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Shell Programming and Scripting

Miscellaneous operators

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# 1  
Old 04-16-2010
Miscellaneous operators
Hi everyone,

I read some shell script code,then I have some issue.

the following code.

Code:
let "t1 = ((5 + 3, 7 - 1, 15 - 4))"
echo "t1 = $t1"

t1=11
Here t1 is set to the result of the last operation.why?
# 2  
Old 04-16-2010
Arithmetic expressions in modern shells use the same syntax as is used in the C programming language. And in C, the value of a numeric expression using the comma operator is the value of the last subexpression, in your case 15 - 4 .
# 3  
Old 04-16-2010
it's just like
Code:
((t1=5+3))   # t1 = 8
((t1=7-1))   # t1 = 6
((t1=15-4))  # t1 = 11

If you have a "modern" shell (like bash), the "let" statement is obsolete.
# 4  
Old 04-16-2010
Quote:
Originally Posted by frans
it's just like
Code:
((t1=5+3))   # t1 = 8
((t1=7-1))   # t1 = 6
((t1=15-4))  # t1 = 11

If you have a "modern" shell (like bash), the "let" statement is obsolete.
Thanks,

I understand your meaning,I use shell of fedora 11 ,I have already tested this issue in my system,and as your result.
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