ifort 中可能存在的错误
Possible bug in ifort
尝试在 Fortran 代码中使用分块对角矩阵类型时，
我在使用以下编译器的以下代码中偶然发现了一个令人惊讶的错误：
GNU Fortran (SUSE Linux) 7.4.0
ifort (IFORT) 18.0.5 20180823
ifort (IFORT) 16.0.1 20151021
如果我编译
gfortran -Wall -Werror --debug ifort_bug.f && valgrind ./a.out
我没有从 valgrind 报告错误。
如果我编译
ifort -warn all,error -debug -stacktrace ifort_bug.f && valgrind ./a.out
我在我的代码中遇到 ifort_18 的分段错误和 ifort_16 的 "only" 内存泄漏。
这是英特尔编译器中的错误，还是 gfortran 默默地修复了我的错误代码？
module blockdiagonal_matrices
implicit none
private
public :: t_blockdiagonal, new, delete,
& blocksizes, operator(.mult.), mult
save
integer, parameter :: dp = kind(1.d0)
type :: t_blockdiagonal
real(dp), allocatable :: block(:, :)
end type
interface new
module procedure block_new
end interface
interface delete
module procedure block_delete
end interface
interface operator (.mult.)
module procedure mult_blocks
end interface
contains
subroutine block_new(blocks, blocksizes)
type(t_blockdiagonal), intent(out) :: blocks(:)
integer, intent(in) :: blocksizes(:)
integer :: i, L
do i = 1, size(blocks)
L = blocksizes(i)
allocate(blocks(i)%block(L, L))
end do
end subroutine
subroutine block_delete(blocks)
type(t_blockdiagonal) :: blocks(:)
integer :: i
do i = 1, size(blocks)
deallocate(blocks(i)%block)
end do
end subroutine
function blocksizes(A) result(res)
type(t_blockdiagonal), intent(in) :: A(:)
integer :: res(size(A))
integer :: i
res = [(size(A(i)%block, 1), i = 1, size(A))]
end function
function mult_blocks(A, B) result(C)
type(t_blockdiagonal), intent(in) :: A(:), B(:)
type(t_blockdiagonal) :: C(size(A))
integer :: i
call new(C, blocksizes=blocksizes(A))
do i = 1, size(A)
C(i)%block = matmul(A(i)%block, B(i)%block)
end do
end function
subroutine mult(A, B, C)
type(t_blockdiagonal), intent(in) :: A(:), B(:)
type(t_blockdiagonal) :: C(:)
integer :: i
do i = 1, size(A)
C(i)%block = matmul(A(i)%block, B(i)%block)
end do
end subroutine
end module blockdiagonal_matrices
program time_blockdiagonal
use blockdiagonal_matrices
integer, parameter :: n_blocks = 2, L_block = 10**2
type(t_blockdiagonal) :: A(n_blocks), B(n_blocks), C(n_blocks)
integer :: i
integer :: start, finish, rate
call system_clock(count_rate=rate)
call new(A, blocksizes=[(L_block, i = 1, n_blocks)])
call new(B, blocksizes=[(L_block, i = 1, n_blocks)])
call new(C, blocksizes=[(L_block, i = 1, n_blocks)])
do i = 1, n_blocks
call random_number(A(i)%block)
call random_number(B(i)%block)
end do
call system_clock(start)
C = A .mult. B
call system_clock(finish)
write(6,*) 'Elapsed Time in seconds:',
& dble(finish - start) / dble(rate)
call system_clock(start)
call mult(A, B, C)
call system_clock(finish)
write(6,*) 'Elapsed Time in seconds:',
& dble(finish - start) / dble(rate)
call delete(A)
call delete(B)
call delete(C)
end program time_blockdiagonal
ifort_18 段错误是：
forrtl: severe (174): SIGSEGV, segmentation fault occurred
Image PC Routine Line Source
a.out 00000000004134BD Unknown Unknown Unknown
libpthread-2.26.s 00007FF720EB6300 Unknown Unknown Unknown
a.out 000000000040ABB8 Unknown Unknown Unknown
a.out 000000000040B029 Unknown Unknown Unknown
a.out 0000000000407113 Unknown Unknown Unknown
a.out 0000000000402B4E Unknown Unknown Unknown
libc-2.26.so 00007FF720B0AF8A __libc_start_main Unknown Unknown
a.out 0000000000402A6A Unknown Unknown Unknown
当我用 gdb 进入它时，结果发现，从函数 mult_blocks.
返回时会引发段错误
blockdiagonal_matrices::mult_blocks (c=..., a=..., b=...) at ifort_bug.f:63
63 do i = 1, size(A)
(gdb) s
64 C(i)%block = matmul(A(i)%block, B(i)%block)
(gdb) s
s
s
65 end do
(gdb) s
64 C(i)%block = matmul(A(i)%block, B(i)%block)
(gdb) s
65 end do
(gdb) s
66 end function
(gdb) s
Program received signal SIGSEGV, Segmentation fault.
0x000000000040abc4 in do_deallocate_all ()
(gdb) q
A debugging session is active.
即使有了这些信息，我也找不到我的代码中的错误。
编辑
我找到了一个修复程序，虽然我不明白它为什么有效。
编译修复
如果我使用 -heap-arrays 进行编译，它就可以工作。所以乍一看好像运行进了一个Whosebug。如果我这样做 ulimit -s unlimited 它 并没有 解决问题。
代码修复
如果我在代码中显式分配它就可以解决问题。
subroutine new(blocks, blocksizes)
type(t_blockdiagonal), allocatable, intent(out) :: blocks(:)
integer, intent(in) :: blocksizes(:)
integer :: i, L
allocate(blocks(size(blocksizes)))
do i = 1, size(blocks)
L = blocksizes(i)
allocate(blocks(i)%block(L, L))
end do
end subroutine
subroutine delete(blocks)
type(t_blockdiagonal), allocatable :: blocks(:)
integer :: i
do i = 1, size(blocks)
deallocate(blocks(i)%block)
end do
deallocate(blocks)
end subroutine
function mult_blocks(A, B) result(C)
type(t_blockdiagonal), intent(in) :: A(:), B(:)
type(t_blockdiagonal), allocatable :: C(:)
integer :: i
call new(C, blocksizes=blocksizes(A))
do i = 1, size(A)
C(i)%block = matmul(A(i)%block, B(i)%block)
end do
end function
恕我直言，这种写法实际上比以前更好，而不是 "dirty hack"。
无法再使用 blocksize 和 blocks.
的不同大小调用 new
未决问题
说 C type(t_blockdiagonal) :: blocks(n)
应该是 float** blocks[n] 所以只是指向指针的指针向量。
实际块的分配也发生在堆上的第一个版本中。因此，我没有得到包含 ca 的向量的 Whosebug。 10 个指针。
Intel® Fortran Compiler - Increased stack usage of 8.0 or higher compilers causes segmentation fault
根据这篇文章，ulimit -s unlimited并不意味着堆栈会字面上的"unlimited":
The size of "unlimited" varies by Linux configuration, so you may need to specify a larger, specific number to ulimit (for example, 999999999). On Linux also note that many 32bit Linux distributions ship with a pthread static library (libpthread.a) that at runtime will fix the stacksize to 2093056 bytes regardless of the ulimit setting.
考虑到 -heap-arrays 解决了问题，很可能您确实遇到了堆栈溢出问题。
尝试在 Fortran 代码中使用分块对角矩阵类型时， 我在使用以下编译器的以下代码中偶然发现了一个令人惊讶的错误：
GNU Fortran (SUSE Linux) 7.4.0
ifort (IFORT) 18.0.5 20180823
ifort (IFORT) 16.0.1 20151021
如果我编译
gfortran -Wall -Werror --debug ifort_bug.f && valgrind ./a.out
我没有从 valgrind 报告错误。
如果我编译
ifort -warn all,error -debug -stacktrace ifort_bug.f && valgrind ./a.out
我在我的代码中遇到 ifort_18 的分段错误和 ifort_16 的 "only" 内存泄漏。
这是英特尔编译器中的错误，还是 gfortran 默默地修复了我的错误代码？
module blockdiagonal_matrices
implicit none
private
public :: t_blockdiagonal, new, delete,
& blocksizes, operator(.mult.), mult
save
integer, parameter :: dp = kind(1.d0)
type :: t_blockdiagonal
real(dp), allocatable :: block(:, :)
end type
interface new
module procedure block_new
end interface
interface delete
module procedure block_delete
end interface
interface operator (.mult.)
module procedure mult_blocks
end interface
contains
subroutine block_new(blocks, blocksizes)
type(t_blockdiagonal), intent(out) :: blocks(:)
integer, intent(in) :: blocksizes(:)
integer :: i, L
do i = 1, size(blocks)
L = blocksizes(i)
allocate(blocks(i)%block(L, L))
end do
end subroutine
subroutine block_delete(blocks)
type(t_blockdiagonal) :: blocks(:)
integer :: i
do i = 1, size(blocks)
deallocate(blocks(i)%block)
end do
end subroutine
function blocksizes(A) result(res)
type(t_blockdiagonal), intent(in) :: A(:)
integer :: res(size(A))
integer :: i
res = [(size(A(i)%block, 1), i = 1, size(A))]
end function
function mult_blocks(A, B) result(C)
type(t_blockdiagonal), intent(in) :: A(:), B(:)
type(t_blockdiagonal) :: C(size(A))
integer :: i
call new(C, blocksizes=blocksizes(A))
do i = 1, size(A)
C(i)%block = matmul(A(i)%block, B(i)%block)
end do
end function
subroutine mult(A, B, C)
type(t_blockdiagonal), intent(in) :: A(:), B(:)
type(t_blockdiagonal) :: C(:)
integer :: i
do i = 1, size(A)
C(i)%block = matmul(A(i)%block, B(i)%block)
end do
end subroutine
end module blockdiagonal_matrices
program time_blockdiagonal
use blockdiagonal_matrices
integer, parameter :: n_blocks = 2, L_block = 10**2
type(t_blockdiagonal) :: A(n_blocks), B(n_blocks), C(n_blocks)
integer :: i
integer :: start, finish, rate
call system_clock(count_rate=rate)
call new(A, blocksizes=[(L_block, i = 1, n_blocks)])
call new(B, blocksizes=[(L_block, i = 1, n_blocks)])
call new(C, blocksizes=[(L_block, i = 1, n_blocks)])
do i = 1, n_blocks
call random_number(A(i)%block)
call random_number(B(i)%block)
end do
call system_clock(start)
C = A .mult. B
call system_clock(finish)
write(6,*) 'Elapsed Time in seconds:',
& dble(finish - start) / dble(rate)
call system_clock(start)
call mult(A, B, C)
call system_clock(finish)
write(6,*) 'Elapsed Time in seconds:',
& dble(finish - start) / dble(rate)
call delete(A)
call delete(B)
call delete(C)
end program time_blockdiagonal
ifort_18 段错误是：
forrtl: severe (174): SIGSEGV, segmentation fault occurred
Image PC Routine Line Source
a.out 00000000004134BD Unknown Unknown Unknown
libpthread-2.26.s 00007FF720EB6300 Unknown Unknown Unknown
a.out 000000000040ABB8 Unknown Unknown Unknown
a.out 000000000040B029 Unknown Unknown Unknown
a.out 0000000000407113 Unknown Unknown Unknown
a.out 0000000000402B4E Unknown Unknown Unknown
libc-2.26.so 00007FF720B0AF8A __libc_start_main Unknown Unknown
a.out 0000000000402A6A Unknown Unknown Unknown
当我用 gdb 进入它时，结果发现，从函数 mult_blocks.
blockdiagonal_matrices::mult_blocks (c=..., a=..., b=...) at ifort_bug.f:63
63 do i = 1, size(A)
(gdb) s
64 C(i)%block = matmul(A(i)%block, B(i)%block)
(gdb) s
s
s
65 end do
(gdb) s
64 C(i)%block = matmul(A(i)%block, B(i)%block)
(gdb) s
65 end do
(gdb) s
66 end function
(gdb) s
Program received signal SIGSEGV, Segmentation fault.
0x000000000040abc4 in do_deallocate_all ()
(gdb) q
A debugging session is active.
即使有了这些信息，我也找不到我的代码中的错误。
编辑
我找到了一个修复程序，虽然我不明白它为什么有效。
编译修复
如果我使用 -heap-arrays 进行编译，它就可以工作。所以乍一看好像运行进了一个Whosebug。如果我这样做 ulimit -s unlimited 它 并没有 解决问题。
代码修复
如果我在代码中显式分配它就可以解决问题。
subroutine new(blocks, blocksizes)
type(t_blockdiagonal), allocatable, intent(out) :: blocks(:)
integer, intent(in) :: blocksizes(:)
integer :: i, L
allocate(blocks(size(blocksizes)))
do i = 1, size(blocks)
L = blocksizes(i)
allocate(blocks(i)%block(L, L))
end do
end subroutine
subroutine delete(blocks)
type(t_blockdiagonal), allocatable :: blocks(:)
integer :: i
do i = 1, size(blocks)
deallocate(blocks(i)%block)
end do
deallocate(blocks)
end subroutine
function mult_blocks(A, B) result(C)
type(t_blockdiagonal), intent(in) :: A(:), B(:)
type(t_blockdiagonal), allocatable :: C(:)
integer :: i
call new(C, blocksizes=blocksizes(A))
do i = 1, size(A)
C(i)%block = matmul(A(i)%block, B(i)%block)
end do
end function
恕我直言，这种写法实际上比以前更好，而不是 "dirty hack"。
无法再使用 blocksize 和 blocks.
new
未决问题
说 C type(t_blockdiagonal) :: blocks(n)
应该是 float** blocks[n] 所以只是指向指针的指针向量。
实际块的分配也发生在堆上的第一个版本中。因此，我没有得到包含 ca 的向量的 Whosebug。 10 个指针。
Intel® Fortran Compiler - Increased stack usage of 8.0 or higher compilers causes segmentation fault
根据这篇文章，ulimit -s unlimited并不意味着堆栈会字面上的"unlimited":
The size of "unlimited" varies by Linux configuration, so you may need to specify a larger, specific number to ulimit (for example, 999999999). On Linux also note that many 32bit Linux distributions ship with a pthread static library (libpthread.a) that at runtime will fix the stacksize to 2093056 bytes regardless of the ulimit setting.
考虑到 -heap-arrays 解决了问题，很可能您确实遇到了堆栈溢出问题。