Question

代码：

#!/bin/bash

declare -i number
# The script will treat subsequent occurrences of "number" as an integer.       

number=3
echo "Number = $number"     # Number = 3

number=three
echo "Number = $number"     # Number = 0
# Tries to evaluate the string "three" as an integer.

当我将字符串number分配给"three"时，我无法弄清楚为什么number发生了变化。我认为number应该保持不变。这真让我感到惊讶。

Answer 1

来自declare的{{1}}部分：

man bash

来自-i The variable is treated as an integer; arithmetic evaluation (see ARITHMETIC EVALUATION) is performed when the variable is assigned a value.的{{1}}部分：

ARITHMETIC EVALUATION

总之，这些明确表明您所看到的行为是预期的行为。在算术上评估字符man bash The value of a variable is evaluated as an arithmetic expression when...a variable which has been given the integer attribute using declare -i is assigned a value. A null value evaluates to 0. t h r时，生成的e值将被评估为e，其中然后被赋值给变量null。

bash中的所有赋值都首先解释为字符串。 0首先将number number=10解释为字符串，将其识别为有效整数，并保持原样。 1在语法和语义上与0一样有效，这就是为什么在将number=three的评估值分配给number=10后，您的脚本会继续而没有任何错误。

当我在Bash脚本中将声明的整数更改为字符串时，为什么整数结果为0

1 个答案: