Ç内存分配（malloc的）

问题描述

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int *p, n, i;
    printf("Enter the size of the array:");
    scanf("%d", &n);
    p = (int *)malloc(n * sizeof(int));
    for (i = 0; i < n; i++)
    {
        printf("\n Enter element %d:", i + 1);
        scanf("%d", &p[i]);
    }
    for (i = 0; i < n; i++)
        printf("\n %d", p[i]);
    return 0;
}

为什么我们需要写＆安培; 在scanf函数，如果它是没有必要的数组？
P 是指向所有内存空间让＆安培; P [I] 应该给指针的地址，但不是我们要存储的数据吗？
此外，如果我们写 * P [I] 在的printf ，它给出了一个错误，点是一个指针，所以我们要尊重它，并把数据存储在内存中的保留空间，但它不工作？即使我编译上面的程序，因为它是停止服用3个值作为输入工作后。

Why do we need to write & in the scanf, if it's an array it's not necessary?p is a pointer to all the memory spaces so &p[i] should give the address of the pointer but not where we want to store the data right?Also if we write *p[i] in printf, it gives an error, p is a pointer so we should deference it and store the data in the reserved space in the memory but it's not working? Even if I compile the above program as it is it stops working after taking 3 values as input.

推荐答案

下标运算符[]也进行提领。这一次进行索引和非关联化。 P [N] 等同于 *（P + N），即它去偏移和间接引用它。因此，＆安培; P [N] 等同于点+ N

The subscript operator [] also performs dereferencing. It performs indexing and dereferencing at once. p[n] is equivalent to *(p+n), i.e. it goes to an offset and dereferences it. Consequently, &p[n] is equivalent to p+n.

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