51

我有d1="11"和d2="07"。我想将d1and转换d2为整数并执行d1-d2. 我如何在 UNIX 中做到这一点？

d1 - d2目前"11-07"为我返回结果。

4

5 回答 5

76

标准溶液：

 expr $d1 - $d2

你也可以这样做：

echo $(( d1 - d2 ))

但请注意，这将被07视为八进制数！（so07与 相同7，但010不同于10）。

于 2012-06-29T21:46:26.713 回答
21

这些中的任何一个都可以从 shell 命令行工作。bc不过，这可能是您最直接的解决方案。

使用bc：

$ echo "$d1 - $d2" | bc

使用awk：

$ echo $d1 $d2 | awk '{print $1 - $2}'

使用perl：

$ perl -E "say $d1 - $d2"

使用Python：

$ python -c "print $d1 - $d2"

全部返回

4
于 2012-06-29T20:27:58.043 回答
6

一个不限于 OP 案例的答案

问题的标题将人们引到这里，所以我决定为其他所有人回答这个问题，因为 OP 描述的案例非常有限。

TL;博士

我终于决定写一个函数。

  1. 如果你想0在非整数的情况下：
int(){ printf '%d' ${1:-} 2>/dev/null || :; }
  1. 如果你想要[empty_string]在非整数的情况下：
int(){ expr 0 + ${1:-} 2>/dev/null||:; }
  1. 如果你想找到第一个 int 或[empty_string]：
int(){ expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null||:; }
  1. 如果你想找到第一个 int 或 0：
# This is a combination of numbers 1 and 2
int(){ expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null||:; }

如果您想在非整数上获得非零状态代码，请删除||:(aka or true) 但保留;

测试

# Wrapped in parens to call a subprocess and not `set` options in the main bash process
# In other words, you can literally copy-paste this code block into your shell to test
( set -eu;
    tests=( 4 "5" "6foo" "bar7" "foo8.9bar" "baz" " " "" )
    test(){ echo; type int; for test in "${tests[@]}"; do echo "got '$(int $test)' from '$test'"; done; echo "got '$(int)' with no argument"; }

    int(){ printf '%d' ${1:-} 2>/dev/null||:; };
    test

    int(){ expr 0 + ${1:-} 2>/dev/null||:; }
    test

    int(){ expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null||:; }
    test

    int(){ printf '%d' $(expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null)||:; }
    test

    # unexpected inconsistent results from `bc`
    int(){ bc<<<"${1:-}" 2>/dev/null||:; }
    test
)

测试输出

int is a function
int ()
{
    printf '%d' ${1:-} 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '0' from '6foo'
got '0' from 'bar7'
got '0' from 'foo8.9bar'
got '0' from 'baz'
got '0' from ' '
got '0' from ''
got '0' with no argument

int is a function
int ()
{
    expr 0 + ${1:-} 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '' from '6foo'
got '' from 'bar7'
got '' from 'foo8.9bar'
got '' from 'baz'
got '' from ' '
got '' from ''
got '' with no argument

int is a function
int ()
{
    expr ${1:-} : '[^0-9]*\([0-9]*\)' 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '6' from '6foo'
got '7' from 'bar7'
got '8' from 'foo8.9bar'
got '' from 'baz'
got '' from ' '
got '' from ''
got '' with no argument

int is a function
int ()
{
    printf '%d' $(expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null) || :
}
got '4' from '4'
got '5' from '5'
got '6' from '6foo'
got '7' from 'bar7'
got '8' from 'foo8.9bar'
got '0' from 'baz'
got '0' from ' '
got '0' from ''
got '0' with no argument

int is a function
int ()
{
    bc <<< "${1:-}" 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '' from '6foo'
got '0' from 'bar7'
got '' from 'foo8.9bar'
got '0' from 'baz'
got '' from ' '
got '' from ''
got '' with no argument

笔记

我被送进了这个兔子洞，因为接受的答案与set -o nounset（又名set -u）不兼容

# This works
$ ( number="3"; string="foo"; echo $((number)) $((string)); )
3 0

# This doesn't
$ ( set -u; number="3"; string="foo"; echo $((number)) $((string)); )
-bash: foo: unbound variable
于 2020-01-17T04:36:22.183 回答
0 
let d=d1-d2;echo $d;

这应该会有所帮助。

于 2016-12-14T22:33:03.573 回答
-8

用这个：

#include <stdlib.h>
#include <string.h>

int main()
{
    const char *d1 = "11";
    int d1int = atoi(d1);
    printf("d1 = %d\n", d1);
    return 0;
}

等等

于 2012-06-29T20:19:53.007 回答