这个循环怎样用bash shell写啊？[基本解决]

fooykid · 发表于 2004-6-6 23:10:10

int num=0;
for ( int i=1; i<1000000; i++ ) num += i;

KornLee · 发表于 2004-6-6 23:15:13

n=0;num=0
while ((n<=1000000));do
((num+=n))
((n+=1))
done;echo $num

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ps:sorry,看错题啦

dearvoid · 发表于 2004-6-7 08:51:07

bash 的 for 语句:

sum=0
for ((i = 0; i < 100; i++))
do
((sum += i))
done

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cnspy · 发表于 2004-6-7 09:16:56

Why use:
((sum += 1))
Not
sum += 1 or (sum += 1)

Thanks in advance.

dearvoid · 发表于 2004-6-7 09:40:34

sum=$((sum + i)) 也是可以的, 我的写法只是简单些, 这是 bash 的语法

dearvoid · 发表于 2004-6-7 09:44:17

ARITHMETIC EVALUATION
The shell allows arithmetic expressions to be evaluated, under certain
circumstances (see the let builtin command and Arithmetic Expansion).
Evaluation is done in fixed-width integers with no check for overflow,
though division by 0 is trapped and flagged as an error. The operators
and their precedence and associativity are the same as in the C lan-
guage. The following list of operators is grouped into levels of
equal-precedence operators. The levels are listed in order of decreas-
ing precedence.
id++ id--
variable post-increment and post-decrement
++id --id
variable pre-increment and pre-decrement
- + unary minus and plus
! ~ logical and bitwise negation
** exponentiation
* / % multiplication, division, remainder
+ - addition, subtraction
<< >> left and right bitwise shifts
<= >= < >
comparison
== != equality and inequality
& bitwise AND
^ bitwise exclusive OR
| bitwise OR
&& logical AND
|| logical OR
expr?expr:expr
conditional evaluation
= *= /= %= += -= <<= >>= &= ^= |=
assignment
expr1 , expr2
comma
Shell variables are allowed as operands; parameter expansion is per-
formed before the expression is evaluated. Within an expression, shell
variables may also be referenced by name without using the parameter
expansion syntax. The value of a variable is evaluated as an arith-
metic expression when it is referenced. A shell variable need not have
its integer attribute turned on to be used in an expression.
Constants with a leading 0 are interpreted as octal numbers. A leading
0x or 0X denotes hexadecimal. Otherwise, numbers take the form
[base#]n, where base is a decimal number between 2 and 64 representing
the arithmetic base, and n is a number in that base. If base# is omit-
ted, then base 10 is used. The digits greater than 9 are represented
by the lowercase letters, the uppercase letters, @, and _, in that
order. If base is less than or equal to 36, lowercase and uppercase
letters may be used interchangably to represent numbers between 10 and
35.
Operators are evaluated in order of precedence. Sub-expressions in
parentheses are evaluated first and may override the precedence rules
above.

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dearvoid · 发表于 2004-6-7 09:45:00

详见 man bash

cnspy · 发表于 2004-6-7 13:52:57

thanks for your kind reply.

dearvoid · 发表于 2004-6-7 14:17:44

更详尽的解释：

Compound Commands
...
[color=red]((expression))[/color]
The expression is evaluated according to the rules described
below under [color=red]ARITHMETIC EVALUATION[/color]. If the value of the expres-
sion is non-zero, the return status is 0; otherwise the return
status is 1. This is exactly equivalent to let "expression".
...
Arithmetic Expansion
Arithmetic expansion allows the evaluation of an arithmetic expression
and the substitution of the result. The format for arithmetic expan-
sion is:
[color=red]$((expression))[/color]
The expression is treated as if it were within double quotes, but a
double quote inside the parentheses is not treated specially. All
tokens in the expression undergo parameter expansion, string expansion,
command substitution, and quote removal. Arithmetic substitutions may
be nested.
The evaluation is performed according to the rules listed below under
[color=red]ARITHMETIC EVALUATION[/color]. If expression is invalid, bash prints a message
indicating failure and no substitution occurs.
[color=red]ARITHMETIC EVALUATION[/color]
The shell allows arithmetic expressions to be evaluated, under certain
circumstances (see the let builtin command and Arithmetic Expansion).
Evaluation is done in fixed-width integers with no check for overflow,
though division by 0 is trapped and flagged as an error. The operators
and their precedence and associativity are the same as in the C lan-
guage. The following list of operators is grouped into levels of
equal-precedence operators. The levels are listed in order of decreas-
ing precedence.
id++ id--
variable post-increment and post-decrement
++id --id
variable pre-increment and pre-decrement
- + unary minus and plus
! ~ logical and bitwise negation
** exponentiation
* / % multiplication, division, remainder
+ - addition, subtraction
<< >> left and right bitwise shifts
<= >= < >
comparison
== != equality and inequality
& bitwise AND
^ bitwise exclusive OR
| bitwise OR
&& logical AND
|| logical OR
expr?expr:expr
conditional evaluation
= *= /= %= += -= <<= >>= &= ^= |=
assignment
expr1 , expr2
comma
Shell variables are allowed as operands; parameter expansion is per-
formed before the expression is evaluated. Within an expression, shell
variables may also be referenced by name without using the parameter
expansion syntax. The value of a variable is evaluated as an arith-
metic expression when it is referenced. A shell variable need not have
its integer attribute turned on to be used in an expression.
Constants with a leading 0 are interpreted as octal numbers. A leading
0x or 0X denotes hexadecimal. Otherwise, numbers take the form
[base#]n, where base is a decimal number between 2 and 64 representing
the arithmetic base, and n is a number in that base. If base# is omit-
ted, then base 10 is used. The digits greater than 9 are represented
by the lowercase letters, the uppercase letters, @, and _, in that
order. If base is less than or equal to 36, lowercase and uppercase
letters may be used interchangably to represent numbers between 10 and
35.
Operators are evaluated in order of precedence. Sub-expressions in
parentheses are evaluated first and may override the precedence rules
above.

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这个循环怎样用bash shell写啊？[基本解决]

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