Cuda char* 变量赋值

Question

这是这篇文章中所选答案的后续问题:Output of cuda program is not what was expected .

虽然以下功能有效:

__global__ void setVal(char **word)
{

    char *myWord = word[(blockIdx.y * gridDim.x) + blockIdx.x];
    myWord[0] = 'H';
    myWord[1] = 'e';
    myWord[2] = 'l';
    myWord[3] = 'l';
    myWord[4] = 'o';
}

为什么这行不通？

__global__ void setVal(char **word)
{

    char *myWord = word[(blockIdx.y * gridDim.x) + blockIdx.x];
    myWord = "Hello\0";

}

Answer 1

您应该开始更加注意编译器的输出。你的第二个内核代码:

__global__ void setVal(char **word)
{
    char *myWord = word[(blockIdx.y * gridDim.x) + blockIdx.x];
    myWord = "Hello\0";
}

编译成空内核，里面什么也没有:

$ nvcc -arch=sm_20 -c nullkernel.cu 
nullkernel.cu(3): warning: variable "myWord" was set but never used

nullkernel.cu(3): warning: variable "myWord" was set but never used

原因是因为你认为的字符串复制赋值实际上只是一个指针赋值，在这种情况下编译器足够聪明，知道 myWord 没有写入内存，所以它只是消除了所有代码并警告您未使用 myWord。

如果我要问一个反问并以这种方式重写代码:

__global__ void setVal(char **word)
{

    char *myWord = word[(blockIdx.y * gridDim.x) + blockIdx.x];
    const char[] mymsg = "Hello\0";
    myWord = mymsg;
}

为什么代码不能编译以及为什么它永远不能“隐式”执行字符串复制赋值，即使它编译了，会更明显吗？