当我只需要大约100 real（8）的内存时，就会发生Fortran运行时错误“内存分配失败”。
我正在用Fortran编写一些代码。我使用了几个（大约十个）可分配的real（8）数组。分配它们时，命令行显示“内存分配失败”。但是我只将它们的大小分配给大约十。我已经对计算机进行了简要检查，但我认为计算机无法处理100个浮点数。
代码是这样的：
module m
contains
function foo (a, b, c) result (...)
implicit none
real(8), allocatable, intent(inout) :: a(:), b(:), c(:)
allocate(a(0:10)) ! Actual size include an integer n passed to here. I'm pretty sure n is no greater then 10
allocate(b(1:10))
allocate(c(0:9))
! Some calculation
a_variable = bar(...)
endfunction foo
function bar (...) result (x)
implicit none
real(8), allocatable :: x(:)
allocate(x(8)) ! This is line X
! Some calculation
endfunction bar
endmodule m
program testm
use m
implicit none
! Some code that calls foo
endprogram testm
当我编译并运行代码时，命令行会告诉我“内存分配失败”，并显示来自操作系统的错误消息（Windows 10）“内存资源不足，无法完成此命令”。这真的很奇怪，因为我只要求大约100个浮点数的内存。
当我尝试确保我没有意外地将大量数字传递给n时，甚至会更加令人困惑。
当我在X行中添加print*, n时，异常仍然存在，并且指示的行仍然是X行，即现在的print代码！
我完全不知道这可能是哪种错误。由于我是Fortran的初学者，所以不确定在这里是否进行了任何未定义的行为。
我正在使用g95在Windows 10上编译.f90文件。
这是完整的代码：
module SplineInterp
! MODULE COMMENT
! Cubic Spline Interpolation Module
! Does Cubic Spline Interpolation with natural boundary conditions
! All calculation is made in real(8). Parameter rk = 8 is defined for real kind specifications.
! In all places where an unexpected zero on the denominator appears, the program will stop with a message,
! usually "The matrix is singular!". A number n is considered to be zero if dabs(n) < e_min, e_min is a
! pre-defined parameter.
integer, parameter :: rk = 8
real(rk), parameter :: e_min = 1d-5
contains
function CH_SOLVE (a, b, c, f, n) result (x)
implicit none
! FUNCTION COMMENT
! Thomas Chasing Method, solves a tri-diagonal-matrix linear equation set Ax=f
! First does LU-decomposition and then solves the upper- and lower-triangular equations
! ARGUMENTS:
! a(2:n), b(1:n), c(1:n-1) :: elements of the coefficient matrix A. See lecture notes for convention
! f(1:n) :: the inhomogeneous term column-vector in the equation
! n :: size of coefficient matrix
! RETURNS:
! x :: solution to the equation
real(rk), allocatable, intent(inout) :: a(:)
real(rk), allocatable, intent(inout) :: b(:)
real(rk), allocatable, intent(in) :: c(:)
real(rk), allocatable, intent(inout) :: f(:)
integer, intent(in) :: n
! Iteration variable
integer :: i
real(rk), allocatable :: x(:)
print*, n
allocate(x(n))
! LU-decomposition
! because L has only one non-trivial diagonal line, we can multiply L^-1 to f during the process
! See lecture notes for formulae used
do i = 2, n
if (dabs(b(i - 1)) < e_min) then
stop "The matrix is singular!"
endif
! Calculate elements of L
a(i) = a(i) / b(i - 1)
! Calculate elements of U
b(i) = b(i) - a(i) * c(i - 1)
! Multiply L^-1 to f
f(i) = f(i) - a(i) * f(i - 1)
enddo
if (dabs(b(n)) < e_min) then
stop "The matrix is singular!"
endif
! Solve Ux=(L^-1 f) using front-iteration method
! See lecture notes for formulae used
x(n) = f(n) / b(n)
do i = n - 1, 1, -1
if (dabs(b(i)) < e_min) then
stop "The matrix is singular!"
endif
x(i) = (f(i) - c(i) * x(i + 1)) / b(i)
enddo
endfunction CH_SOLVE
function interp (x, y, n, xi) result (yi)
implicit none
! FUNCTION COMMENT
! Interpolates the values of y at points xi based on the given data points (x, y)
! Using Cubic-Spline Interpolation, solving the intermediate equation set with Thomas Method
! ARGUMENTS:
! x, y :: Known data points, should be one-dimensional arrays of same shape (0:n)
! n :: size of given datas points
! xi :: The points at which the interpolation is to be evaluated (assumed to be in increasing order)
! RETURNS:
! yi :: Interpolation results, an array with the same size of xi
real(rk), allocatable, intent(in) :: x(:), y(:), xi(:)
integer, intent(in) :: n
real(rk), allocatable :: yi(:)
! Intermediate variables
! h(0:n-1) :: length of intervals
! M(0:n) :: to be solved, second-order-derivatives of spline function at data points
! mu(2:n-1), l(1:n-2), diag2(1:n-1) :: coefficients in the tri-diagonal-matrix
! d(1:n-1) :: non-homogeneous term in the linear equation set
real(rk), allocatable :: h(:), M(:), mu(:), l(:), diag2(:), d(:)
! Iteration variable
integer :: i, j
print*, n
allocate(yi(size(xi)))
allocate(h(0 : n-1))
allocate(M(0 : n))
allocate(mu(2 : n-1))
allocate(l(1 : n-2))
allocate(diag2(1 : n-1))
allocate(d(1 : n-1))
! Construction of linear equation set for M
! Calculation of h, m & d, see lecture notes for formulae used
h(1) = x(2) - x(1)
do i = 2, n-1
h(i) = x(i + 1) - x(i)
! Used a trick here, calculate d(i) in two parts to save calculation time
d(i+1) = - (y(i + 1) - y(i)) / h(i)
d(i) = (d(i) - d(i + 1)) * 6 / (h(i - 1) + h(i))
mu(i) = h(i - 1) / (h(i - 1) + h(i))
if (i <= n-2) then
l(i) = h(i) / (h(i - 1) + h(i))
endif
enddo
! Do Thomas Method solving
diag2 = 2 ! Diagonal elements of coefficient matrix are all 2
M(1 : n - 1) = CH_SOLVE(mu, diag2, l, d, n-1)
! Boundary conditions
M(0) = 0
M(n) = 0
! Do evaluation
! Here j represents the index of interval where the current xi points is
j = 0
! Interpolation point is invalid if it's smaller than the smallest x
if (xi(1) < x(0) - e_min) then
stop "Invalid interpolation point!"
endif
do i = 1, size(xi)
! Locate the interval where xi(i) is
do while (xi(i) > x(j + 1) + e_min)
j = j + 1
! Invalid interpolation point if xi(i) > x(n)
if (j >= n) then
stop "Invalid interpolation point!"
endif
enddo
! See lecture notes for this formula
yi(i) = (x(j+1)-xi(i))**3*M(j)/(6*h(j))&
+(xi(i)-x(j))**3*M(j+1)/(6*h(j))&
+(y(j)-M(j)*h(j)**2/6)*(x(j+1)-xi(i))/h(j)&
+(y(j+1)-M(j+1)*h(j)**2/6)*(xi(i)-x(j))/h(j)
enddo
endfunction interp
endmodule SplineInterp
program DoSplineInterp
use SplineInterp
implicit none
! PROGRAM COMMENT
! A program to use to above algorithm to interpole the profile of an airplane wing
real(rk), allocatable :: x(:), y(:), xi(:), yi(:)
! Iteration variable
integer :: i
integer, parameter :: n = 9
allocate(x(0:9))
allocate(y(0:9))
allocate(xi(151))
allocate(yi(151))
x = (/0d0, 3d0, 5d0, 7d0, 9d0, 11d0, 12d0, 13d0, 14d0, 15d0/)
y = (/0d0, 1.2d0, 1.7d0, 2d0, 2.1d0, 2d0, 1.8d0, 1.2d0, 1d0, 1.6d0/)
! xi are evenly-distributed points with interval 0.1 between 0 and 15
do i = 1, 151
xi(i) = (i - 1) * 0.1
enddo
yi = interp(x, y, n, xi)
! Write results to file
open (1, file = "Spline.txt", status = "new")
write (1, *) "x:"
write (1, *) x
write (1, *) "y:"
write (1, *) y
write (1, *) "xi:"
write (1, *) xi
write (1, *) "yi:"
write (1, *) yi
close(1)
endprogram DoSplineInterp