Why Does My C Code Print the Last Line of a Text File Twice When Reading Until EOF?

Reading from Text File Until EOF: Solving the Last Line Duplication

The provided C code reads integers from a text file until it encounters the end of file (EOF) indicator. However, it mistakenly prints the last integer on the file twice. This is due to how the EOF indicator is handled within the program's loop structure.

During execution, the program enters a loop that continues until the EOF indicator is detected. As the loop iterates through the file, it fetches an integer "x" from the file and prints it to the console. However, after the last integer in the file is read, the EOF indicator is encountered.

The problem arises because the loop condition evaluates EOF after fetching the last integer "x". This means that the loop continues with the same value of "x" that was loaded in the previous iteration, effectively printing it twice.

To resolve this issue, the loop condition should be modified to check for the EOF indicator before attempting to fetch an integer. This can be achieved by using a slightly different loop structure:

while (iFile) {
    int x;
    iFile >> x;
    if (iFile) {
        cerr << x << endl;
    }
}

In this modified loop, the "iFile" object is used as the loop condition, which implicitly checks for the EOF indicator. If "iFile" is true (indicating that the file is not at EOF), the program fetches and prints an integer. If "iFile" is false (indicating EOF), the loop is exited.

By implementing this adjustment, the program reads integers from the text file until it reaches EOF and prints each integer once, resolving the issue of the last line being printed twice.

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