C中的主下标值错误

Question

我是 C 新手，我试图从某些值（px 和 py）计算动能，但我的例程总是指责主要错误 下标值既不是数组也不是指针向量也不是。任何人都可以帮助我，因为我知道我已经尝试过。

动力学函数出现错误

谢谢

我的日常

#include <stdio.h>
#include <stdlib.h>

#ifndef RAN2_H_
#define RAN2_H_

void UniformBox(long, long *, double, double, double *, double *, double *, double *);

void kinetic(long, double, double, double); 

#endif /* RAN2_H_ */

#include "ran2.h"

float ran2(long*);

void UniformBox(long n, long *idum, double L, double p0, double *rx, double *ry, double *px, double *py){


    long i;

    for(i = 0 ; i < n ; i++){
        rx[i] = ((double)ran2(idum))*L;
        ry[i] = ((double)ran2(idum))*L;
        px[i] = ((double)ran2(idum) -.5)*2*p0;
        py[i] = ((double)ran2(idum) -.5)*2*p0;
    }



    return;
}


void kinetic(long n, double x, double y, double kint){

    long i;

    for(i = 0 ; i < n ; i++){

        kint[i] = (x*x)+(y*y);
    }

    return;
 }


void UniformBox(long, long *, double, double, double *, double *, double *, double *);

void kinetic(long n, double x, double y, double kint);


int main(){

    long i, n, seed, idum;
    double *rx, *ry, *px, *py;
    double L, p0, kint;

    n = 1000;
    L = 2.0;
    p0 = 1.22;
    seed = 10;
    idum = -seed;

    FILE *init = fopen("initialPosition.dat", "w");

    rx = (double *) malloc((double) n * sizeof(double));
    ry = (double *) malloc((double) n * sizeof(double));
    px = (double *) malloc((double) n * sizeof(double));
    py = (double *) malloc((double) n * sizeof(double));

    UniformBox(n, &idum, L, p0, rx, ry, px, py);
    kinetic(n, px, py, kint);
    for(i = 0 ; i < n ; i++){
        printf("%lf\t%lf\t%lf\t%lf\t%lf\n", rx[i], ry[i], px[i], py[i], kint[i]);
        fprintf(init,"%lf\t%lf\t%lf\t%lf\n", rx[i], ry[i], px[i], py[i]);
    }

    fclose(init);

    free(rx);
    free(ry);
    free(px);
    free(py);
    return 0;
}

Answer 1

我已经修复了编译错误。有几点需要注意。

我删除了多个不兼容的函数原型。

除了kinetic 的kint 需要是指针之外，x 和y 需要是指针。请注意，这是我的推测，基于代码的整体。还要注意循环内部从x 到x[i] 的变化。否则，[对我来说]事情没有意义。

在main 中，我为kint 添加了malloc

无论如何，这里是代码[请原谅无偿的风格清理]：

#include <stdio.h>
#include <stdlib.h>

#ifndef RAN2_H_
#define RAN2_H_

float ran2(long *);

#endif /* RAN2_H_ */

//#include "ran2.h"

void
UniformBox(long n, long *idum, double L, double p0, double *rx, double *ry,
    double *px, double *py)
{

    long i;

    for (i = 0; i < n; i++) {
        rx[i] = ((double) ran2(idum)) * L;
        ry[i] = ((double) ran2(idum)) * L;
        px[i] = ((double) ran2(idum) - .5) * 2 * p0;
        py[i] = ((double) ran2(idum) - .5) * 2 * p0;
    }

    return;
}

void
kinetic(long n, double *x, double *y, double *kint)
{

    long i;

    for (i = 0; i < n; i++) {
        kint[i] = (x[i] * x[i]) + (y[i] * y[i]);
    }

    return;
}

int
main()
{

    long i, n;
    long seed, idum;
    double *rx, *ry;
    double *px, *py;
    double L, p0;
    double *kint;

    n = 1000;
    L = 2.0;
    p0 = 1.22;
    seed = 10;
    idum = -seed;

    FILE *init = fopen("initialPosition.dat", "w");

    rx = (double *) malloc((double) n * sizeof(double));
    ry = (double *) malloc((double) n * sizeof(double));
    px = (double *) malloc((double) n * sizeof(double));
    py = (double *) malloc((double) n * sizeof(double));
    kint = (double *) malloc((double) n * sizeof(double));

    UniformBox(n, &idum, L, p0, rx, ry, px, py);
    kinetic(n, px, py, kint);

    for (i = 0; i < n; i++) {
        printf("%lf\t%lf\t%lf\t%lf\t%lf\n", rx[i], ry[i], px[i], py[i], kint[i]);
        fprintf(init, "%lf\t%lf\t%lf\t%lf\n", rx[i], ry[i], px[i], py[i]);
    }

    fclose(init);

    free(rx);
    free(ry);
    free(px);
    free(py);

    return 0;
}

Answer 2

在 C 中，您必须通过引用将数组传递给函数。因此，你的函数头应该是

void kinetic(long n, double x, double y, double * kint);

您也可以使用 cmets 中建议的 [] 表示法。