Почему вывод типа F# не может обработать это?

Question

Почему вывод типа F# не может обработать это?

Вы можете написать переменную функцию и передать параметры в vsnprintf():

#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>

void display(int foo, int bar, char const *format, ...)
{
    va_list arglist;
    va_start(arglist, format);

    int length = vsnprintf(NULL, 0, format, arglist);
    char *buffer = malloc(length * sizeof *buffer); 
    vsnprintf(buffer, length, format, arglist);
    va_end(arglist);

    puts(buffer);
    free(buffer);
}

int main(void)
{
    display(42, 13, "%s %d %f", "Hello", 99, 100.13);
}

13

f# type-inference

задан kinghajj 10 May 2009 в 05:07

1 ответ

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f# type-inference

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score 24 · Accepted Answer

Type inference works left-to-right. This is where the pipeline operator is useful; if you already know the type of 'fis', then write it as

fis |> Seq.map (fun fi -> fi.Name)

and the inference works for you.

(In general, expressions of the form

o.Property
o.Method args

require the type of 'o' to be known a priori; for most other expressions, when a type is not pinned down the inference system can 'float a constraint' along that can be solved later, but for these cases, there are no constraints of the form 'all types with a property named P' or 'all types with a method named M' (like duck typing) that can be postponed and solved later. So you need that info now, or inference fails immediately.)

See also an overview of type inference in F#.