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Why Does My `std::fstream` Fail to Create a File?

Barbara Streisand
Release: 2024-10-29 23:28:29
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 Why Does My `std::fstream` Fail to Create a File?

File Creation Oversights in std::fstream

When attempting to manipulate files using the C IO library, std::fstream offers convenient file access and input/output capabilities. However, encountering issues with file creation can hinder progress.

One common pitfall lies in the use of incorrect flags when opening a file. Let's consider the following snippet:

<code class="cpp">std::fstream my_stream;
my_stream.open("my_file_name", std::fstream::binary | std::fstream::in | std::fstream::out);
if (!my_stream) std::cout << "error" << strerror(errorno);</code>
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In this code, the attempt is made to create a file named "my_file_name" for writing. However, the inclusion of std::fstream::in in the mode argument has unintended consequences:

<code class="cpp">// In effect, the above code is equivalent to:
my_stream.open("my_file_name", std::fstream::binary | std::fstream::in);</code>
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The std::fstream::in flag explicitly specifies the need for an existing file, which contradicts the intention of creating a file that potentially does not exist.

To rectify this issue, several solutions are available:

Solution 1: Remove std::fstream::in

Simply remove the std::fstream::in flag from the mode argument. This will allow the stream to create the file if it does not exist:

<code class="cpp">my_stream.open("my_file_name", std::fstream::binary | std::fstream::out);</code>
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Solution 2: Specify std::fstream::out and std::fstream::trunc

If the file already exists, you may want to truncate it instead of appending to it. To achieve this, add std::fstream::trunc to the mode argument:

<code class="cpp">my_stream.open("my_file_name", std::fstream::binary | std::fstream::out | std::fstream::trunc);</code>
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Understanding the appropriate use of flags and their implications is crucial for effectively managing files with std::fstream.

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