FILE*One= fopen("1.txt", "rb"); // указывает на 1.txt i=0; while(!feof(One)){ if(i<a||i>number){ fin_1.getline(buff,30); fout_2<<buff<<endl; } else{ fin_3.getline(buff,30); fout_2<<buff<<endl; } i++; } 1 answer
Although you are told that you are reading one file and checking another ... in any case, checking
while(!feof(file)) { ... } incorrect, because the end of file attribute is set only after an unsuccessful attempt to read past the end of the file .
- when feof makes sense to cause looks like a frequent question. Perhaps it makes sense in one place popular with examples to present, and other questions to close as duplicates. - jfs
- @jfs has I once wrote such a question-answer, such an array of critics flew in that I thought it would be best to remove it and promised myself that I would not do this anymore. Today I make another (if it does not work out - certainly the last one ....) attempt. - Harry
- It is worth remembering that the rules for questions / answers remain the same, regardless of who is the author. What is the procedure for creating canonical questions? . Problems can, of course, arise from scratch. In such cases, you can discuss a specific example on Mete. - jfs
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fin_1), and on reaching the end you check another (One). - ߊߚߤߘкак из этих потоков сделать один- choose one of these two streams and work only with it. The second one is removed as extra. - ߊߚߤߘ