Question

我需要创建一个程序，使用fork（）

计算先前通过 main 参数声明的一个的阶乘数。

对于葡萄牙语中的printfs，请忽略它们，重点是除去在参数上设置的数字并使用两个子项进行计算，之后使用父进程打印结果。但是我无法继续处理我对孩子们的影响，因为他们是不同的过程，任何人都可以帮我解决这个问题吗？

遵循我到目前为止的代码：

#include <stdio.h> 
#include <stdlib.h>
#include <sys/types.h>

void filho_01();
void filho_02();

int num=0, total1=1,total2=1, cont=0,num2=0,valor,resultado=0;

int main(int agrc, char** argv) { 
    pid_t var1, var2;
    printf("Calculo de Fatorial com Fork \n\n");
    printf("Processo-Pai: Iniciando Execucao.\n\n");
    printf("Valor Obtido atraves do Parametro\n");

    num = atoi(argv[1]); 
    num2=num/2;
    valor=num;

    var1=fork();
    if ( var1 == 0 ) filho_01();
    var2=fork();
    if ( var2 == 0 ) filho_02();
    waitpid(var1,&total1,0);
    waitpid(var2,&total2,0);

    printf("%d",total1);
    resultado=total1*total2;


    printf("\nProcesso-Pai: Encerrando Execucao.\n");  
    printf(" \n   Fatorial de %d = %d \n",valor, resultado); 

}

void filho_01() {
printf("\n\n Filho 1 iniciando:\n");
printf("Calculando...\n");
sleep(3);


    for (cont = num; num >num2; num--) {    
        printf("  filho01: valor = %d \n", num);
        total1=total1*num;
    }

         exit(0);
}


void filho_02() {
printf("\n\n Filho 2 iniciando:\n");
printf("Calculando...\n");
sleep(5);


    for (cont = num2; num2 >=1; num2--) {    
        printf("  filho02: valor = %d \n", num2);
        total2=total2*num2;
    }
       exit(0);
}

Answer 1

所以几天后我做到了！我没有评论代码，因为我想它很容易理解（对于那个很抱歉）。我确实使用管道进行进程通信。

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

int product = 1; // global scope variable

//  prototypes
int factorial(int, int);

int factorial(int start, int n) {
  int product_ = 1;
  
  for ( int i = start; i <= n; i++ ) {
    product_ *= i;
  }

  return product_;
}

int main( void ) {
    siginfo_t infop;
    pid_t firstSon, secondSon;
    __id_t fSon = firstSon, sSon_ = secondSon;
    int fd[2], n, fStatus, sStatus;
    int middle, factor;

    printf( "%s", "Enter a number: " );
    scanf( "%d", &n );

    if ( n < 2 ) {
        printf( "\nFactorial of %d is %s\n", n, "1" );
        return EXIT_SUCCESS;
    }
    if ( pipe( fd ) < 0 ) {
        fprintf( stderr, "\nerror pipe" );
        exit( EXIT_FAILURE );
    }
    else if ( (firstSon = fork()) < 0 || (secondSon = fork()) < 0 ) {
        fprintf( stderr, "\nfork error\n" );
        exit( EXIT_FAILURE  );
    }
    else if ( firstSon == 0 ) {
        middle = n / 2;
        factor = factorial( 1, middle );
        printf( "\n%s: %d", "First process result", factor );
        close( fd[0] );
        write( fd[1], &factor, sizeof( int ) );
    }
    else if ( secondSon == 0 ) {
        middle = n / 2 + 1;
        factor = factorial( middle, n );
        printf( "\n%s: %d", "Second process result", factor );
        close( fd[0] );
        write( fd[1], &factor, sizeof( int ) );
    }
    else {
        waitid(P_PID, fSon, &infop, WEXITED);
        close( fd[1] );
        read( fd[0], &factor, sizeof( int ) );
        product *= factor;
        waitid(P_PID, sSon_, &infop, WEXITED);
        close( fd[1] );
        read( fd[0], &factor, sizeof( int ) );
        product *= factor;
        printf( "\nFactorial of %d is %d\n", n, product );
    }

    return EXIT_SUCCESS;
}

Answer 2

儿童不能修改他们的父母。在子项中，total变量与父项中的变量分开。你根本无法以这种方式与父母沟通。使用显式共享内存或其他一些RPC机制。

使用C语言中的fork（）进行因子计算

2 个答案: