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In c program, how to use Linux tee to redirect the output of itself [duplicate]

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0

This question already has an answer here:

• 
  How can I implement 'tee' programmatically in C?
  
  5 answers
  
  

I can't use the Linux command tee at the time the C program gets executed since I can't find which line of the code in a big project precisely that runs this program. I am struggling to come up with a way to call the tee command inside the C program to redirect its printf statement to a log file also. The problem is that I can't call tee without an executable like "./exe | tee some.log" since the program is already executed. I did some research on how to get the stdout of a running process and saw some answers suggesting to check out /proc//fd/1. But I don't know why "1"(stdout) file is empty, and I was expecting the file should store the printf output of the process.

Sincerely appreciate for any help!

c linux tee

share|improve this question

asked Mar 7 at 21:51

Ian HuangIan Huang

32

marked as duplicate by jww, Community♦ Mar 8 at 15:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Using command-line tools within C programs is sort of missing the point of C. There's easy ways to write a logging function that outputs to one or more filehandles, like stdout and/or a file.
  
  – tadman
  Mar 7 at 21:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  ..since the program is already executed. - what? Then how do you intend to "call" tee from within the program which is already executed?
  
  – Eugene Sh.
  Mar 7 at 21:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. You can call tee using the system().
  
  – Ian Huang
  Mar 7 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @tadman If I understand correctly, are you suggesting to for example customize a my_printf()? If so, it wouldn't work for me since then I will need to change every printf in the project, and it is just undoable...
  
  – Ian Huang
  Mar 7 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  EugeneSh is saying that if the program is already running, then making changes to it won't affect the running process; you'd need to recompile and start the process again. And if you're restarting it anyway, you can just pipe it to tee instead of modifying the source.
  
  – Ray
  Mar 7 at 22:10
  
  

  
  
  
  

 | 
show 4 more comments

0

This question already has an answer here:

• 
  How can I implement 'tee' programmatically in C?
  
  5 answers
  
  

I can't use the Linux command tee at the time the C program gets executed since I can't find which line of the code in a big project precisely that runs this program. I am struggling to come up with a way to call the tee command inside the C program to redirect its printf statement to a log file also. The problem is that I can't call tee without an executable like "./exe | tee some.log" since the program is already executed. I did some research on how to get the stdout of a running process and saw some answers suggesting to check out /proc//fd/1. But I don't know why "1"(stdout) file is empty, and I was expecting the file should store the printf output of the process.

Sincerely appreciate for any help!

c linux tee

share|improve this question

asked Mar 7 at 21:51

Ian HuangIan Huang

32

marked as duplicate by jww, Community♦ Mar 8 at 15:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Using command-line tools within C programs is sort of missing the point of C. There's easy ways to write a logging function that outputs to one or more filehandles, like stdout and/or a file.
  
  – tadman
  Mar 7 at 21:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  ..since the program is already executed. - what? Then how do you intend to "call" tee from within the program which is already executed?
  
  – Eugene Sh.
  Mar 7 at 21:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @EugeneSh. You can call tee using the system().
  
  – Ian Huang
  Mar 7 at 22:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @tadman If I understand correctly, are you suggesting to for example customize a my_printf()? If so, it wouldn't work for me since then I will need to change every printf in the project, and it is just undoable...
  
  – Ian Huang
  Mar 7 at 22:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  EugeneSh is saying that if the program is already running, then making changes to it won't affect the running process; you'd need to recompile and start the process again. And if you're restarting it anyway, you can just pipe it to tee instead of modifying the source.
  
  – Ray
  Mar 7 at 22:10
  
  

  
  
  
  

 | 
show 4 more comments

This question already has an answer here:

• 
  How can I implement 'tee' programmatically in C?
  
  5 answers
  
  

I can't use the Linux command tee at the time the C program gets executed since I can't find which line of the code in a big project precisely that runs this program. I am struggling to come up with a way to call the tee command inside the C program to redirect its printf statement to a log file also. The problem is that I can't call tee without an executable like "./exe | tee some.log" since the program is already executed. I did some research on how to get the stdout of a running process and saw some answers suggesting to check out /proc//fd/1. But I don't know why "1"(stdout) file is empty, and I was expecting the file should store the printf output of the process.

Sincerely appreciate for any help!

c linux tee

share|improve this question

asked Mar 7 at 21:51

Ian HuangIan Huang

32

share|improve this question

asked Mar 7 at 21:51

Ian HuangIan Huang

32

share|improve this question

asked Mar 7 at 21:51

Ian HuangIan Huang

32

asked Mar 7 at 21:51

Ian HuangIan Huang

32

asked Mar 7 at 21:51

Ian HuangIan Huang

32

1 Answer
1

active

oldest

votes

2

A shell sets up a pipeline using pipe and dup2 before invoking the program. There's no reason why you couldn't do it after:

void pipeto(char** cmd) 
 int fds[2];
 pipe(fds);
 if(fork()) 
 // Set up stdout as the write end of the pipe
 dup2(fds[1], 1);
 close(fds[0]);
 close(fds[1]);
 else 
 // Double fork to not be a direct child
 if(fork()) exit(0);
 // Set up stdin as the read end of the pipe, and run tee
 dup2(fds[0], 0);
 close(fds[0]);
 close(fds[1]);
 execvp(cmd[0], cmd);
 


int main() 
 char* cmd[] = "tee", "myfile.txt", NULL ;
 pipeto(cmd);
 printf("This is a testn");

Result:

$ ./foo
This is a test

$ cat myfile.txt
This is a test

Note that since the shell is no longer aware that there's a tee being piped to, it will not wait for it. This means that output can get a bit jumbled if the program exits and the shell redraws the prompt before tee has finished writing. Output like this is purely a cosmetic problem:

user@host $ ./foo
user@host $ This is a test

It does not mean that the command is hanging, the prompt still works and you can type commands or just hit Enter to redraw.

share|improve this answer

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems like exactly what I am looking for, thank you! But please allow me some time to take a look at it, and I will come back and accept it at latest tomorrow! Anyway, thank you again for your answer!
  
  – Ian Huang
  Mar 7 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can avoid the premature exiting problem by having the program wait for the child before it exits...
  
  – Chris Dodd
  Mar 8 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ChrisDodd Yes, though that assumes having an extra child does not interfere with the program. I conservatively assumed it would and double-forked
  
  – that other guy
  Mar 8 at 3:17
  

  
  
  
  

add a comment | 

2

A shell sets up a pipeline using pipe and dup2 before invoking the program. There's no reason why you couldn't do it after:

void pipeto(char** cmd) 
 int fds[2];
 pipe(fds);
 if(fork()) 
 // Set up stdout as the write end of the pipe
 dup2(fds[1], 1);
 close(fds[0]);
 close(fds[1]);
 else 
 // Double fork to not be a direct child
 if(fork()) exit(0);
 // Set up stdin as the read end of the pipe, and run tee
 dup2(fds[0], 0);
 close(fds[0]);
 close(fds[1]);
 execvp(cmd[0], cmd);
 


int main() 
 char* cmd[] = "tee", "myfile.txt", NULL ;
 pipeto(cmd);
 printf("This is a testn");

Result:

$ ./foo
This is a test

$ cat myfile.txt
This is a test

Note that since the shell is no longer aware that there's a tee being piped to, it will not wait for it. This means that output can get a bit jumbled if the program exits and the shell redraws the prompt before tee has finished writing. Output like this is purely a cosmetic problem:

user@host $ ./foo
user@host $ This is a test

It does not mean that the command is hanging, the prompt still works and you can type commands or just hit Enter to redraw.

share|improve this answer

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems like exactly what I am looking for, thank you! But please allow me some time to take a look at it, and I will come back and accept it at latest tomorrow! Anyway, thank you again for your answer!
  
  – Ian Huang
  Mar 7 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can avoid the premature exiting problem by having the program wait for the child before it exits...
  
  – Chris Dodd
  Mar 8 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ChrisDodd Yes, though that assumes having an extra child does not interfere with the program. I conservatively assumed it would and double-forked
  
  – that other guy
  Mar 8 at 3:17
  

  
  
  
  

add a comment | 

2

A shell sets up a pipeline using pipe and dup2 before invoking the program. There's no reason why you couldn't do it after:

void pipeto(char** cmd) 
 int fds[2];
 pipe(fds);
 if(fork()) 
 // Set up stdout as the write end of the pipe
 dup2(fds[1], 1);
 close(fds[0]);
 close(fds[1]);
 else 
 // Double fork to not be a direct child
 if(fork()) exit(0);
 // Set up stdin as the read end of the pipe, and run tee
 dup2(fds[0], 0);
 close(fds[0]);
 close(fds[1]);
 execvp(cmd[0], cmd);
 


int main() 
 char* cmd[] = "tee", "myfile.txt", NULL ;
 pipeto(cmd);
 printf("This is a testn");

Result:

$ ./foo
This is a test

$ cat myfile.txt
This is a test

Note that since the shell is no longer aware that there's a tee being piped to, it will not wait for it. This means that output can get a bit jumbled if the program exits and the shell redraws the prompt before tee has finished writing. Output like this is purely a cosmetic problem:

user@host $ ./foo
user@host $ This is a test

It does not mean that the command is hanging, the prompt still works and you can type commands or just hit Enter to redraw.

share|improve this answer

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This seems like exactly what I am looking for, thank you! But please allow me some time to take a look at it, and I will come back and accept it at latest tomorrow! Anyway, thank you again for your answer!
  
  – Ian Huang
  Mar 7 at 22:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can avoid the premature exiting problem by having the program wait for the child before it exits...
  
  – Chris Dodd
  Mar 8 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ChrisDodd Yes, though that assumes having an extra child does not interfere with the program. I conservatively assumed it would and double-forked
  
  – that other guy
  Mar 8 at 3:17
  

  
  
  
  

add a comment | 

A shell sets up a pipeline using pipe and dup2 before invoking the program. There's no reason why you couldn't do it after:

void pipeto(char** cmd) 
 int fds[2];
 pipe(fds);
 if(fork()) 
 // Set up stdout as the write end of the pipe
 dup2(fds[1], 1);
 close(fds[0]);
 close(fds[1]);
 else 
 // Double fork to not be a direct child
 if(fork()) exit(0);
 // Set up stdin as the read end of the pipe, and run tee
 dup2(fds[0], 0);
 close(fds[0]);
 close(fds[1]);
 execvp(cmd[0], cmd);
 


int main() 
 char* cmd[] = "tee", "myfile.txt", NULL ;
 pipeto(cmd);
 printf("This is a testn");

Result:

$ ./foo
This is a test

$ cat myfile.txt
This is a test

Note that since the shell is no longer aware that there's a tee being piped to, it will not wait for it. This means that output can get a bit jumbled if the program exits and the shell redraws the prompt before tee has finished writing. Output like this is purely a cosmetic problem:

user@host $ ./foo
user@host $ This is a test

It does not mean that the command is hanging, the prompt still works and you can type commands or just hit Enter to redraw.

share|improve this answer

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124

void pipeto(char** cmd) 
 int fds[2];
 pipe(fds);
 if(fork()) 
 // Set up stdout as the write end of the pipe
 dup2(fds[1], 1);
 close(fds[0]);
 close(fds[1]);
 else 
 // Double fork to not be a direct child
 if(fork()) exit(0);
 // Set up stdin as the read end of the pipe, and run tee
 dup2(fds[0], 0);
 close(fds[0]);
 close(fds[1]);
 execvp(cmd[0], cmd);
 


int main() 
 char* cmd[] = "tee", "myfile.txt", NULL ;
 pipeto(cmd);
 printf("This is a testn");

$ ./foo
This is a test

$ cat myfile.txt
This is a test

user@host $ ./foo
user@host $ This is a test

share|improve this answer

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124

answered Mar 7 at 22:08

that other guythat other guy

74.5k886124