if statement

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:

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//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;
}

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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 :D

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

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

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

Quote:

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Originally Posted by junkeat90

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

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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

Quote:

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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).