Question

我正在尝试完成一项任务，要求我编写一个函数a，其中b和n是指向长度为#include<stdio.h> #define N 10 double inner_product(const double *a, const double *b, int n) { int sum=0; int row= (int)&a; int column=(int)&b; const double ar[row][column]; const double *p; for(p=&ar[0][0]; p<&ar[n][n]; p++) {sum += *p;} return sum; } int main() { const double *a, *b; int row= (int)&a; int column=(int)&b; int N; double x; const double *p, ar[row][column]; printf("Enter the %d elements of each row and column seperated by a comma:\n",N); for(p=&ar[0][0];p<&ar[N][N];p++) {scanf("%d%d",&row, &column);} x=inner_product(a,b, N); printf("the inner product of the matrix is: %lf", x); return 0; }的数组的指针。此函数将仅使用指针算法找到定义为$ \ sum_ {i = 0} ^ n i ^ 2 = a_i * b_i $的内积。

以下是我提出的代码。它不编译并给出错误。有人可以帮我修复此错误并建议对此代码进行任何改进。

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Answer 1

正如我在评论中所说，你需要从头开始。这是一个重写的程序来完成这项工作。 assessment = getIntent().getExtras().getParcelable("assessment");函数只是因为它有调试打印代码（注释掉）和三种可选的算术方法（其中两个也被注释掉），因此适度大。核心非常简单。

inner_product()

打印功能很有用;您可以设计每五个数字输出换行符的变体，或每行输出任意数量的数字。打印测试数据很重要;它确保程序正在处理您期望的数据。

输出（启用调试）：

#include <stdio.h>

#define N 10

static double inner_product(const double *a, const double *b, int n)
{
    double sum = 0;
    for (int i = 0; i < n; i++)
    {
        /*printf("%8.3f * %8.3f = %14.6f  ", *a, *b, *a * *b);*/
        // Alternative ways of computing the sum
        // sum += *(a+i) * *(b+i);
        // sum += a[i] * b[i];
        sum += *a++ * *b++;
        /*printf("CS = %15.6f\n", sum);*/
    }
    return sum;
}

static void print_vector(const char *tag, const double *v, int n)
{
    printf("Vector %s (%d):\n", tag, n);
    for (int i = 0; i < n; i++)
        printf("%9.3f", v[i]);
    putchar('\n');
}

int main(void)
{
    // random -n 10 -F '%7.3f' -- -1000 1000 | commalist -B 8 -n 5 -W 8
    double a[N] =
    {
         -78.533,   52.153, -632.825, -196.897,  157.031,
        -804.630, -787.420, -281.817, -963.641,  561.922,
    };
    double b[N] =
    {
         112.700,  -83.580, -294.587,  320.394,   -8.366,
         218.917,  148.115,  421.533, -691.338, -741.578,
    };

    print_vector("A", a, N);
    print_vector("B", b, N);

    double x = inner_product(a, b, N);
    printf("The inner product of the vectors is: %lf\n", x);

    return 0;
}

输出（禁用调试）：

Vector A (10):
  -78.533   52.153 -632.825 -196.897  157.031 -804.630 -787.420 -281.817 -963.641  561.922
Vector B (10):
  112.700  -83.580 -294.587  320.394   -8.366  218.917  148.115  421.533 -691.338 -741.578
 -78.533 *  112.700 =   -8850.669100  CS =    -8850.669100
  52.153 *  -83.580 =   -4358.947740  CS =   -13209.616840
-632.825 * -294.587 =  186422.018275  CS =   173212.401435
-196.897 *  320.394 =  -63084.617418  CS =   110127.784017
 157.031 *   -8.366 =   -1313.721346  CS =   108814.062671
-804.630 *  218.917 = -176147.185710  CS =   -67333.123039
-787.420 *  148.115 = -116628.713300  CS =  -183961.836339
-281.817 *  421.533 = -118795.165461  CS =  -302757.001800
-963.641 * -691.338 =  666201.641658  CS =   363444.639858
 561.922 * -741.578 = -416708.992916  CS =   -53264.353058
The inner product of the vectors is: -53264.353058

我通过编写脚本来驱动Vector A (10): -78.533 52.153 -632.825 -196.897 157.031 -804.630 -787.420 -281.817 -963.641 561.922 Vector B (10): 112.700 -83.580 -294.587 320.394 -8.366 218.917 148.115 421.533 -691.338 -741.578 The inner product of the vectors is: -53264.353058重复计算来测试它;它给出了相同的答案（在我调试脚本之后）。

二维数组：内积编译错误

1 个答案: