自适应积分问题 [英] Adaptive Quadrature issues
问题描述
#include< stdio.h>
#include< math.h>
double f（ double x）
{
return （1 /（x * sin（x * x）））;
// return（2 * x * x）;
}
int main（）
{
int i = 1 ，num = 20 ，N;
double b，APP，TOL [num]，a [num]，h [num]，FA [num]，FB [num]，FC [ num]，S [num]，S1，S2，L [num]，FD，FE，v [num];
printf（ 左端点：）;
scanf（ ％lf，& a）;
printf（ 右端点：）;
scanf（ ％lf，& b）;
printf（ Desired Tolerence：）;
scanf（ ％lf，& TOL）;
printf（ 级别数：）;
scanf（ ％d，& N）;
APP = 0 ;
TOL [num] = 10 .0 * TOL;
a [num] = a;
h [num] =（b-a）/ 2;
FA [num] = f（a）;
FC [num] = f（a + h [num]）;
FB [num] = f（b）;
S [num] = h [num] *（FA [num] + 4 * FC [num] + FB [num]）/ 3; // 辛普森一家近似
L [num] = 1 ;
while （i> 0）
{
FD = f（a [num] + h [num] / 2）;
FE = f（a [num] + 3 * h [num] / 2）;
S1 = h [num] *（FA [num] + 4 * FD + FC [num]）/ 6;
S2 = h [num] *（FC [num] + 4 * FE + FB [num]）/ 6;
v [ 1 ] = a [num];
v [ 2 ] = FA [num];
v [ 3 ] = FC [num];
v [ 4 ] = FB [num];
v [ 5 ] = h [num];
v [ 6 ] = TOL [num];
v [ 7 ] = S [num];
v [ 8 ] = L [num];
i = i-1;
if （abs（S1 + S2 - v [ 7 ]）< ; v [ 6 ]）
{
APP = APP +（S1 + S2）;
}
其他
{
如果（v [ 8 ] > = N）
{
printf（ 等级超过）;
return ;
}
else
{
i ++;
a [num] = v [ 1 ] + v [ 5 ];
FA [num] = v [ 3 ];
FC [num] = FE;
FB [num] = v [ 4 ];
h [num] = v [ 5 ] / 2;
TOL [num] = v [ 6 ] / 2;
S [num] = S2;
L [num] = v [ 8 ] + 1 ;
i ++;
a [num] = v [ 1 ];
FA [num] = v [ 2 ];
FC [num] = FD;
FB [num] = v [ 3 ];
h [num] = h [num-1];
TOL [num] = TOL [num-1];
S [num] = S1;
L [num] = L [num-1];
}
}
}
printf（ 积分近似是％lf。，APP）;
}
构建消息：
C：\Users\Mal \Desktop \MTH 415 \自适应正交2.c ||功能'main'：|
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 34 |错误：二进制操作数无效*（有'double'和'double *'）|
C：\ Users \ Maltion \Desktop \ MTH 415 \自适应正交2.c | 35 |错误：从类型'double *'分配类型'double'时出现不兼容的类型
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 36 |错误：二进制操作数无效 - （有'double'和'double *' ）|
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 38 |错误：'f'参数1的不兼容类型|
C：\ Users \ Maltion \Desktop \ MTH 415 \自适应正交2.c | 10 |注意：预期'双'但参数是类型'double *'|
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 39 |错误：无效操作数到二进制+（有' double *'和'double'）|
|| ===构建完成：5个错误，0个警告（0分0秒）=== |
之前我没有使用过数组，这可能是我问题的重要部分。我相信我至少走了很长一段路，因为在本学期之前我还不知道任何编程。谢谢你的帮助。
[edit]
我编辑了你所描述的程序：
#include < stdio.h >
＃ include < math.h >
＃define num 20
double f（ double x）
{
return （1 /（x * sin（x * x）））;
// return（2 * x * x）;
}
int main（）
{
int i = 1 ，N;
double b，APP，TOL [num]，a [num]，h [num]，FA [num]，FB [num]，FC [ num]，S [num]，S1，S2，L [num]，FD，FE，v [num];
printf（ 左端点：）;
scanf（ ％lf，& a）;
printf（ 右端点：）;
scanf（ ％lf，& b）;
printf（ Desired Tolerence：）;
scanf（ ％lf，& TOL）;
printf（ 级别数：）;
scanf（ ％d，& N）;
APP = 0 ;
TOL [num] = 10 。 0 * TOL;
a [num] = a;
h [num] =（b-a）/ 2;
FA [num] = f（a）;
FC [num] = f（a + h [num]）;
FB [num] = f（b）;
S [num] = h [num] *（FA [num] + 4 * FC [num] + FB [num]）/ 3; // 辛普森一家近似
L [num] = 1 ;
while （i> 0）
{
FD = f（a [num] + h [num] / 2）;
FE = f（a [num] + 3 * h [num] / 2）;
S1 = h [num] *（FA [num] + 4 * FD + FC [num]）/ 6;
S2 = h [num] *（FC [num] + 4 * FE + FB [num]）/ 6;
v [ 1 ] = a [num];
v [ 2 ] = FA [num];
v [ 3 ] = FC [num];
v [ 4 ] = FB [num];
v [ 5 ] = h [num];
v [ 6 ] = TOL [num];
v [ 7 ] = S [num];
v [ 8 ] = L [num];
i = i- 1 ;
if （abs（S1 + S2 - v [ 7 ]）< ; v [ 6 ]）> = N）
{
printf（ 等级超过）;
return ;
}
else
{
i ++;
a [num] = v [ 1 ] + v [ 5 ];
FA [num] = v [ 3 ];
FC [num] = FE;
FB [num] = v [ 4 ];
h [num] = v [ 5 ] / 2;
TOL [num] = v [ 6 ] / 2;
S [num] = S2;
L [num] = v [ 8 ] + 1 ;
i ++;
a [num] = v [ 1 ];
FA [num] = v [ 2 ];
FC [num] = FD;
FB [num] = v [ 3 ];
h [num] = h [num- 1 ];
TOL [num] = TOL [num- 1 ];
S [num] = S1;
L [num] = L [num- 1 ];
}
}
}
printf（ 积分近似是％lf。，APP）;
}
仍然获得：
C：\Users\Mal \Desktop \MTH 415 \自适应正交2.c ||在函数'main'中：
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 35 |错误：无效操作数到二进制*（有'double'和'double *'）|
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 36 |错误：从类型'double *'分配类型'double'时出现不兼容的类型
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 37 |错误：二进制操作数无效 - （有'double'和'double *'）|
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 39 |错误：'f'参数1的不兼容类型
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 11 |注意：预期'双'但参数类型为'double *'|
C：\ Users \ Maltion \Desktop \MTH 415 \自适应正交2.c | 40 |错误：二进制操作数无效+（有'double *'和'double'）|
|| ===构建完成：5个错误，0个警告（0分0秒）=== |< /math.h>< /stdio.h>
[/ edit]
int i = 1 ，num = 20 ，N;
double b，APP，TOL [num]，a [num]，h [num]，FA [num]，FB [num]，FC [ num]，S [num]，S1，S2，L [num]，FD，FE，v [num];
在这种情况下你不能使用变量来声明一个数组，你必须使用一个常量值，例如：
< pre lang =c ++> ＃define NUM $
int i = 1 ，N;
double b，APP，TOL [NUM]，[NUM]，h [NUM]，FA [NUM]，FB [NUM]，FC [ NUM]，S [NUM]，S1，S2，L [NUM]，FD，FE，v [NUM];
#include <stdio.h>
#include <math.h>
double f(double x)
{
return (1/(x*sin(x*x)));
//return (2*x*x);
}
int main()
{
int i=1, num=20, N;
double b, APP, TOL[num], a[num], h[num], FA[num], FB[num], FC[num], S[num], S1, S2, L[num], FD, FE, v[num];
printf("Left Endpoint: ");
scanf("%lf", &a);
printf("Right Endpoint: ");
scanf("%lf", &b);
printf("Desired Tolerence: ");
scanf ("%lf", &TOL);
printf("Number of Levels: ");
scanf("%d", &N);
APP = 0;
TOL[num] = 10.0*TOL;
a[num] = a;
h[num] = (b-a)/2;
FA[num] = f(a);
FC[num] = f(a+h[num]);
FB[num] = f(b);
S[num] = h[num]*(FA[num]+4*FC[num]+FB[num])/3; //Simpsons approximation
L[num] = 1;
while (i>0)
{
FD = f(a[num] + h[num]/2);
FE = f(a[num] + 3*h[num]/2);
S1 = h[num]*(FA[num]+4*FD+FC[num])/6;
S2 = h[num]*(FC[num]+4*FE+FB[num])/6;
v[1] = a[num];
v[2] = FA[num];
v[3] = FC[num];
v[4] = FB[num];
v[5] = h[num];
v[6] = TOL[num];
v[7] = S[num];
v[8] = L[num];
i=i-1;
if (abs(S1 + S2 - v[7])<v[6])
{
APP = APP + (S1 + S2);
}
else
{
if (v[8]>=N)
{
printf("Level Exceeded");
return;
}
else
{
i++;
a[num] = v[1] + v[5];
FA[num] = v[3];
FC[num] = FE;
FB[num] = v[4];
h[num] = v[5]/2;
TOL[num] = v[6]/2;
S[num] = S2;
L[num] = v[8] + 1;
i++;
a[num] = v[1];
FA[num] = v[2];
FC[num] = FD;
FB[num] = v[3];
h[num] = h[num-1];
TOL[num] = TOL[num-1];
S[num] = S1;
L[num] = L[num-1];
}
}
}
printf("The Integral Approximation is %lf." , APP);
}
Build Messages:
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c||In function 'main':|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|34|error: invalid operands to binary * (have 'double' and 'double *')|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|35|error: incompatible types when assigning to type 'double' from type 'double *'|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|36|error: invalid operands to binary - (have 'double' and 'double *')|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|38|error: incompatible type for argument 1 of 'f'|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|10|note: expected 'double' but argument is of type 'double *'|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|39|error: invalid operands to binary + (have 'double *' and 'double')|
||=== Build finished: 5 errors, 0 warnings (0 minutes, 0 seconds) ===|
I have not used arrays before, which is probably a big part of my problem. I believe I have at least come a long way since I did not know any programming before this semester. Thanks for your help.
[edit]
I edited my program as you described:
#include <stdio.h>
#include <math.h>
#define num 20
double f(double x)
{
return (1/(x*sin(x*x)));
//return (2*x*x);
}
int main()
{
int i=1, N;
double b, APP, TOL[num], a[num], h[num], FA[num], FB[num], FC[num], S[num], S1, S2, L[num], FD, FE, v[num];
printf("Left Endpoint: ");
scanf("%lf", &a);
printf("Right Endpoint: ");
scanf("%lf", &b);
printf("Desired Tolerence: ");
scanf ("%lf", &TOL);
printf("Number of Levels: ");
scanf("%d", &N);
APP = 0;
TOL[num] = 10.0*TOL;
a[num] = a;
h[num] = (b-a)/2;
FA[num] = f(a);
FC[num] = f(a+h[num]);
FB[num] = f(b);
S[num] = h[num]*(FA[num]+4*FC[num]+FB[num])/3; //Simpsons approximation
L[num] = 1;
while (i>0)
{
FD = f(a[num] + h[num]/2);
FE = f(a[num] + 3*h[num]/2);
S1 = h[num]*(FA[num]+4*FD+FC[num])/6;
S2 = h[num]*(FC[num]+4*FE+FB[num])/6;
v[1] = a[num];
v[2] = FA[num];
v[3] = FC[num];
v[4] = FB[num];
v[5] = h[num];
v[6] = TOL[num];
v[7] = S[num];
v[8] = L[num];
i=i-1;
if (abs(S1 + S2 - v[7])<v[6])>=N)
{
printf("Level Exceeded");
return;
}
else
{
i++;
a[num] = v[1] + v[5];
FA[num] = v[3];
FC[num] = FE;
FB[num] = v[4];
h[num] = v[5]/2;
TOL[num] = v[6]/2;
S[num] = S2;
L[num] = v[8] + 1;
i++;
a[num] = v[1];
FA[num] = v[2];
FC[num] = FD;
FB[num] = v[3];
h[num] = h[num-1];
TOL[num] = TOL[num-1];
S[num] = S1;
L[num] = L[num-1];
}
}
}
printf("The Integral Approximation is %lf." , APP);
}
Still Obtaining:
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c||In function 'main':|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|35|error: invalid operands to binary * (have 'double' and 'double *')|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|36|error: incompatible types when assigning to type 'double' from type 'double *'|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|37|error: invalid operands to binary - (have 'double' and 'double *')|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|39|error: incompatible type for argument 1 of 'f'|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|11|note: expected 'double' but argument is of type 'double *'|
C:\Users\Mal\Desktop\MTH 415\Adaptive Quadrature 2.c|40|error: invalid operands to binary + (have 'double *' and 'double')|
||=== Build finished: 5 errors, 0 warnings (0 minutes, 0 seconds) ===|</math.h></stdio.h>
[/edit]
int i=1, num=20, N; double b, APP, TOL[num], a[num], h[num], FA[num], FB[num], FC[num], S[num], S1, S2, L[num], FD, FE, v[num];
You cannot use a variable in this case to declare an array, you must use a constant value, like:
#define NUM 20 int i=1, N; double b, APP, TOL[NUM], a[NUM], h[NUM], FA[NUM], FB[NUM], FC[NUM], S[NUM], S1, S2, L[NUM], FD, FE, v[NUM];
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