如果我有一个空指针，我怎么把一个int成吗？ [英] If I have a void pointer, how do I put an int into it?

问题描述

我有任意值的数组，所以我把它定义为空指针数组，所以我可以指向任何类型的信息（如 INT ，性格阵列等）。然而，实际上，我怎么分配 INT 呢？

就拿这些初始化：

  void *的数据[10];
INT X = 100;
 

我的直觉会想到这一点，但是这给编译错误：

 数据[0] =的malloc（sizeof的（INT））;
*（数据[0]）= X;
 

我也想过用＆放大器; X ，但我会采取的局部变量，它（我的理解）将在程序退出后清除的地址。所以，如果我有一个局部变量 X ，我怎么会让它变成一个空指针的变量类型是否正确？

解决方案

  *（（INT *）数据[0]）= X;
 

将做到这一点。

您可能要考虑使用一个联盟。事情是这样的：

 工会myvalues
{
    INT I;
    双D;
    长升;
};
 

然后，您可以有

 工会myvalues​​ * foo的[10];
富[0] =的malloc（sizeof的（工会myvalues​​））;
富[0]  -  I标记= X;
 

您也可以的typedef 工会。 的sizeof（工会myvalues​​）将sizeof的成员的最大的。所以，如果你有 INT I; 和字符C [40] 在工会，的sizeof （联合myvalues​​）将40写入 I 然后将覆盖第4角色C （假设你的int为4个字节）。

I have an array of arbitrary values, so I have defined it as an array of void pointers, so I can point to any kind of information (like int, character arrays, etc). However, how do I actually assign an int to it?

Take for example these initializations:

void* data[10];
int x = 100;

My intuition would think this, but this gives a compile error:

data[0] = malloc(sizeof(int));
*(data[0]) = x;

Also I thought about using &x, but I would take the address of a local variable, which (to my understanding) would be cleared after exiting from the procedure. So if I have a local variable x, how would I get it into a void pointer type of variable correctly?

解决方案

*((int*)data[0])=x;

will do it.

You might want to consider using a union. Something like this:

union myvalues
{
    int i;
    double d;
    long l;
};

You could then have

union myvalues *foo[10];
foo[0] = malloc(sizeof(union myvalues));
foo[0]->i = x;

You can also typedef the union. sizeof(union myvalues) will be the maximum of sizeof the members. So if you have int i; and char c[40] in the union, sizeof(union myvalues) will be 40. Writing to i will then overwrite the first 4 characters in c (assuming your ints are 4 bytes).

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