julia - 为什么 Julia 不在条件中打印 `false`？

Question

Julia 1.4.2
难倒我的行为发生在第一种和第四种情况:

julia> if (val = (1 == 2))
           println(val)
       end

julia> if (val = (1 == 1))
           println(val)
       end
true

julia> if (val = (1 == 1))
           println(val + 5)
       end
6

julia> if (val = (1 == 2))
           println(val + 5)
       end

julia> 1 == 2
false

julia> 1 == 1
true

julia> println(1==2)
false

julia> println(1==1)
true

我在 REPL 和 Jupyter 中观察到了这一点。
我的问题是:

为什么会发生这种情况？

如何检索 val 的值，最好是“真实”值，在这两种情况下？

Answer 1

让我们更详细地看一下第一个示例

# initially, `val` does not exist
julia> val
ERROR: UndefVarError: val not defined

# 1==2 is false
# the assignment returns the value of the right-hand side -> false
julia> if (val = (1 == 2))
           # therefore, this branch is not taken
           println(val)
       else
           # but this one is
           println("in the else clause")
       end
in the else clause

# `if` returns whatever the branch taken evaluates to.
# In this case, `println` returns nothing, so the `if` block does not
# return anything either
julia> @show ans
ans = nothing

# `if` doesn't introduce a new scope, so the newly created `val` variable
# can still be inspected
julia> val
false