C++ 收到错误“fopen”：此函数或变量可能不安全。编译时
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Getting an error "fopen': This function or variable may be unsafe." when compling
提问by SeverusSwan
I'm receiving this error when compiling:
我在编译时收到此错误：
'fopen': This function or variable may be unsafe.
Consider using fopen_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS.
I'm new to C++ and open CV, therefore please help me to get rid of this error.
我是 C++ 新手并打开 CV，因此请帮助我摆脱这个错误。
Thanks
谢谢
void _setDestination(const char* name)
{
if (name==NULL) {
stream = stdout;
}
else {
stream = fopen(name,"w");
if (stream == NULL) {
stream = stdout;
}
}
}
回答by nvoigt
This is not an error, it is a warning from your Microsoft compiler.
这不是错误，而是来自 Microsoft 编译器的警告。
Select your project and click "Properties" in the context menu.
选择您的项目并在上下文菜单中单击“属性”。
In the dialog, chose Configuration Properties-> C/C++-> Preprocessor
在对话框中，选择Configuration Properties-> C/C++->Preprocessor
In the field PreprocessorDefinitions add ;_CRT_SECURE_NO_WARNINGSto turn those warnings off.
在 PreprocessorDefinitions 字段中添加;_CRT_SECURE_NO_WARNINGS以关闭这些警告。
回答by Blacktempel
This is a warning for usual. You can either disable it by
这是对平常的警告。您可以通过以下方式禁用它
#pragma warning(disable:4996)
or simply use fopen_s like Microsoft has intended.
或者像微软那样简单地使用 fopen_s 。
But be sure to use the pragma before other headers.
但一定要在其他标头之前使用编译指示。
