Bash "not"：反转命令的退出状态答案

【问题标题】：Bash "not": inverting the exit status of a commandBash "not"：反转命令的退出状态
【发布时间】：2013-02-10 22:55:53
【问题描述】：

我知道我能做到……

if diff -q $f1 $f2
then
    echo "they're the same"
else
    echo "they're different"
fi

但是如果我想否定我正在检查的条件怎么办？即像这样的东西（显然不起作用）

if not diff -q $f1 $f2
then
    echo "they're different"
else
    echo "they're the same"
fi

我可以做这样的事情......

diff -q $f1 $f2
if [[ $? > 0 ]]
then
    echo "they're different"
else
    echo "they're the same"
fi

这里我检查上一条命令的退出状态是否大于0。不过这样感觉有点别扭。有没有更惯用的方法来做到这一点？

【问题讨论】：

【解决方案1】：

if ! diff -q "$f1" "$f2"; then ...

【讨论】：

【解决方案2】：

如果你想否定，你正在寻找!：

if ! diff -q $f1 $f2; then
    echo "they're different"
else
    echo "they're the same"
fi

或（简单地反转 if/else 操作）：

if diff -q $f1 $f2; then
    echo "they're the same"
else
    echo "they're different"
fi

或者，尝试使用 cmp 执行此操作：

if cmp &>/dev/null $f1 $f2; then
    echo "$f1 $f2 are the same"
else
    echo >&2 "$f1 $f2 are NOT the same"
fi

【讨论】：

【解决方案3】：

否定使用if ! diff -q $f1 $f2;。记录在man test:

! EXPRESSION
      EXPRESSION is false

不太清楚为什么需要否定，因为你处理这两种情况......如果你只需要处理它们不匹配的情况：

diff -q $f1 $f2 || echo "they're different"

【讨论】：