c - C中需要将全局变量转换为局部变量

Question

基本上，我草率地编写了一个 OpenCL 程序来使用这些全局变量进行分配:

int devType = CL_DEVICE_TYPE_GPU;

cl_int err;         /* Error code returned from api calls.  */
size_t global;      /* Global domain size for our calculation.  */
size_t local;           /* Local domain size for our calculation.  */

cl_platform_id cpPlatform;  /* openCL platform.  */
cl_device_id device_id; /* Compute device id.  */
cl_context context;     /* Compute context.  */
cl_command_queue commands;  /* Compute command queue.  */
cl_program program;     /* Compute program.  */
cl_kernel kernel;       /* Compute kernel.  */

/* Create data for the run.  */
float *data = NULL;     /* Original data set given to device.  */
float *results = NULL;  /* Results returned from device.  */
unsigned int correct;       /* Number of correct results returned.  */
cl_mem input;           /* Device memory used for the input array.  */
cl_mem output;      /* Device memory used for the output SUM.  */

int rc = EXIT_FAILURE;

现在我正在尝试将它们全部设为本地以整理程序。

我通过将全局变量 N 从上面的变量移到 main() 函数中来转换它。然后我更新了每个使用 N 的函数头，将“int N”作为参数，并将 N 传递给任何需要它作为参数的函数调用。该程序按预期运行。

所以我想我要问的是，对于这些变量的其余部分，会这么简单吗？我了解按引用和值传递的概念，并意识到某些函数可能会更改变量，因此我需要使用指针引用/取消引用。我担心的是我的指针理论有点粗糙，我担心会遇到问题。我也不确定我定义的函数是否可以采用所有这些 cl 变量。

另外，在函数中使用相同的变量名有什么问题吗？

编辑:

正如我担心的那样，在尝试本地化 device_id 时，以下函数确实出现了问题:

void deviceSetup(int devType) {
    cl_platform_id cpPlatform;  /* openCL platform.  */

    /* Connect to a compute device.  */
    if (CL_SUCCESS != clGetPlatformIDs (1, &cpPlatform, NULL))
        die ("Error: Failed to find a platform!");

    /* Get a device of the appropriate type.  */
    if (CL_SUCCESS != clGetDeviceIDs (cpPlatform, devType, 1, &device_id, NULL))
        die ("Error: Failed to create a device group!");
}

/* Create a compute context.  */
void createContext(cl_int err){
    context = clCreateContext (0, 1, &device_id, NULL, NULL, &err);
    if (!context || err != CL_SUCCESS)
        die ("Error: Failed to create a compute context!");
}

/* Create a command commands.  */
void createCommandQueue(cl_int err) {
    commands = clCreateCommandQueue (context, device_id, 0, &err);
    if (!commands || err != CL_SUCCESS)
        die ("Error: Failed to create a command commands!");
}       

void createAndCompile(cl_int err){
    /* Create the compute program from the source buffer.  */
    program = clCreateProgramWithSource (context, 1,
                                         (const char **) &KernelSource,
                                         NULL, &err);
    if (!program || err != CL_SUCCESS)
        die ("Error: Failed to create compute program!");

    /* Build the program executable.  */
    err = clBuildProgram (program, 0, NULL, NULL, NULL, NULL);
    if (err != CL_SUCCESS)
    {
        size_t len;
        char buffer[2048];

        clGetProgramBuildInfo (program, device_id, CL_PROGRAM_BUILD_LOG,
                               sizeof (buffer), buffer, &len);
        die ("Error: Failed to build program executable!\n%s", buffer);
    }
}

Answer 1

您确实回答了您自己的问题。是的，这真的就是它的全部。您可能需要考虑将大量相关 变量组合到一个结构中，如果您发现您已经为您的函数生成了大量参数列表，那么只传递一个指向该结构的指针，但仅此而已。 (与您传递给任何函数的参数数量有关的性能考虑很小，但我认为现在这是您可以避免的不必要的复杂程度!)

虽然 C 中的指针(通过引用传递的唯一方法)是无法避免的，因此像这样的小项目很可能是加强这些知识的理想时机!

好吧，让我们举个例子，这样解释生活总是更好。

我们有:

int cheddar;
int montereyjack;
int brie;

void print_cheeses(void)
{
    printf("I have %d cheddar %d montereyjack and %d brie\n", cheddar, montereyjack, brie);
}

void add_cheeses(void)
{
   cheddar = cheddar + 1;
   montereyjack = montereyjack + 1;
   brie = brie + 1;
   print_cheeses();
}

int main(int argc, char *argv[])
{
    add_cheeses();
    printf ("Now I have %d cheddars %d jacks %d bries\n", cheddar, montereyjack, brie);
}

我们需要做的是:

// By value here because we're not changing anything
void print_cheeses(int cheds, int jacks, int bries)
{
    printf("I have %d cheddar %d montereyjack and %d brie\n", cheds, jacks, bries);
}

// Pointers here because we need to change the values in main
void add_cheeses(int *cheese_one, int *cheese_two, int *cheese_three)
{
   *cheese_one = *cheese_one + 1; // We're following the pointer to get to the data we want to change
   *cheese_two = *cheese_two + 1;
   *cheese_three = *cheese_three + 1;
   print_cheeses(*cheese_one, *cheese_two, *cheese_three); // We're following the pointer to get to the data we want to print
}

int main(int argc, char *argv[])
{
    int cheddar = 0;
    int montereyjack = 0;
    int brie = 0;

    add_cheeses(&cheddar, &montereyjack, &brie);

    printf ("Now I have %d cheddars %d jacks %d bries\n", cheddar, montereyjack, brie);
}

但每次都传递所有三个值可能会很痛苦，并且由于它们是相关的，您可以将它们捆绑在一个结构中，然后只传递一个指向该结构的指针。