如何让 gfortran 进行 INTEGER - LOGICAL 转换

发布于 2024-11-05 00:45:21 字数 537 浏览 0 评论 0原文

根据 this，gfortran 可以进行整数逻辑转换，但我收到此错误：

                  if (.not.bDropped.and.(zz_unif01() .lt. (1 - (Test_Dru
                           1
Error: Operand of .not. operator at (1) is INTEGER(4)

我知道最好将代码更改为.not.bDropped 到 (bDropped.eq.0)，但这并不简单，因为它是生成的代码。

我尝试了各种 -std=xxx 标志，但它们没有任何区别。

原文

According to this, gfortran can do integer-logical conversion, but I'm getting this error:

                  if (.not.bDropped.and.(zz_unif01() .lt. (1 - (Test_Dru
                           1
Error: Operand of .not. operator at (1) is INTEGER(4)

I know it would be better to change the code from .not.bDropped to (bDropped.eq.0), but that would not be simple because it's generated code.

I tried various -std=xxx flags but they made no difference.

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喵星人汪星人 2024-11-12 00:45:21

您链接到的页面上的最后一行内容

However, there is no implicit conversion of INTEGER values in if-statements, [...].

是我猜想这与它有关。

编辑：这似乎不是全部事实。简单地做l1 = .not。 0（当l1 是逻辑变量时）给出与您收到的相同错误。所以在这种情况下也没有隐式转换。

The last line on the page you linked to reads

However, there is no implicit conversion of INTEGER values in if-statements, [...].

I'd guess that has something to do with it.

Edit: This seems not to be the entire truth. Simply doing l1 = .not. 0 (when l1 is a logical variable) gives the same error you received. So there is no implicit conversion in this case either.

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