Fortran双精度程序,带有简单的MKL BLAS程序

Question

在尝试在一个简单的程序中混合精度 - 使用真实和双重 - 并使用BLAS的ddot例程时,我想出了双精度部分的错误输出.这是代码:

program test

!! adding this statement narrowed the issue down to ddot being considered real(4)
implicit none

integer, parameter :: dp = kind(1.0d0)

!! The following 2 lines were added for the calls to the BLAS routines.
!! This fixed the issue.
real(dp), external :: ddot
real, external :: sdot

real, dimension(3) :: a,b
real(dp), dimension(3) :: d,e

integer :: i

do i = 1,3
    a(i) = 1.0*i
    b(i) = 3.5*i
    d(i) = 1.0d0*i
    e(i) = 3.5d0*i
end do

write (*,200) "sdot real(4) = ", sdot(3,a,1,b,1)  ! should work and return 49.0
write (*,200) "ddot real(4) = ", ddot(3,a,1,b,1)  ! should not work

write (*,200) "sdot real(8) = ", sdot(3,d,1,e,1)  ! should not work
write (*,200) "ddot real(8) = ", ddot(3,d,1,e,1)  ! should work and return 49.0

200 format(a,f5.2)

end program test

我尝试使用MKL BLAS库对gfortran和ifort进行编译,如下所示:

ifort -lmkl_intel_lp64 -lmkl_sequential -lmkl_core

gfortran -lmkl_intel_lp64 -lmkl_sequential -lmkl_core main.f90

输出是:

sdot real(4) = 49.00
ddot real(4) =  0.00
sdot real(8) =  4.10
ddot real(8) =  0.00

如何让ddot例程正确处理双精度值？

此外,添加-autodouble标志(ifort)或-fdefault-real-8(gfortran)标志会使两个ddot例程都起作用,但sdot例程会失败.

编辑: 我添加了隐式none语句,以及ddot和sdot函数的两个类型语句.如果没有为函数调用指定的类型,则ddot将被隐式键入为单精度实数.

Answer 1

我没有使用MKL,但是你可能需要一个"use"语句,以便编译器知道函数的接口？或以其他方式声明功能.它们未被声明,因此编译器可能假设ddot的返回是单精度并且错误地解释了这些位.

打开警告选项会使编译器告诉您该问题.使用gfortran,尝试:-fimplicit-none -Wall -Wline-truncation -Wcharacter-truncation -Wsurprising -Waliasing -Wimplicit-interface -Wunused-parameter -fwhole-file -fcheck = all -std = f2008 -pedantic -fbacktrace