Re: Question about std.conv.parse

[Jacky via Digitalmars-d-learn]

On Tuesday, 30 January 2018 at 09:29:22 UTC, Jonathan M Davis 
wrote:
On Tuesday, January 30, 2018 09:19:22 Jacky via 
Digitalmars-d-learn wrote:

[...]


The first one passes an lvalue. The second one passes an 
rvalue. parse takes its argument by ref so that what is parsed 
is removed from the input. As such, it requires an lvalue.



[...]


You don't. parse requires that you pass it a variable.

std.conv.to does not take its argument by ref, so you can use 
that instead, but it converts the entire argument instead of 
just the front portion that matches the requested type. So, if 
you're trying to convert the entire argument, then you can use 
to, but if you're trying to just convert the front, then you 
have to use parse, and that means passing a variable.


- Jonathan M Davis


Thank you very much！I think i should dig deeper into this 
interesting language. :)

Re: Question about std.conv.parse

[Jonathan M Davis via Digitalmars-d-learn]

On Tuesday, January 30, 2018 09:19:22 Jacky via Digitalmars-d-learn wrote:
> Hello everyone.I'm a newbie on the D language.When i use the
> library 'std.conv' ,i met some problem.
> This is what I have:
>
>  static import std.conv;
>  string aaa = "123456789";
>  uint idx = 5;
>  string bbb = aaa[0 .. idx];
>
>  uint work = std.conv.parse!(uint)(bbb); // this works
>
>  uint didnotwork = std.conv.parse!(uint)(aaa[0 .. idx]);
> //but here's a error
>  //template std.conv.parse cannot deduce function from
> argument types !(uint)(string)
>
> So my questions are:
> 1) What is the difference between these two lines?

The first one passes an lvalue. The second one passes an rvalue. parse takes
its argument by ref so that what is parsed is removed from the input. As
such, it requires an lvalue.

> 2) How to correct the second without assign a new variable?

You don't. parse requires that you pass it a variable.

std.conv.to does not take its argument by ref, so you can use that instead,
but it converts the entire argument instead of just the front portion that
matches the requested type. So, if you're trying to convert the entire
argument, then you can use to, but if you're trying to just convert the
front, then you have to use parse, and that means passing a variable.

- Jonathan M Davis

Question about std.conv.parse

[Jacky via Digitalmars-d-learn]

Hello everyone.I'm a newbie on the D language.When i use the 
library 'std.conv' ,i met some problem.

This is what I have:

static import std.conv;
string aaa = "123456789";
uint idx = 5;
string bbb = aaa[0 .. idx];

uint work = std.conv.parse!(uint)(bbb); // this works

uint didnotwork = std.conv.parse!(uint)(aaa[0 .. idx]);   
//but here's a error
//template std.conv.parse cannot deduce function from 
argument types !(uint)(string)


So my questions are:
1) What is the difference between these two lines?
2) How to correct the second without assign a new variable?

Cheers,
Jacky