我的流程编程老师让我在C上编写了一个程序，它创建了四个孩子并让他们分别计算一系列数字的第一，第二，第三和第四个四分之一，给父母他们所有的素数。 / p>
我正确编码了第一个子季度，但是当我添加第二个孩子时，程序的行为无法控制。我的老师和我花了大约2个小时查看代码的深度，但我们没有发现问题。
代码是这样的，因为我现在拥有它：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void main(){
unsigned long long a=500000,b,c; // not used yet -> d,e,i;
pid_t pid1;
unsigned long long fin = 0; //This is used in each child to write if it has finished the prime number calculation.
unsigned long long fin1 = 0, fin2 = 0; //This is used on the parent to check if a child has finished.
int primo = 0; //This is used to know if a number is a prime number.
int fd1[2]; //Pipe which communicates the parent with the first child.
int fd2[2]; //Pipe which communicates the parent with the second child.
// int fd3[2]; //Not used yet
// int fd4[4]; //Not used yet
pipe(fd1); //First child pipe
pipe(fd2); //Second child pipe
// pipe(fd3); //Not used yet
// pipe(fd4); //Not used yet
pid1 = fork(); //Creating first child
switch (pid1){
case -1: //Error
printf("Error creating child.");
exit(-1);
case 0: //First child
close(fd1[0]); //Input close
for(b=100;b<(a/4);b++){ //
for(i=2;i<b/2;i++){ // These loops check each number from 100 to 125000
if(b%i==0){ // and if it is NOT a prime number, it breaks and tries
primo=0; // to check the next number.
break; //
} //
primo=1; //
}
if(primo==1){ //If it IS a prime number, it's written on the pipe
write(fd1[1], &b, sizeof(b)); //and sent to the parent.
}
}
fin=1; //The child sets it has finished calculating and writes it in the pipe to tell his parent.
write(fd1[1], &fin, sizeof(fin));
close(fd1[1]); //Output closing
break; //First child ends
default: //Parent
pid1 = fork(); //Creating second child
switch (pid1) {
case -1: //error
printf("Error");
exit(-1);
case 0: //Sencond child
close(fd2[0]); //This behavior is EXACTLY equals to the first child behavior
for(c=(a/4);c<(a/2);c++){ //
q for(i=2;i<c/2;i++){ //
if(c%i==0){ //
primo=0; //
break; //
} //
primo=1; //
} //
if(primo==1){ //
write(fd2[1], &c, sizeof(c)); //
} //
} //
fin=1; //
write(fd2[1], &fin, sizeof(fin)); //
close(fd2[1]); //
break;
default: //Parent
//HERE WOULD COME THE CODE FOR THIRD AND FOURTH CHILDS.
break;
} //second child switch close
//Parent reads answers from childs
close(fd1[1]); //First child output closing
close(fd2[1]); //Second child output closing
for(;;){ //Infinite loop
if(fin1==0){ //If first child HAS NOT finished (As it sends a 1 if it does)
read(fd1[0], &b, sizeof(b)); //Read the prime number
if(b==1){ //If it is a 1, then the child has finished.
fin1=1; //We set the first child has finished
close(fd1[0]); //First child input closing
}else{
printf("%llu es primo\n", b); //Otherwise it is a prime number, then it's printed to console.
}
}
if(fin2==0){ //Same behavior as with first child
read(fd2[0], &c, sizeof(c));
if(c==1){
fin2=1;
close(fd2[0]);
}else{
printf("%llu es primo\n", c);
}
}
if(fin1==1&&fin2==1){ //If both childs have finished, then we exit.
exit(0);
}
}
break;
}
exit(0);
}
它似乎是正确的，但它无法正常工作。当第二个孩子完成计算其数字范围（从125000到249999）时，它会阻止第一个孩子，第一个孩子就会停止。
然后程序进入读取和打印管道内容的无限循环 它看起来像这样：
[first child last calculated number] es primo
250000 es primo
[first child last calculated number] es primo
250000 es primo
[first child last calculated number] es primo
250000 es primo
[first child last calculated number] es primo
250000 es primo
等等。因此，我们询问如何将250000写入管道并从父母处读取，以及为什么第二个孩子完成阻止第一个孩子。
问候。
答案 0 :(得分：0)
在第67行，你的代码突破了第二个switch语句（第45行）。执行从第77行恢复。因此子进程2试图关闭（fd2 [1]）两次。子进程2正在执行以父进程为目标的代码。
您可以尝试用
替换第67行exit(0);