使用MINLOC与Fortran语言：在分配不兼容行列0和1 [英] Using MINLOC with Fortran: Incompatible ranks 0 and 1 in assignment

问题描述

program hello
   integer a(9)
   integer index; ! note no dimension here
   a=(/1, 3, 4, 5, 6, 7, 8, 9, 10/)
   index = MINLOC(a, MASK=(a > 5))
   Print *, index
end program Hello

错误信息

main.f95：5.3：

Error message

main.f95:5.3:

指数= MINLOC（一，MASK =（> 5））
   1
错误：不兼容行列0和1的分配在（1）

index = MINLOC(a, MASK=(a > 5)) 1 Error: Incompatible ranks 0 and 1 in assignment at (1)

program hello
   integer a(9)
   integer index(1) ! note dimension 1 here which looks redundant at first
   a=(/1, 3, 4, 5, 6, 7, 8, 9, 10/)
   index = MINLOC(a, MASK=(a > 5))
   Print *, index
end program Hello

搜索

这里我能找到相关的讨论，但我不找不到它足够详细，我理解上的差异。

Search

Here I could find related discussion but I don't find it sufficiently verbose for me to understand the difference.

推荐答案

您可以修复的第一个版本，获得从 minloc A标回报，通过使用 DIM 参数：

You can fix the first version, obtaining a scalar return from minloc, by using the DIM argument:

index = MINLOC(a, DIM=1, MASK=(a > 5))

P.S。无需分号，除非你把每行多条语句结束语句。 Fortran语言不是C

P.S. No need for semicolons to end statements unless you place multiple statements per line. Fortran isn't C.

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