c - C语言中简洁的动态内存分配

用伪代码，我想替换为：

ra             =  (double *)  realloc(ra,   Num*sizeof(*ra));
dec            =  (double *)  realloc(dec,  Num*sizeof(*dec));
zobs           =  (double *)  realloc(zobs, Num*sizeof(*zobs));
M_B            =  (double *)  realloc(M_B,  Num*sizeof(*M_B));

与

例如mem_allocate（＆ra，＆dec，＆zobs，＆M_B）

哪里

mem_allocate(arg1, arg2,..., arg4){

for argi in arg1 to arg4{
     argi = (double *)  realloc(argi, Num*sizeof(*argi))

}

}

最佳答案

为了完整起见，我举了一个例子

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

bool mem_allocate (double *vars[], size_t N_vars, size_t newWidth)
{
    double *temp;

    printf("\n");

    for (size_t i=0; i<N_vars; i++)
    {
        temp = realloc(vars[i], newWidth*sizeof(*temp));

        if (temp != NULL)
        {
            printf ("temp: %p\n", (void *)temp);
            vars[i] = temp;
        }
        else
        {
            return false;
        }
    }

    printf("\n");

    return true;
}

int main (void)
{
    double *var1 = malloc(sizeof(*var1));
    double *var2 = malloc(sizeof(*var2));
    double *var3 = malloc(sizeof(*var3));
    double *var4 = malloc(sizeof(*var4));

    printf ("var1: %p\n", (void *)var1);
    printf ("var2: %p\n", (void *)var2);
    printf ("var3: %p\n", (void *)var3);
    printf ("var4: %p\n", (void *)var4);

    double *vars[4];

    vars[0] = var1;
    vars[1] = var2;
    vars[2] = var3;
    vars[3] = var4;

    if (mem_allocate (vars, sizeof(vars)/sizeof(vars[0]), 200) == true)
    {
        var1 = vars[0];
        var2 = vars[1];
        var3 = vars[2];
        var4 = vars[3];

        printf ("var1: %p\n", (void *)var1);
        printf ("var2: %p\n", (void *)var2);
        printf ("var3: %p\n", (void *)var3);
        printf ("var4: %p\n", (void *)var4);
    }

    return 0;
}

在函数mem_allocate中更改varx意味着3星实现，我真的不喜欢。

显然，必须检查所有malloc返回值。

关于c - C语言中简洁的动态内存分配，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42116162/