Question

我试图创建使用realloc完成的动态增长数组。我在下面有示例，但是我不了解使用malloc的代码是如何工作的。

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

int main(void) {
    int cnt = 0;
    double *numbers = NULL;
    double newnum;
    while (scanf("%lf", &newnum) == 1 && newnum != -1) {
        double *newarr = (double*) malloc(sizeof(double) * (cnt+1)); 
        for (int i = 0; i < cnt; ++i)
            newarr[i] = numbers[i]; 
        free(numbers); 
        numbers = newarr;
        numbers[cnt] = newarr; 
        ++cnt;
    }

    for (int i = cnt-1; i >= 0; --i) {
        printf("%f\n", numbers[i]);
    }

    free(numbers);

    return 0;
}

Answer 1

realloc()与malloc()，memcpy()，free()都是同一功能（*）。

代码中的循环有效地替换了memcpy()

        // memcpy(newarr, numbers, cnt * sizeof *newarr);
        for (int i = 0; i < cnt; ++i)
            newarr[i] = numbers[i];

^{（*） realloc()可能很聪明，可以避免使用memcpy()和free()并重新使用内存。}

如何在我的代码中实现呢？

它有助于将指针和全部/已使用元素封装在结构中。

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

struct DynArray {
    double *data;
    size_t m; // total
    size_t n; // used
};

void growarray(struct DynArray *x) {
    size_t newsize = x->m * 13 / 8 + 1; // use phi aproximation 13/8
    double *newarr = realloc(x->data, newsize * sizeof *x->data);
    if (!newarr) exit(EXIT_FAILURE);
    fprintf(stderr, "INFO: realloc'ed with %d elements.\n", (int)newsize);
    x->data = newarr;
    x->m = newsize;
}

int main(void) {
    struct DynArray numbers = { 0 };
    double newnum;
    while (scanf("%lf", &newnum) == 1 && newnum != -1) {
        if (numbers.n == numbers.m) growarray(&numbers);
        numbers.data[numbers.n] = newnum;
        numbers.n++;
    }

    for (int i = numbers.n - 1; i >= 0; --i) {
        printf("%f\n", numbers.data[i]);
    }

    free(numbers.data);

    return 0;
}

See code running on ideone或previous version without growarray function或previous version with cnt

C用malloc动态增长数组

1 个答案: