Searched refs:nfields (Results 1 – 1 of 1) sorted by relevance
9 split(string, fields, nfields, sep) in split() argument12 int nfields; /* number of entries available in fields[] */42 fn = nfields;50 return(nfields - fn);54 fn = nfields;66 fn = nfields;74 return(nfields - fn);84 fn = nfields;100 if (fn == nfields+1)111 if (fn < nfields)[all …]
Completed in 4 milliseconds