C语言 创建4个子进程,分别求2500个浮点数之和

可以回答这个问题。C语言可以使用fork()函数创建子进程，然后在子进程中使用循环计算每个子进程需要计算的2500个浮点数之和，最后将结果返回给父进程。具体实现可以参考以下代码：

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/wait.h>

#define NUM_CHILDREN 4
#define NUM_FLOATS 2500

float floats[NUM_CHILDREN][NUM_FLOATS];

float sum(float *arr, int size) {
    float s = 0.0;
    for (int i = 0; i < size; i++) {
        s += arr[i];
    }
    return s;
}

int main() {
    pid_t pid;
    int status;
    float total = 0.0;

    // Generate random floats
    for (int i = 0; i < NUM_CHILDREN; i++) {
        for (int j = 0; j < NUM_FLOATS; j++) {
            floats[i][j] = (float) rand() / RAND_MAX;
        }
    }

    // Fork child processes
    for (int i = 0; i < NUM_CHILDREN; i++) {
        pid = fork();
        if (pid == 0) {
            // Child process
            float s = sum(floats[i], NUM_FLOATS);
            exit(s);
        } else if (pid < 0) {
            // Fork error
            perror("fork");
            exit(1);
        }
    }

    // Wait for child processes to finish
    for (int i = 0; i < NUM_CHILDREN; i++) {
        pid = wait(&status);
        if (WIFEXITED(status)) {
            // Child process exited normally
            total += WEXITSTATUS(status);
        } else {
            // Child process exited abnormally
            fprintf(stderr, "Child process %d exited abnormally\n", pid);
        }
    }

    printf("Total sum: %f\n", total);

    return 0;
}