请教：对perl进制转换函数oct、hex的疑问

shangtian2007

如下图所示，
为什么第三条打印的是 0？

图好像有问题，写下来吧：
> print oct(10)
8
> print oct("010")
8
> print oct(010)
0


对hex有同样疑问：第三条打印的是 22


> print hex(10)
16
> print hex("0x10")
16
> print hex(0x10)
22

请教：Perl是怎么解释第三条语句的？

Perl_Er

回复 1# shangtian2007

1. $ perl -e '$a=010; print $a,"-",oct($a)'
   
2. 8-0
   
3. James_Zhu@WN7X64-JYM9WM1 ~/PERL
   
4. $ perl -e '$a=0x10; print $a,"-",hex($a)'
   
5. 16-22
   

复制代码

because 010&0x10 are treated as a decimal and perl feed the decimal to oct/hex

Perl understands numbers specified in binary (base-2), octal (base-, and hexadecimal (base-16) notation only when they occur as literals in your programs. If they come in as data—such as by reading from files or environment variables, or when supplied as command-line arguments—no automatic conversion takes place.