问题描述
我想打印某些顺序的N个元素,例如1,2,4,8 ...,而我试图使用N个子进程和N-1个管道来实现.因此,当进程"i"计算"a [i]"时,它将该值传递给进程i + 1,这样他就可以计算a [i + 1],依此类推...
I want to print N elements of some sequence, like 1,2,4,8... and I'm trying to do that with N child processes and N-1 pipes.So, when process "i" calculates "a[i]" it passes that value to process i+1 so he can calculate a[i+1] and so on...
我写了这个:
int main(){
int a = 1;
int i,j;
int cev[N-1][2];
for(i=0; i<N-1; i++) pipe(cev[i]);
for(i=0; i<N; i++){
if(fork() == 0){ // child
if(i>0){
read(cev[i][READ],&a,sizeof(int));
a = f(a); // calculate next element
}
printf("%d ",a); fflush(stdout);
if(i!=N-1) write(cev[i+1][WRITE],&a,sizeof(int));
// closing copies of pipes
for(j=0; j<N-1; j++){
close(cev[j][READ]);
close(cev[j][WRITE]);
}
exit(0);
}
}
对我来说似乎正确,但是我得到的N = 5的序列是2 1 2 48.Somebody.help.me.
it seems right to me,but the sequence I get for N=5 is 2 1 2 4 8. Somebody.help.me.
推荐答案
最后一个子进程( i == N-1
)读取 cev [N-1] [0]
超出了 cev
的末尾.类似地,进程 N-2
写入结束后的 cev [N-1] [1]
.您需要将所有管道索引偏移-1:
The last child process (i == N - 1
) reads cev[N - 1][0]
which is past the end of cev
. Similarly process N - 2
writes to cev[N - 1][1]
which is also past the end. You need to offset all of the pipe indices by -1:
if(i>0) {
read(cev[i-1][READ],&a,sizeof(int));
a = f(a); // calculate next element
}
printf("%d ",a); fflush(stdout);
if(i!=N-1) write(cev[i][WRITE],&a,sizeof(int));
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