ç的sizeof一个传递的数组 [英] C sizeof a passed array

问题描述

  

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/492384/how-to-find-the-sizeof-a-pointer-pointing-to-an-array\">How找到的sizeof（指针指向一个数组）

据我所知，sizeof操作符进行评估，并在编译时常数取代。鉴于此，怎么能一个功能，而在不同的点在程序中通过不同的阵列，有它的尺寸计算的？我可以把它作为一个参数的功能，但我宁愿没有添加其他的参数，如果我没有绝对要。

下面是说明我在问什么一个例子：

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;的#define SIZEOF（一）（sizeof的一个/ sizeof的一个[0]）
无效printarray（双X []，int）;在诠释的main（）
{
        双ARRAY1 [100];
        的printf（的数组1 =％LD \\ n的大小。SIZEOF（数组1））;
        的printf（N的数组1 =％LD \\的大小。，sizeof的数组1）;
        的printf（数组1的大小[0] =％LD \\ n \\ n，sizeof的ARRAY1 [0]）;        printarray（数组1，SIZEOF（数组1））;        返回EXIT_SUCCESS;
}
无效printarray（双P []，int类型）
{
        INT I;
        //这是没有按T工作，所以其结果是，我通过
        //大小作为第二个参数，这是我最好不做。
        的printf（的对计算值=％LD \\ n的大小。SIZEOF（P））;
        的printf（的P =％LD \\ n的大小。，sizeof的P）;
        的printf（P [0] =％大小LD \\ n，sizeof的P [0]）;        对于（i = 0; I＆LT; S;我++）
                的printf（％Eelement：D =％LF \\ n，I，P [I]）;        返回;
}
 

解决方案

有没有神奇的解决方案。 C不是反射语言。对象不自动知道它们是什么。

但是，你有很多选择：

1. 显然，添加参数


2. 编写一个宏来包装调用，并自动添加参数


3. 使用更复杂的对象。定义包含动态数组，也数组的大小的结构。然后，通过结构的地址。

Possible Duplicate:
How to find the sizeof( a pointer pointing to an array )

I understand that the sizeof operator is evaluated and replaced with a constant at compile time. Given that, how can a function, being passed different arrays at different points in a program, have it's size computed? I can pass it as a parameter to the function, but I'd rather not have to add another parameter if I don't absolutely have to.

Here's an example to illustrate what I'm asking:

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

#define SIZEOF(a) ( sizeof a / sizeof a[0] )


void printarray( double x[], int );

int main()
{
        double array1[ 100 ];


        printf( "The size of array1 = %ld.\n", SIZEOF( array1 ));
        printf( "The size of array1 = %ld.\n", sizeof array1 );
        printf( "The size of array1[0] = %ld.\n\n", sizeof array1[0] );

        printarray( array1, SIZEOF( array1 ) );

        return EXIT_SUCCESS;
}


void printarray( double p[], int s )
{
        int i;


        // THIS IS WHAT DOESN"T WORK, SO AS A CONSEQUENCE, I PASS THE 
        // SIZE IN AS A SECOND PARAMETER, WHICH I'D RATHER NOT DO.
        printf( "The size of p calculated = %ld.\n", SIZEOF( p ));
        printf( "The size of p = %ld.\n", sizeof p );
        printf( "The size of p[0] = %ld.\n", sizeof p[0] );

        for( i = 0; i < s; i++ )
                printf( "Eelement %d = %lf.\n", i, p[i] );

        return;
}

解决方案

There is no magic solution. C is not a reflective language. Objects don't automatically know what they are.

But you have many choices:

1. Obviously, add a parameter
2. Write a macro to wrap the call, and automatically add a parameter
3. Use a more complex object. Define a structure which contains the dynamic array and also the size of the array. Then, pass the address of the structure.

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