Question

我试图在Octave中使用递归来实现和自适应正交算法，并且我得到一个对我来说没有意义的数组越界错误。

function integral = adaptive(f, a, b, TOL, count)
max_count = 20;
h = (b-a)/2;
i1_midpoints = a:h:b; 
i2_midpoints = a:(h/2):b;
i1 = f(i1_midpoints(1)) + 4*f(i1_midpoints(2)) + f(i1_midpoints(3)); //line 15
i2 = f(i2_midpoints(1)) + 4*f(i2_midpoints(2)) + 2*f(i2_midpoints(3)) + 4*f(i2_midpoints(4)) + f(i2_midpoints(5));
if count == max_count
    integral = i1;
    return;
endif
disp(length(i1_midpoints));
disp(length(i2_midpoints));
disp(i1_midpoints);
disp(i2_midpoints);
if abs(i1 - i2) < (10 * TOL)
    integral = i1;
    return;
endif

count = count + 1;
adaptive(f, a, b/2, TOL, count); //line 31
adaptive(f, b/2, b, TOL, count); //line 32

错误讯息：

error: adaptive: A(I): index out of bounds; value 1 out of bound 0
error: called from:
error:   C:\Users\ADam\Desktop\Adam\Numerical_Analysis\adaptive.m at line 15, column 4
error:   C:\Users\ADam\Desktop\Adam\Numerical_Analysis\adaptive.m at line 31, column 1
error:   C:\Users\ADam\Desktop\Adam\Numerical_Analysis\adaptive.m at line 32, column 1

31行和32行不访问数组，只是进行递归调用。第15行只是使用访问数组的不同元素，其长度为5，因此，为什么要么给出超出范围的错误也没有意义。

Octave中的自适应正交实现 - 越界误差

0 个答案: