在函数中更改数组不会更改它在 main 中的值

Question

我尝试用介于 -0.8 和 0.8 之间的随机数填充一个 vector (我必须分配)。我的问题是为什么在主函数中当我调用函数 setvector() 时不返回 vector 并且我仍然使用零初始化？多谢。这是我所做的

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

void allocate(double **a, int size) {
    *a = malloc(size);
}

double setvector(double *v){
    int i, seed, send_size;

    send_size = 10;

    allocate(&v, send_size * sizeof(double)); // allocate memory for the vector
    seed = time(NULL);
    srand(seed);
    for (i = 0; i < send_size; i++)
    {
        v[i] = 1.6*((double) rand() / (double) RAND_MAX) - 0.8;
    }
    printf("Inside function the vector is:\n\n");
    for (i = 0; i < 10; i++)
    {
        printf("The %d element has the random %4.2f\n", i, v[i]);
    }
    return *v;
}

int main(){
    double *v = NULL;
    setvector(v);
    printf("\nThe vector from main is:\n\n");
    printf("The 1st element of v is %4.2f\n", &v[0]);
    printf("The 1st element of v is %4.2f\n", &v[1]);

    return 0;
}

这是我的屏幕输出:

Inside function the vector is:

The 0 element has the random -0.79
The 1 element has the random -0.34
The 2 element has the random 0.48
The 3 element has the random -0.67
The 4 element has the random -0.70
The 5 element has the random 0.61
The 6 element has the random -0.67
The 7 element has the random -0.66
The 8 element has the random -0.44
The 9 element has the random -0.36

The vector from main is:

The 1st element of v is 0.00
The 1st element of v is 0.00

Answer 1

在 main 中，您将数组元素的地址传递给 printf:

printf("The 1st element of v is %4.2f\n",&v[0]);
printf("The 1st element of v is %4.2f\n",&v[1]);

应该是 printf(..., v[0]);。

进一步:

double setvector(double *v){

不会改变 main 中的 v，所以 v 在那里保持 NULL。您应该让 setvector 获取一个 double**，例如 allocate，并将 v 的地址传递给它。