if statement

**junkeat90** · 01-10-2008

Hi guys, it's me the newbie again =) I've a new question. Here's a piece of code I copied from a book:

Code:

//This programme shows an example of the AND && expression
#include <iostream>


int main()
{
    using namespace std;
    
    short number1 = 10;
    char word = 't';
    
    if (number1 == 10 && word == 't')
    cout<< "Both sides of the AND expression are True";
    cin.get();
    return 0;
}

It seems that when I didn't end this line of command:

Code:

if (number1 == 10 && word == 't')

with a semi-colon, the whole script can still run and the main function is working. Why is that so ??? Thanks in advance guys

**laserlight** · 01-10-2008

I suggest reading the book more for the explanation. If you still do not understand read the tutorial on if statements.

**matsp** · 01-10-2008

That is correct.

If you indent your code correctly, it becomes more clear:

Code:

    if (number1 == 10 && word == 't')
       cout<< "Both sides of the AND expression are True";

If you do this:

Code:

    if (number1 == 10 && word == 't');
    cout<< "Both sides of the AND expression are True";

The compiler will happily produce code, but you will ALWAYS get the output from the cout line.

--
Mats

**junkeat90** · 01-10-2008

Oh I see...thanks matsp =) You're a real good programmer. Wonder how many years have you been learning C++?

**matsp** · 01-10-2008

Originally Posted by junkeat90

Oh I see...thanks matsp =) You're a real good programmer. Wonder how many years have you been learning C++?

I've been working with C++ for about half a year, but first time I used C was more than 15 years ago, and the above "problem" would be identical in C, only that you wouldn't use cout, but printf to print the text.

--
Mats

**CornedBee** · 01-10-2008

char word = 't';

Is that really what's in the book? If so, beware, because it's misleading: a char stores a single character, not a word in the traditional sense (you need a string for that), nor a machine word (the CPU's "most comfortable" integer type), nor an x86 word (a 16-bit integer).

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