This is not only a condition, but it also creates an array element code[$3] with an empty value
Indeed and that's exactly what I find weird. With code[$3] in the second block I was expecting awk to *evaluate* the value of code[$3] *not* to assign any value to it, albeit NULL.
Code:
awk 'foo="bar"{print "block 1"} END{print foo}' f1
Returns bar.
foo="bar" assigns "bar" to foo and returns a TRUE. No problem with that. But in the condition of the second block code[$3] there is no assignment sign and it still assigns a value. I can't stop finding it weird.
Furthermore, if you look to my code above and its return.
Code:
FNR == NR && /file1_l1/ {
code[$2] = 1
next
}
code[$3] {
print
}
The instruction next should make the program to loop on the first file until it reaches the end of file1. Then it continues with the second file, right? I understand that the condition of the second block assigns a value while evaluating code[$3] but how come that it assigns values from the first file as the pointer NR is already on the second file? See my point?
Last edited by ripat; 09-11-2013 at 02:36 PM..
Reason: More confusion...
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Discussion started by: Kulasekar
2 Replies
LEARN ABOUT PHP
print
PRINT(3) 1 PRINT(3)print - Output a stringSYNOPSIS
int print (string $arg)
DESCRIPTION
Outputs $arg.
print is not actually a real function (it is a language construct) so you are not required to use parentheses with its argument list.
PARAMETERS
o $arg
- The input data.
RETURN VALUES
Returns 1, always.
EXAMPLES
Example #1
print examples
<?php
print("Hello World");
print "print() also works without parentheses.";
print "This spans
multiple lines. The newlines will be
output as well";
print "This spans
multiple lines. The newlines will be
output as well.";
print "escaping characters is done "Like this".";
// You can use variables inside a print statement
$foo = "foobar";
$bar = "barbaz";
print "foo is $foo"; // foo is foobar
// You can also use arrays
$bar = array("value" => "foo");
print "this is {$bar['value']} !"; // this is foo !
// Using single quotes will print the variable name, not the value
print 'foo is $foo'; // foo is $foo
// If you are not using any other characters, you can just print variables
print $foo; // foobar
print <<<END
This uses the "here document" syntax to output
multiple lines with $variable interpolation. Note
that the here document terminator must appear on a
line with just a semicolon no extra whitespace!
END;
?>
NOTES
Note
Because this is a language construct and not a function, it cannot be called using variable functions.
SEE ALSO echo(3), printf(3), flush(3), Heredoc syntax.
PHP Documentation Group PRINT(3)