指针到指针的语法问题
可以说我有以下内容：
void init_gpu(cuComplex* d_hhBuff)
{
cutilSafeCall(cudaMalloc((void **)&d_hhBuff, memsize));
}
我用类似的方式调用它
cuComplex *my_buff;
init_gpu(my_buff);
，当 init_gpu 返回时，它并不指向 cudaMalloc 分配的设备内存。
如何修改它，以便 init_gpu 的调用者将 my_buff 指向 cudaMalloc 创建的修改后的 d_hhBuff？
Lets say I have the following:
void init_gpu(cuComplex* d_hhBuff)
{
cutilSafeCall(cudaMalloc((void **)&d_hhBuff, memsize));
}
and I call it with something like
cuComplex *my_buff;
init_gpu(my_buff);
Well, when init_gpu returns, it is NOT pointing to the device memory that cudaMalloc allocated.
How do I modify this so that the caller of init_gpu will have my_buff pointing to the modified d_hhBuff that cudaMalloc creates?
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问题是您正在按值传递指针。将函数头更改为
The problem is that you are passing the pointer by value. Change the function header to
您的
d_hhBuf是本地副本。你应该做的是通过引用传递指针：Your
d_hhBufis a local copy. What you should do is pass the pointer by reference:假设您 typedef：
并编写一个函数：
并调用它：
通过查看 call 和函数（并且不知道 typedef），您会推断出什么？您可能会说，
ptr(cptr) 是按值传递的，而不是按引用/指针传递的，并且不会更改。现在考虑这个调用：
这与上面完全相同，您实际上传递的是
ComplexPointer类型 - 它只是按值传递。最后了解更改后的签名：
它仍然是相同的 - 您通过 VALUE 传递 type
cuComplex*，而不是通过引用！Assume you typedef:
And write a function:
And call it:
What would you deduce by looking at call and the function (and not knowing the typedef) ? You would say, the
ptr(cptr) is passed by value, and not by reference/pointer and won't change.Now consider this call:
This is exactly same as above, you are actually passing type
ComplexPointer- which just being passed by value.Finally understand the changed signature:
And it is still the same - you are pass the type
cuComplex*by VALUE, and not by reference!