数组衰变成指针 [英] Arrays decaying into pointers

问题描述

请帮助我了解下面的节目。

 ＃包括LT＆;＆stdio.h中GT;
诠释的main（）
{
    诠释一个[7];
    一个[0] = 1976;
    一个[1] = 1984;
    的printf（一个内存位置：％P，一）;
    的printf（内存位置％，p值％D，A，* A）;
    的printf（内存位置％，P值为％d，＆安培; A [1]，A [1]）;
    返回0;
}
 

＆放大器;一个[1] 和＆安培A + 1 。他们是相同或不同的？

 的#include＆LT;＆stdio.h中GT;
诠释的main（）
{
    INT v [10];
    INT ** P;
    为int *一个[5];
    v [0] = 1234;
    v [1] = 5678;
    一个[0] = v的
    一个[1] = V + 1;
    的printf（％d个\\ t％d个\\ t％d个\\ t％d个\\ N，* A [0]，*一[1]中，[0] [0]，**一）;
    的printf（％d个\\ N的sizeof（V））;
    返回0;
}
 

我想知道如何 *一个[5] 重新$ P $内存psented。为 *一个基类指针指向 A [0]，A [1]，A [2]中，[3]，A [ 4] ？

 ＃包括LT＆;＆stdio.h中GT;诠释的main（）
{
    INT v [10];
    INT ** P;
    INT（*一）[10];
    A =＆放大器;伏;
    的printf（％d个\\ N，*一）;
        返回0;
}
 

A = V; //提示错误，为什么？确实 v 这里衰变为 * V 。然后执行＆放大器; v 获取衰变成（*）[]垂直？ ＆安培;意味着常量指针。在这里，怎么可能设置一个const指向一个非const指针，未作类型转换？

在什么地方得到的数组存储在内存中。它得到存储到存储器的数据段。

 ＃包括LT＆;＆stdio.h中GT;诠释的main（）
{
    INT CARRAY [5] = {1,2,3,4,5};
    的printf（％d个\\ N，CARRAY [0]）;
    的printf（％d个\\ t％d个\\ t％d个\\ N的sizeof（CARRAY）的sizeof（安培; CARRAY）的sizeof（安培; CARRAY [0]））;
    返回0;
}
 

编辑：

我已经经历了其中一些规定的物品不见了，凡数组名称不能decyed到指针只有两种可能的情况是的sizeof 和＆安培; 。但是，在上述程序的sizeof（安培; CARRAY）给出大小为4和＆放大器; CARRAY 衰变为（*）[] CARRAY 作为它的右值。

然后就是数组名不能得到腐朽为指针两个条件语句的sizeof 和＆安培; 成为这里假

解决方案

  

＆放大器;一个[1] 和＆安培A + 1 。他们是相同或不同的？

不同。 ＆放大器;一个[1] 相同（A + 1）。一般情况下， X [Y] 是定义等同于 *（X + Y）。

  

我想知道如何 *一个[5] 重新$ P $内存psented。请问 *一个是一个基
  指针指向 A [0]，A [1]，A [2]，A [3]，A [4]

在你的第二个例子， A 是指针数组。 * A [I] 是对象的值，其地址被存储为I ^个元素的数组中为止。 *一个在这种情况下是一样的 A [0] ，这是你的数组中的第一个元素（是一个指针）。

  

A = V //为什么这给误差

由于 A （在你的最后一个例子）是一个指向数组的指针。要分配到 A ，那么你需要指定数组 v （或任何其他数组的地址正确的尺寸）;

  A =安培;伏;
 

这是非常好的，你已经COMMITED理解的事情，但没有什么可以帮助你比的一个好的C书。

希望这有助于。

Please help me understand the programs below.

#include<stdio.h>
int main()
{
    int a[7];
    a[0] = 1976;
    a[1] = 1984;
    printf("memory location of a: %p", a); 
    printf("value at memory location %p is %d", a, *a); 
    printf("value at memory location %p is %d", &a[1], a[1]);
    return 0;
}

&a[1] and &a+1. Are they same or different?

#include <stdio.h> 
int main() 
{
    int v[10];
    int **p; 
    int *a[5];
    v[0] = 1234;
    v[1] = 5678;
    a[0] = v; 
    a[1] = v+1;
    printf("%d\t%d\t%d\t%d\n", *a[0],*a[1],a[0][0],**a);
    printf("%d\n", sizeof(v));
    return 0;
} 

I wanted to know how *a[5] is represented in memory. Is *a a base pointer that points to a[0],a[1],a[2],a[3],a[4]?

#include<stdio.h>

int main()
{
    int v[10];
    int **p;
    int (*a)[10];
    a=&v;
    printf("%d\n",*a);
        return 0;
}

a=v; // gives error why? does v here decay into *v. Then does &v get decayed into (*)[]v? & means const pointer. Here, how is it possible to set a const pointer to a non-const pointer without a typecast?

Where does the array get stored in the memory. Does it get stored onto the data segment of the memory.

#include<stdio.h>

int main()
{
    int carray[5]={1,2,3,4,5};
    printf("%d\n",carray[0]);
    printf("%d\t%d\t%d\n",sizeof(carray),sizeof(&carray),sizeof(&carray[0]));
    return 0;
}

EDITED:

I have gone through some of the articles which stated that the only two possible situations where an array name cannot be decyed into pointer is the sizeof and &. But in the above program sizeof(&carray) gives the size as 4. and &carray decays into (*)[]carray as its an rvalue.

Then the statement that array name cannot get decayed into pointers on two conditions sizeof and & becomes false here.

解决方案

&a[1] and &a+1. Are they same or different?

Different. &a[1] is the same as (a+1). In general, x[y] is by definition equivalent to *(x+y).

I wanted to know how *a[5] is represented in memory. Does *a is a base pointer that points to a[0],a[1],a[2],a[3],a[4].

In your second example, a is an array of pointers. *a[i] is the value of the object, the address of which is stored as the i^th element in your array. *a in this case is the same as a[0], which is the first element in your array (which is a pointer).

a=v //why this gives error

Because a (in your last example) is a pointer to an array. You want to assign to a, then you need to assign the address of the array v (or any other array with correct dimensions);

a = &v;

This is very good that you've commited to understanding things, but nothing will help you better than a good C book.

Hope this helps.

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