () { --x; return *this; }.
friend ostream & operator << (ostream& os, X& x) { return (os << x.x);}
};

int main () {
X x;.
++++++++++++++++++ x ++++++++++++++++++;.
------ x ------;.
--++ x --++++;
cout << x << "\n";.
}

[dwhitney67]

No I am not wrong. Y value is the same and he wanted to know if that expression (++x) is equal to (1 + x) which is correct.
:-)

Yes, that is correct. However in the context of the original question, and the response doled out by WildeBeest, it fell out of context what I was stating. At any rate, I originally was replying to WildeBeest.

[SledgeHammer_999]

One quick question that's been puzzling me for quite a time. I just never needed an answer for my amateurish/hobby programming fun. What is the difference between "++x" and "x++"?

[dwhitney67]

I could give you my $0.02 worth of knowledge, but the folks that replied to this thread (http://stackoverflow.com/questions/2115682/pre-increment-operator-vs-post-increment-operator) already posted the relevant information you seek.

http://stackoverflow.com/questions/2115682/pre-increment-operator-vs-post-increment-operator

[akvino]

OR here


http://ubuntuforums.org/showthread.php?t=1388284

[mdurham]

x += 3; -> x = x +3
y = ++x % 4; -> y = (x + 1) % 4
z = y--; -> z = (y - 1)
I think the 3rd line is incorrect.
Should it be: z = y--; -> z = y do you think?

[Ravenshade]

++x increments before the number
x++ gets the value then increments it.

It's like writing 1 + x, x + 1

or at least that's what my ZNotation lecturer taught me.

[mdurham]

++x increments before the number
x++ gets the value then increments it.

It's like writing 1 + x, x + 1

or at least that's what my ZNotation lecturer taught me.
I don't quite unerstand what you are saying here. "++x increments before the number" What does that mean?
Likewise, "x++ gets the value then increments it" is a bit meaningless.

[Ravenshade]

In other words there's no difference.

++x = 1+ x

x++ = x + 1

[omeraygor]

x = 15;
x += 3;
y = ++x % 4;
z = y--;
cout << “x = “ << x << “ y = “ << y << “ z = “ << z;

x+=3; ====> x=x+3 x=15+3=18;
y=++x % 4; ===> y=1+(x%4); 18mod(4)=2 y=1+2=3;
z=y--; ===> z= (y=y-1); z=3-1; z=2;

so as result. when you use like this.
y = ++x % 4;
that mean is: y=1+(x%4);
example x=18 y=3;
Otherwise
y = x++ % 4;
that mean is: y=(x+1)%4;
example x=19 y=2;

[Ravenshade]

My idea was right, but my implementation was wrong. Omeraygor has what I was trying to say.

[mdurham]

In other words there's no difference.

++x = 1+ x

x++ = x + 1

So are you saying that:

x = ++y is the same as x = y++?
If so I suggest that you try it, printing the result of x. You might be surprised.

y = 10;
x = ++y;
printf("%d %d\n", x, y);

y = 10;
x = y++;
printf("%d %d\n", x, y);

[omeraygor]

i tried.

#include <stdio.h>

int main(){
int x,y;
y = 10;
x = ++y;
printf("%d %d\n", x, y);

y = 10;
x = y++;
printf("%d %d\n", x, y);

return 0;
}

result:
11 11
10 11

So there are difference...

[mdurham]

i tried.

#include <stdio.h>

int main(){
int x,y;
y = 10;
x = ++y;
printf("%d %d\n", x, y);

y = 10;
x = y++;
printf("%d %d\n", x, y);

return 0;
}

result:
11 11
10 11

So there are difference...

You are correct, there is a difference omeraygor, well done.

[dwhitney67]

This is an example of pre-increment


++x

This is an example of post-increment:


x++

If you have statements like:


x = 18;
y = ++x % 4;

Then the result for y is 3. WHY? Well, first x is incremented by 1 to become 19; then the modulo takes place.

For the following:


x = 18;
y = x++ % 4;

The result of y is 2. WHY? Well, first the modulo takes place; then the increment of x takes place. Thus the resulting value of x is (once again) 19.

[omeraygor]

the C language has sequential structure. so.
x = 18;
y = ++x;

increment x before assigning

1.x=18;
2.x=x+1;
3.y=x;

x = 18;
y = x++;

increment x after assigning
1.x=18;
2.y=x;
3.x=x+1;




#include <stdio.h>
int main(){
int x,y;
x = 18;
y = ++x;
printf("%d %d\n", x, y);

x = 18;
y = x++;
printf("%d %d\n", x, y);

return 0;
}

19 19
19 18

[smdawson]

not to change the subject--but, can I compile c++ programs from the terminal? Or some command window in netbeans? I was told that

g++ -o filename sourcecode.cpp
is the command. But I am confused about if I am using the terminal or what. Because if I use the terminal then don't I have to specify the whole path?

[MadCow108]

if your in the same directory as the source file you don't need a path. It searches in the current directory for the file.
but of course you can use relative or absolute paths if you wish, as with every other terminal command

[smdawson]

How can I initialize a variable, ask the user a question and for input, and then have the program recognize a string of characters ( a word or two ) and then make a decision based on the answer?

Not just true or false, but if I wanted the user to choose a color and have the program recognize it like 'blue' or 'red' and then had the program execute different code based on the users input

Would the data type be string? I tried making an if/else loop where each answer would illicit a different response, but it did not work. I realize this is wrong, I'm just showing my work so someone can correct me.(kindly)
My thoughts followed these lines



#include <iostream>
using namespace std;

int main()
{
string color1;
string color2;
string input1;

color1 = "blue"
color2 = "red"

cout << "What is your favoite color? (red/blue)" << endl;
cin >> input1;

if ( input1 = color1 )
cout << "The color of the sea." << endl;
else if (input1 = color2 )
cout << "The color of tomatoes." << endl;

return 0;
}

[dwhitney67]

The single-equals character is used for assigning a value; a double-equals is used for comparisons.



if (input == red)
{
}
else if (input == blue)
{
}
else
{
// error
}

[smdawson]

Ha. I feel stupid. Thank you.

[smdawson]

If I am outputing long lines of text through

cout << ".........multiple lines vertically......" << endl;
is there a way I can keep words from splitting up between different lines of text? Right now I am getting half of a word on a line, and then as it continues down one line the other half of the word follows. this makes for very difficult reading.

[dwhitney67]

If I am outputing long lines of text through

cout << ".........multiple lines vertically......" << endl;
is there a way I can keep words from splitting up between different lines of text? Right now I am getting half of a word on a line, and then as it continues down one line the other half of the word follows. this makes for very difficult reading.

I'm not sure what you are asking; perhaps a trick question?

Anyhow, maybe this will assist you in finding enlightenment...


#include <iostream>

int main(void)
{
using namespace std;

cout << "This is my first line\n"
<< "This is my second line\n"
<< "And this is my very long third line "
"which I will continue here."
<< endl;

return 0;
}

[smdawson]

That answers my question. I will just use "\n" to keep my words from splitting up from one line of text to the next. Thanks