pirnt打印时对于列表的解释

zhuzhzh

pirnt (1..5), "\n";
和
@num=(1..5);
print @num, "\n";
执行时不一样

前者第一个列表打印后就结束了
而后面的方式print知道列表是（@num, "\n")

这是不是perl的bug啊？

xfoucs

挖坑吧？

1. 
   
2. print +(1..5),"\n";
   

复制代码

zhuzhzh

回复 2# xfoucs

是真不知道
刚刚才知道这被解释成逗号表达式了

lqianc

回复 3# zhuzhzh


    我还是不懂，哈哈，被解释成逗号表达式了，啥意思？ 前面多个+号又是啥意思~~

yinyuemi

回复 4# lqianc


   

Be careful not to follow the print keyword with a left parenthesis unless you want the corresponding right parenthesis to terminate the arguments to the print; put parentheses around all arguments (or interpose a + , but that doesn't look as good).

Binary "," is the comma operator. In scalar context it evaluates its left argument, throws that value away, then evaluates its right argument and returns that value. This is just like C's comma operator.

zhlong8

这就是可省略括号的代价

wangzhe19950128

我记得在书上看到过了，在前面多一个+好像是，先让后面的计算吧，在Perl中看着像是函数调用，它就是函数调用 print();就是一个函数调用，会存在返回值的