如何从无效转换*回诠释 [英] How do I convert from void * back to int

问题描述

如果我有

  int类型的= 5;
长B = 10;
INT COUNT0 = 2;
无效** args0;
args0 =（无效**）的malloc（COUNT0 *的sizeof（无效*））;
args0 [0] =（无效*）及一个;
args0 [1] =（无效*）和b;
 

我怎样才能从ARGS [0]和args0 [1]回int和长转换？
例如：

  INT C =（IM东西丢失）args0 [0]
长D =（IM东西丢失）args1 [0]
 

解决方案

假设你和放大器; A0和＆放大器; B0被认为是与放大器A和和B，以及你的意思是args0 [1]设立长D，已存储的指针，在args0 [0]和指针args0 [1]为b。这意味着你需要将它们转换为正确的指针类型。

  INT C = *（（INT *）args0 [0]）;
INT D = *（（*长）args0 [1]）;
 

if I have

int a= 5;
long b= 10;
int count0 = 2;
void ** args0;
args0 = (void **)malloc(count0 * sizeof(void *));
args0[0] = (void *)&a;
args0[1] = (void *)&b;

how can I convert from args[0] and args0[1] back to int and long? for example

int c=(something im missing)args0[0]
long d=(something im missing)args1[0]

解决方案

Assuming that your &a0 and &b0 are supposed to be &a and &b, and that you mean args0[1] for setting up long d, you have stored a pointer to a in args0[0] and a pointer to b in args0[1]. This means you need to convert them to the correct pointer types.

int c = *((int *)args0[0]);
int d = *((long *)args0[1]);

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