void *字面上是float，如何转换？ [英] void* is literally float, how to cast?

问题描述

所以我在我的C ++应用程序中使用这个C库，其中一个函数返回一个void *。现在我不是纯C的最尖锐的，但听说一个void *可以投射到几乎任何其他*类型。我也知道，我期望在这个函数的某个地方有一个浮点。

So I'm using this C library in my C++ app, and one of the functions returns a void*. Now I'm not the sharpest with pure C, but have heard that a void* can be cast to pretty much any other *-type. I also know that I expect a float at the end somewhere from this function.

所以我将void *转换为float *，并解除引用float *，crash。 darn！
我调试代码，在gdb让它评估（float）voidPtr 和低，看看这个值是我期望和需要！

So I cast the void* to a float* and dereference the float*, crash. darn!. I debug the code and in gdb let it evaluate (float)voidPtr and low and behold the value is what I expect and need!

但是等待，在编译期间是不可能的。如果我写 float number =（float）voidPtr; 它不编译，这是可以理解的。

But wait, it's impossible to the same during compile time. If I write float number = (float)voidPtr; it doesn't compile, which is understandable.

现在的问题，我如何让我的浮动出这个fricking void *？

So now the question, how do I get my float out of this fricking void*?

编辑：感谢H2CO3这被解决了，但我看到很多答案和评论和dissappering不相信我可以做（float）voidPtr在gdb。

Thanks to H2CO3 this was solved, but I see lots of answers and comments appearing and dissappering not believing that I could do (float)voidPtr in gdb. here is the screenshot.

推荐答案

尝试使用指针：

void *theValueAsVoidPtr = // whatever

float flt = *(float *)&theValueAsVoidPtr;

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