[英]Multiplication doesn't work well inside of for loops
它不起作用,我做了所有修復工作,但是每次我得到相同的答案時。我的功能運行良好,如我所見。 但是,當我嘗試將它們乘以for循環時,它不起作用,我得到的噸數為零。
我已經修復了5天以上,我不想更改函數中的數組,但是如果不能解決,那么我刪除數組,並添加更多函數
#include <stdio.h>
#include <conio.h>
#include <math.h>
int main(void) {
int noktalar[4][4][3] = {
{ {30, 50, 1}, {130, 40, 1}, {200, 40, 1}, {240, 45, 1} },
{ {35, 90, 1}, {100, 95, 1}, {220, 95, 1}, {245, 90, 1} },
{ {25, 160, 1}, {80, 170, 1}, {240, 150, 1}, {260, 140, 1} },
{ {30, 250, 1}, {130, 200, 1}, {220, 220, 1}, {300, 300, 1} }};
int i, j, a, b, k = 1;
float X, Y, Z;
float tX = 0, tY = 0, tZ = 0;
for (a = 0; a <= 10; a++) {
for (b = 0; b <= 10; b++) {
for (i = 0; i <= 3; i++) {
for (j = 0; j <= 3; j++) {
X = bezierx(a, b, i, j) * noktalar[i][j][2];
Y = beziery(a, b, i, j) * noktalar[i][j][2]; // this part doesnt work well...
Z = bezierz(a, b, i, j) * noktalar[i][j][2];
}//--j for--
}//--i for--
printf("%d::::: %f -- %f -- %f \n", k, X, Y, Z);
k++;
}// --b for--
}// --a for--
getch();
return 0;
}
// -----------------main-------------------
int bezierx(int a, int b, int i, int j) {
float Xx;
float u = (float)a / 10, w = (float)b / 10;
float Uu = 1 - u, Ww = 1 - w;
float uu[4] = { (float)pow(Uu, 3), (float)(3 * pow(Uu, 2) * u), (float)(3 * Uu * pow(u, 2)), (float)pow(u, 3) };
float ww[4] = { (float)pow(Ww, 3), (float)(3 * pow(Ww, 2) * w), (float)(3 * Ww * pow(w, 2)), (float)pow(w, 3) };
Xx = uu[i] * ww[j];
// printf(" x : %f -- \n ", Xx);
return (float)Xx;
}
int beziery(int a, int b, int i, int j) {
float Yx;
float u = (float)a / 10, w = (float)b / 10;
float Uu = 1 - u, Ww = 1 - w;
float uu[4] = { (float)pow(Uu, 3), (float)(3 * pow(Uu, 2) * u), (float)(3 * Uu * pow(u, 2)), (float)pow(u, 3) };
float ww[4] = { (float)pow(Ww, 3), (float)(3 * pow(Ww, 2) * w), (float)(3 * Ww * pow(w, 2)), (float)pow(w, 3) };
Yx = uu[i] * ww[j];
// printf("y : %f -- ", Yx);
return (float)Yx;
}
int bezierz(int a, int b, int i, int j) {
float Zx;
float u = (float)a / 10, w = (float)b / 10;
float Uu = 1 - u, Ww = 1 - w;
float uu[4] = { (float)pow(Uu, 3), (float)(3 * pow(Uu, 2) * u), (float)(3 * Uu * pow(u, 2)), (float)pow(u, 3) };
float ww[4] = { (float)pow(Ww, 3), (float)(3 * pow(Ww, 2) * w), (float)(3 * Ww * pow(w, 2)), (float)pow(w, 3) };
Zx = uu[i] * ww[j];
// printf("z : %f \n", Zx);
return (float)Zx;
}
您必須將bezierx
和friends定義為返回float
,然后在main
函數之前聲明它們或將其定義移到main
函數之前。
請注意,您可以簡化代碼,應免費使用double
精度算術而不是float
來提高精度。 為了避免精度損失,最好為這些簡單的整數冪寫x*x
和x*x*x
而不是使用pow
。
仔細閱讀后發現, bezierx
, beziery
和bezierz
沒有區別。 使用相同的功能。
這是修改后的版本:
#include <stdio.h>
#include <conio.h>
#include <math.h>
double bezier(int a, int b, int i, int j);
int main(void) {
int noktalar[4][4][3] = {
{ {30, 50, 1}, {130, 40, 1}, {200, 40, 1}, {240, 45, 1} },
{ {35, 90, 1}, {100, 95, 1}, {220, 95, 1}, {245, 90, 1} },
{ {25, 160, 1}, {80, 170, 1}, {240, 150, 1}, {260, 140, 1} },
{ {30, 250, 1}, {130, 200, 1}, {220, 220, 1}, {300, 300, 1} }};
int i, j, a, b, k = 1;
double X, Y, Z;
double tX = 0, tY = 0, tZ = 0;
for (a = 0; a <= 10; a++) {
for (b = 0; b <= 10; b++) {
for (i = 0; i <= 3; i++) {
for (j = 0; j <= 3; j++) {
X = bezier(a, b, i, j) * noktalar[i][j][0];
Y = bezier(a, b, i, j) * noktalar[i][j][1];
Z = bezier(a, b, i, j) * noktalar[i][j][2];
}
}
printf("%d::::: %f -- %f -- %f \n", k, X, Y, Z);
k++;
}
}
getch();
return 0;
}
double bezier(int a, int b, int i, int j) {
double Xx;
double u = a / 10.0, w = b / 10.0;
double Uu = 1 - u, Ww = 1 - w;
double uu[4] = { Uu * Uu * Uu, 3 * Uu * Uu * u, 3 * Uu * u * u, u * u * u };
double ww[4] = { Ww * Ww * Ww, 3 * Ww* Ww * w, 3 * Ww * w * w, w * w * w };
Xx = uu[i] * ww[j];
//printf(" bezier : %f -- \n ", Xx);
return Xx;
}
該代碼應進一步簡化: bezier
計算8個系數,其中只有2個系數用於最終結果。
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