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How to read the expression (i = j) || (j = k)

This is a discussion on How to read the expression (i = j) || (j = k) within the C Programming forums, part of the General Programming Boards category; I was practicing some exercise from the book, so I wrote down the program below. Code: #include<stdio.h> int main(void) { ...

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    How to read the expression (i = j) || (j = k)

    I was practicing some exercise from the book, so I wrote down the program below.

    Code:
    #include<stdio.h>
    
    int main(void)
    {
        int i = 7, j = 8, k = 9;
        
        printf("%d ", (i = j) || (j = k));
        printf("%d %d %d", i, j, k);
        
        getch();
        return 0;
    }
    The result should be 1 8 8 9, but I don't know how this expression work.
    How to interpret the expression (i = j) || (j = k)?
    In the first parentheses (i = j), did the program assign 8 to i?
    If so, why didn't it assign 9 to j as well, in the second parentheses (j = k)?
    Later on, why the i remain as 8, when was it saved?

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    Ok, this is about the way C evaluates logical or ...

    It will look at (i = j) || (j = K) only far enough to decide if one OR the other is true... left to right i = j will assign the value of j to i ... this will always result in true, so it never tests the second set of braces and thus never assigns j = k.

    The I will remain 8 because you used assignment (single equal sign) not comparison (double equal sign). That is to say that ...
    Code:
    if (i = j) 
     printf ("Hello");
    ... assigns the value of j to i and will always be true (because it succeeded) and it will always print "Hello".

    However...
    Code:
    if (i == j)
     printf("Hello");
    ... only if the value of i is the same as the value of j ... neither value is changed.

    Try your original code with double equals ( == ) in that first printf() call ... it should print 1789 instead.
    Last edited by CommonTater; 08-30-2011 at 01:23 AM.

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    Why is i = j will always be true?

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    Quote Originally Posted by Roger Lo View Post
    Why is i = j will always be true?
    It's a standard set by ANSI C. Anything not 0 is true. This world is full of standards and conventions which sometimes make sense, sometimes don't.

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    > ... assigns the value of j to i and will always be true (because it succeeded) and it will always print "Hello".
    It only succeeds because j is non-zero.
    If j was zero, then the assignment would still happen, but the logical side effect (the implied (expr) != 0) is false.
    CommonTater likes this.
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    Quote Originally Posted by Roger Lo View Post
    Why is i = j will always be true?
    See Salem's post... I should qualify my statement to say that "in your example" it will always be true.
    As Salem points out, if j is 0, it will evaluate to false...
    Another way of breaking it down to understand it is...
    Code:
    i = j;
    if ( i != 0 ) 
      Printf("Hello");

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    understood, thanks for everyone!

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