using parse with string slice

[Quentin Ladeveze via Digitalmars-d-learn]

Hi,

I try to parse some hexadecimal numbers stocked in a string. So I 
use this code



---
import std.conv;

string s = "B";

int value = parse!int(s, 16);

assert(value == 11);
---

But when I have several hexadecimal numbers in the string, and I 
slice it, the compiler can't deduce which version of parse to use 
:


---
import std.conv;

string s = "1B2A";

int value = parse!int(s[0..2], 16); //template std.conv.parse 
cannot deduce function from argument types !(int)(string, int)

---

Does someone have an idea of why it happens ? The version of 
parse that is used here is :


---
std.conv.parse(Target, Source)(ref Source s, uint radix) if 
(isSomeChar!(ElementType!Source) && isIntegral!Target && 
!is(Target == enum))

---

And I can't see which template constraint is not respected when I 
call it with a slice.


Thanks

Re: using parse with string slice

[Adam D. Ruppe via Digitalmars-d-learn]

On Saturday, 5 December 2015 at 22:05:11 UTC, anonymous wrote:
You can use std.conv.to instead or assign the slice to a 
variable first.


This is a bit of a FAQ I think because people don't realize you 
can use to and parse to do the same thing.


The big difference is parse will advance the slice you pass it 
past the item it reads, whereas to will try to convert the entire 
input.


That's why parse wants a ref slice, so it can move it forward. 
But since you're slicing it yourself, to is the way to go.

Re: using parse with string slice

[Quentin Ladeveze via Digitalmars-d-learn]

On Saturday, 5 December 2015 at 22:05:11 UTC, anonymous wrote:

On 05.12.2015 22:59, Quentin Ladeveze wrote:

---
import std.conv;

string s = "1B2A";

int value = parse!int(s[0..2], 16); //template std.conv.parse 
cannot

deduce function from argument types !(int)(string, int)
---

Does someone have an idea of why it happens ? The version of 
parse that

is used here is :

---
std.conv.parse(Target, Source)(ref Source s, uint radix) if
(isSomeChar!(ElementType!Source) && isIntegral!Target && 
!is(Target ==

enum))
---

And I can't see which template constraint is not respected 
when I call

it with a slice.


It's the `ref` part. A slice expression cannot be passed in a 
ref parameter. You can use std.conv.to instead or assign the 
slice to a variable first.


Thank you for your quick answer, learned something today :)

Re: using parse with string slice

[anonymous via Digitalmars-d-learn]

On 05.12.2015 22:59, Quentin Ladeveze wrote:

---
import std.conv;

string s = "1B2A";

int value = parse!int(s[0..2], 16); //template std.conv.parse cannot
deduce function from argument types !(int)(string, int)
---

Does someone have an idea of why it happens ? The version of parse that
is used here is :

---
std.conv.parse(Target, Source)(ref Source s, uint radix) if
(isSomeChar!(ElementType!Source) && isIntegral!Target && !is(Target ==
enum))
---

And I can't see which template constraint is not respected when I call
it with a slice.


It's the `ref` part. A slice expression cannot be passed in a ref 
parameter. You can use std.conv.to instead or assign the slice to a 
variable first.