生成一段C++多线程流水线模型推理代码

以下是一段C多线程流水线模型推理代码的示例：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>

#define NUM_THREADS 4

typedef struct {
int *input;
int *output;
int size;
} PipelineData;

void *stage1(void *arg) {
PipelineData *data = (PipelineData *) arg;
for (int i = 0; i < data->size; i++) {
data->output[i] = data->input[i] * 2;
}
pthread_exit(NULL);
}

void *stage2(void *arg) {
PipelineData *data = (PipelineData *) arg;
for (int i = 0; i < data->size; i++) {
data->output[i] = data->input[i] + 1;
}
pthread_exit(NULL);
}

void *stage3(void *arg) {
PipelineData *data = (PipelineData *) arg;
for (int i = 0; i < data->size; i++) {
data->output[i] = data->input[i] - 3;
}
pthread_exit(NULL);
}

void *stage4(void *arg) {
PipelineData *data = (PipelineData *) arg;
for (int i = 0; i < data->size; i++) {
data->output[i] = data->input[i] / 2;
}
pthread_exit(NULL);
}

int main() {
int input[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int output[10];

pthread_t threads[NUM_THREADS];
PipelineData data[NUM_THREADS];

for (int i = 0; i < NUM_THREADS; i++) {
data[i].input = (i == 0) ? input : output;
data[i].output = (i == NUM_THREADS - 1) ? output : (int *) malloc(sizeof(int) * 10);
data[i].size = 10;

pthread_create(&threads[i], NULL, (i == 0) ? stage1 : (i == 1) ? stage2 : (i == 2) ? stage3 : stage4, (void *) &data[i]);
}

for (int i = 0; i < NUM_THREADS; i++) {
pthread_join(threads[i], NULL);
if (i != NUM_THREADS - 1) {
free(data[i].output);
}
}

for (int i = 0; i < 10; i++) {
printf("%d ", output[i]);
}
printf("\n");

return 0;
}