Re: Unable to open filename passed as command line argument
On Thursday, 3 September 2020 at 11:12:49 UTC, user1234 wrote:
On Thursday, 3 September 2020 at 10:47:04 UTC, Curious wrote:
otherwise what you get as args are D dynamic arrays (a payload
made of .ptr and .length) so you can use std.file or std.stdio
to open a file using the "D main" arguments (it's not the like
"C main").
---
void main(string[] args)
{
import std.stdio;
File f = File(args[1], "r");
}
---
Okay. Got it!
Thanks
Re: Unable to open filename passed as command line argument
On Thursday, 3 September 2020 at 10:47:04 UTC, Curious wrote:
Given the following:
=a==
void main(string[] args)
{
FILE* fp = fopen(args[1].ptr, "r");
if (!fp) throw new Exception("fopen");
}
=b==
void main(string[] args)
{
FILE* fp = fopen(args[1].dup.ptr, "r");
if (!fp) throw new Exception("fopen");
}
Why does a fail but b succeed?
version b works by accident/UB. You need to null terminate your
filename if you use the C library functions:
---
void main(string[] args)
{
FILE* fp = fopen((args[1] ~ '\0').ptr, "r");
if (!fp) throw new Exception("fopen");
}
---
otherwise what you get as args are D dynamic arrays (a payload
made of .ptr and .length) so you can use std.file or std.stdio to
open a file using the "D main" arguments (it's not the like "C
main").
---
void main(string[] args)
{
import std.stdio;
File f = File(args[1], "r");
}
---
Re: Unable to open filename passed as command line argument
On 9/3/20 1:47 PM, Curious wrote:
Given the following:
=a==
void main(string[] args)
{
FILE* fp = fopen(args[1].ptr, "r");
if (!fp) throw new Exception("fopen");
}
=b==
void main(string[] args)
{
FILE* fp = fopen(args[1].dup.ptr, "r");
if (!fp) throw new Exception("fopen");
}
Why does a fail but b succeed?
try `toStringz`:
```D
import std.string : toStringz;
void main(string[] args)
{
FILE* fp = fopen(args[1].toStringz, "r");
if (!fp) throw new Exception("fopen");
}
```
The reason is that args are D strings (that contains no terminating 0)
but `fopen` gets C string (null terminated) so your `a` variant fails
because the filename becomes wrong as there is no terminating 0. Your
`b` variant works in fact accidentally because after duplication in new
memory after filename 0 appears due to random reason (for example all
that memory area zeroed by allocator).
Unable to open filename passed as command line argument
Given the following:
=a==
void main(string[] args)
{
FILE* fp = fopen(args[1].ptr, "r");
if (!fp) throw new Exception("fopen");
}
=b==
void main(string[] args)
{
FILE* fp = fopen(args[1].dup.ptr, "r");
if (!fp) throw new Exception("fopen");
}
Why does a fail but b succeed?
