2012-02-17 68 views

回答

4

是的;

print "Result = ", $a+$b, "\n"; 
+0

Thx很多。这很简单。我应该能够想出来 – 2012-02-17 13:23:22

+1

或者'print'Result = $ {\($ a + $ b)} \ n“;' – flesk 2012-02-17 13:31:59

+1

或者缩短一个字符,'print'Result = @ {[$ a + $ b]} \ n“;'或'my $ str = sprintf'结果=%s \ n”,$ a + $ b“\ n”;' – chepner 2012-02-17 13:44:56

4
print "Result=", $a+$b, "\n";   # Multi-arg print 
print "Result=".($a+$b)."\n";   # Concatenation 
print "Result=${\($a+$b)}\n";  # ref-deref trick (scalar context) 
print "[email protected]{[ $a+$b ]}\n";  # ref-deref trick (list context) 
printf "Result=%s\n", $a+$b;   # printf 
print sprintf("Result=%s\n", $a+$b); # sprintf 
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