（警告 - 不是完美的英语！）
在拼贴画上，他们给了我一个任务，我在几个小时内挣扎，无法解决它。 我将数据从一个进程发送到另一个进程时出现问题。在下面的代码中，我将输入数据发送到子进程但我没有从子进程接收数据。你们可以帮帮忙吗？
程序描述：我的工作是从用户那里获取输入，使用管道将其发送到子进程，进行一些计算然后将此计算进一步发送到下一个进程。上一个进程对传入数据求和并将其分配给最终结果。更具体地说，第一个子进程计算Maclaurin系列的单个术语并将其发送到对其求和的进程。这是代码：
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <fcntl.h>
#include <string.h>
#include <math.h>
#include <sys/stat.h>
int mkfifo(const char *s, mode_t m) {
return 0;
}
int i = 1;
double FINAL_RESULT = 0 ;
double factorial(int N){
if(N==1)return 1.0;
return N*factorial(N - 1);
}
double Maclaurine(int i , double x , double eps){
double result;
result = (pow(x,(double)i) / (double)factorial(i));
if( result < eps) return 0;
return result;
}
void print(char * tab, int size){
printf("\n");
for( int i = 0; i < size ; i++){
printf("%c" , tab[i]);
}
printf("\n");
}
int main(){
int pd;
if(mkfifo("tmp/myfifo1", 0666) == -1){
printf("Failed to create myfifo2\n");
}
if(mkfifo("tmp/myfifo2", 0666) == -1){
printf("Failed to create myfifo2\n");
}
pd = open("tmp/myfifo1", O_WRONLY);
double x1,eps1;
printf("X Value : \n");
scanf("%lf", &x1);
printf("EPS Value : \n");
scanf("%lf", &eps1);
char tab1[128], tab2[128];
//converting to char*
sprintf(tab1,"%lf",x1);
sprintf(tab2,"%lf",eps1);
//printing to standard output
print(tab1, strlen(tab1));
print(tab2, strlen(tab2));
//creating new process
if(fork() == 0){
//creating another process
if(fork() == 0){
int pd3;
pd3 = open("tmp/myfifo2", O_RDONLY);
char tab5[128];
double x3;
//reading data from pipe , myfifo2
read(pd3,tab5,sizeof(tab5));
//converting char* to float
x3 = atof(tab5);
FINAL_RESULT += x3;
printf("\nCurrent result = %lf" , FINAL_RESULT);
}
int pd2;
//opening myfifo1 for reading
pd2 = open("tmp/myfifo1", O_RDONLY);
if(pd2 == - 1){
printf("\nFailed to open myfifo1 for reading!\n");
}
double x2, eps2;
char tab3[128], tab4[128];
//reading inputed parameters
read(pd2,tab3, sizeof(tab3) );
read(pd2,tab4, sizeof(tab4) );
//converting to floats
x2 = atof(tab3);
eps2 = atof(tab4);
printf("\nReceived initial data x=%f, eps = %f \n" , x2, eps2);
double singleTerm = 0;
int pd4;
char wyraz[128];
//opening myfifo2 for writing
pd4 = open("tmp/myfifo2", O_WRONLY);
if(pd4 == -1 ){
printf("\nfailed to open myfifo2\n");
}
do{
//calculate single term of Maclaurin series
singleTerm = Maclaurine(i,x2,eps2);
printf("\n%d. Series term = %lf\n", i,singleTerm);
sprintf(wyraz,"%lf",singleTerm);
if(write(pd4, wyraz, sizeof(wyraz)) == -1 ){
printf("\nfailed to send singleTerm to myfifo2\n");
}
i++;
//if its less than eps, loop ends
}while(singleTerm);
printf("\nFinal Result : %lf", FINAL_RESULT);
close(pd4);
close(pd2);
unlink("tmp/myfifo2");
}
//write data to myfifo1
write(pd, tab1, sizeof(tab1));
write(pd, tab2, sizeof(tab2));
close(pd);
unlink("tmp/myfifo1");
return 0;
}
这里是代码的结果：
pawel@IMPERIUMVB:~$ gcc pipes.c -lm -o x
pawel@IMPERIUMVB:~$ ./x
X Value :
10
EPS Value :
0.001
10.000000
0.001000
pawel@IMPERIUMVB:~$
Failed to open myfifo1 for reading!
Received initial data x=0.000000, eps = 0.000000
failed to open myfifo2
1. Series term = 0.000000
failed to send singleTerm to myfifo2
Final Result : 0.000000
Current result = 0.000000
Failed to open myfifo1 for reading!
Received initial data x=0.000000, eps = 0.000000
failed to open myfifo2
1. Series term = 0.000000
failed to send singleTerm to myfifo2
Final Result : 0.000000
答案 0 :(得分：0)
好的，我解决了这个问题，程序给了我预期的结果。 第一个关键因素是擦除
int mkfifo(const char *s, mode_t m) { return 0; }
就像Jonathan Leffler所说*“在你的程序中包含int mkfifo（const char s，mode_t m）{return 0;}意味着你不会创建FIFO而不能使用它们。你需要删除它以使用系统调用。“
第二件事是在lseek（）函数单次使用后反复开始我的fifo开始。此外，我必须在使用后使用memset（）清除所有缓冲区。
现在是我的代码：
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <fcntl.h>
#include <string.h>
#include <math.h>
#include <sys/stat.h>
int i = 0;
double FINAL_RESULT = 1.0 ;
double factorial(int N){
if(N==1 || N == 0)return 1.0;
return N*factorial(N - 1);
}
double Maclaurine(int i , double x , double eps){
double result;
result = (pow(x,(double)i) / (double)factorial(i));
if( result < eps) return -1.0;
return result;
}
void print(char * tab, int size){
printf("\n");
for( int i = 0; i < size ; i++){
printf("%c" , tab[i]);
}
printf("\n");
}
int main(){
umask(0);
if(mkfifo("myfifo1", 0666) == -1){
printf("Failed to create myfifo2\n");
}
if(mkfifo("myfifo2", 0666) == -1){
printf("Failed to create myfifo2\n");
}
if(fork() == 0){
int pd = open("myfifo1", O_WRONLY);
double x1,eps1;
char tab1[256];
//assigning tab1 data to 0's
lseek(pd,0,0);
memset(tab1, 0, 256);
printf("X Value : \n");
scanf("%lf", &x1);
sprintf(tab1,"%lf",x1);
//write data to myfifo1
write(pd, tab1, strlen(tab1));
printf("EPS Value : \n");
scanf("%lf", &eps1);
lseek(pd,0,0);
memset(tab1, 0, 256);
//converting to char*
sprintf(tab1,"%lf",eps1);
write(pd, tab1, strlen(tab1));
lseek(pd,0,0);
memset(tab1, 0, 256);
close(pd);
unlink("myfifo1");
//printing to standard output
//print(tab1, strlen(tab1));
//print(tab2, strlen(tab2));
//creating new process
}
else if(fork() == 0){
int pd2, pd3;
//opening myfifo1 for reading
pd2 = open("myfifo1", O_RDONLY);
//opening myfifo2 for writing
pd3 = open("myfifo2", O_WRONLY);
if(pd2 == - 1){
printf("\nFailed to open myfifo1 for reading!\n");
}
if(pd3 == -1 ){
printf("\nfailed to open myfifo2\n");
}
double x2, eps2;
char tab2[256];
lseek(pd2,0,0);
memset(tab2, 0, 256);
//reading inputed parameters
read(pd2,tab2, sizeof(tab2) );
//converting to floats
x2 = atof(tab2);
lseek(pd2,0,0);
memset(tab2, 0, 256);
read(pd2,tab2, sizeof(tab2) );
//converting to floats
eps2 = atof(tab2);
lseek(pd2,0,0);
memset(tab2, 0, 256);
//printf("\nReceived initial data x=%f, eps = %f \n" , x2, eps2);
double singleTerm = 0.0;
char wyraz[256];
while(1){
//calculate single term of Maclaurin series
singleTerm = Maclaurine(i,x2,eps2);
if(singleTerm > 0.0)printf("\n%d. Term of Mac Series = %lf", i+ 1,singleTerm);
sprintf(wyraz,"%lf",singleTerm);
lseek(pd3,0,0);
if(write(pd3, wyraz, sizeof(wyraz)) == -1 ){
printf("\nfailed to send singleTerm to myfifo2\n");
}
if(singleTerm < 0.0){
//printf("\nI just broke the 1 loop\n");
memset(wyraz, 0, 256);
break;
}
i=i+1;
}
sleep(1);
close(pd3);
close(pd2);
unlink("myfifo2");
unlink("myfifo1");
}//creating another process
else if(fork() == 0){
int pd4;
pd4 = open("myfifo2", O_RDONLY);
char tab3[256];
double x3 = 0.0;
while(1){
x3=0.0;
lseek(pd4,0,0);
memset(tab3, 0, 256);
//reading data from pipe , myfifo2
read(pd4,tab3,sizeof(tab3));
//converting char* to float
x3 = atof(tab3);
FINAL_RESULT += x3;
if(x3<=0.0){
printf("\nFinal Result : %lf", FINAL_RESULT );
break;}
//else
//printf("\n x = %lf, Current result = %lf" ,x3, FINAL_RESULT);
}
sleep(1);
printf("\nFinal Result : %lf", FINAL_RESULT);
close(pd4);
unlink("myfifo2");
}
sleep(7);
unlink("myfifo1");
unlink("myfifo2");
while(1)pause();
return 0;
}
现在我可以通过使用Maclaurin系列看到e ^ x函数的数学估计之美。结果如下：
pawel@IMPERIUMVB:~$ gcc rury.c -lm -o zad1
pawel@IMPERIUMVB:~$ ./zad1
X Value :
8
EPS Value :
0.0001
1. Term of Mac Series = 1.000000
2. Term of Mac Series = 8.000000
3. Term of Mac Series = 32.000000
4. Term of Mac Series = 85.333333
5. Term of Mac Series = 170.666667
6. Term of Mac Series = 273.066667
7. Term of Mac Series = 364.088889
8. Term of Mac Series = 416.101587
9. Term of Mac Series = 416.101587
10. Term of Mac Series = 369.868078
11. Term of Mac Series = 295.894462
12. Term of Mac Series = 215.195972
13. Term of Mac Series = 143.463982
14. Term of Mac Series = 88.285527
15. Term of Mac Series = 50.448873
16. Term of Mac Series = 26.906065
17. Term of Mac Series = 13.453033
18. Term of Mac Series = 6.330839
19. Term of Mac Series = 2.813706
20. Term of Mac Series = 1.184718
21. Term of Mac Series = 0.473887
22. Term of Mac Series = 0.180529
23. Term of Mac Series = 0.065647
24. Term of Mac Series = 0.022834
25. Term of Mac Series = 0.007611
26. Term of Mac Series = 0.002436
27. Term of Mac Series = 0.000749
Final Result : 2980.957900 // after adding all of above
e ^ 8~2980.9579870417
估计值令人满意：）