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Top Forums UNIX for Dummies Questions & Answers Bash loops and variable scope Post 302503767 by Corona688 on Friday 11th of March 2011 03:02:41 PM
Old 03-11-2011
Yes, the problem is not the loop but the pipe; things behind a pipe are executed in a separate subshell and don't modify variables in your original shell. Temporary files are actually a pretty good way to solve it.
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io_pipe(3)						     Library Functions Manual							io_pipe(3)

NAME
io_pipe - create a Unix pipe SYNTAX
#include <io.h> int io_pipe(int64 pfd[2]); DESCRIPTION
io_pipe creates a new UNIX ``pipe.'' The pipe can receive data and provide data; any bytes written to the pipe can then be read from the pipe in the same order. A pipe is typically stored in an 8192-byte memory buffer; the exact number depends on the UNIX kernel. Bytes are written to the end of the buffer and read from the beginning of the buffer. Once a byte has been read, it is eliminated from the buffer, making space for another byte to be written; readers cannot ``rewind'' a pipe to read old data. Once 8192 bytes have been written to the buffer, the pipe will not be ready for further writing until some of the bytes have been read. Once all the bytes written have been read, the pipe will not be ready for further reading until more bytes are written. io_pipe sets d[0] to the number of a new descriptor reading from the pipe, and sets d[1] to the number of a new descriptor writing to the pipe. It then returns 1 to indicate success. If something goes wrong, io_pipe returns 0, setting errno to indicate the error; in this case it frees any memory that it allocated for the new pipe, and it leaves d alone. SEE ALSO
io_readfile(3), io_createfile(3), io_socketpair(3) io_pipe(3)
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