fopen

YehudaSinger · 11-24-2016, 02:52 PM

Dear sir/Mrs.

I use the Red-Hat 6.2 on an 4 cores of Intel. The IDE is Eclipse.
The code is:
std::string filename_str;
FILE * fp ;

fp = fopen ( file_name_str.c_str() , "a+" ) ;
if (fp==NULL)
{

cout << " Error in openingfile" ;
}

in run-time the file is created. the value of fp is not null. However, the if statement is executed and the cout statement works.

Best regards,
Yehuda

dogpatch · 11-24-2016, 03:47 PM

filename_str != file_name_str

pan64 · 11-25-2016, 01:01 AM

This is impossible:

Quote:

the value of fp is not null. However, the if statement is executed

if that cout statement was executed than fp must be null.
Checking compiler warnings you could see/recognize the different variable names.

BW-userx · 11-25-2016, 07:27 AM

Of course it is going to be NULL you only just created it. Nothing was put into it, therefore it is NULL. You'd be better off it check to see if it has been created, if not, then throw an error.

sundialsvcs · 11-25-2016, 01:44 PM

You've been given some unintended mis-information here. By far the easiest thing to do is to add a printf statement that will print the value of the pointer, cast as an integer.

Also, check the value of the errno variable.

The fact that the message appears, indicates definitively that the value returned was NULL.

YehudaSinger · 11-27-2016, 01:33 AM

fopen

Dear sir/Mrs.

I use the Red-Hat 6.2 on an 4 cores of Intel. The IDE is Eclipse.
The code is:
std::string file_name_str;
FILE * fp ;

fp = fopen ( file_name_str.c_str() , "a+" ) ;
if (fp==NULL)
{

cout << " Error in openingfile" ;
}

in run-time the file is created. The value of fp is not null(examined in the debugger). However, the if statement is executed and the cout statement works.

Best regards,
Yehuda

pan64 · 11-27-2016, 09:09 AM

as it was already told that is impossible. You may have mixed something. Try to use perror instead of that cout to check what's happened.

sundialsvcs · 11-27-2016, 09:38 AM

Instead of relying on a debugger, add print-statements throughout the code to display the values and to trace execution through the various program paths.

I (almost ...) never use a debugger.

For one thing, debuggers are fairly contrary to the actual behavior of any decent optimizing compiler, which will pass values directly in registers whenever possible and which will not (for instance) actually store a value into a variable if it sees that it doesn't have to. (Remember: the generated object-code will be functionally equivalent to, but not necessarily identical to, what your source-code said.)

YehudaSinger · 11-27-2016, 02:03 PM

Thanks for your answer. When I use the debugger, I set the optimization to NULL. Otherwise, you are right.
Yehuda