@sprintf format string cannot be a String variable or related

[JakeZw]

In working with @sprintf the formatting string must be a string within the quotes within the macro call. It cannot be a String variable or a function that compiles to a string.

What I tried is

julia> a = [1,2,3,4,5,6]
6-element Vector{Int64}:
 1
 2
 3
 4
 5
 6

julia> @sprintf(" %12.5E %12.5E %12.5E %12.5E %12.5E %12.5E", a...)
"  1.00000E+00  2.00000E+00  3.00000E+00  4.00000E+00  5.00000E+00  6.00000E+00"

julia> af = " %12.5E"^length(a)
" %12.5E %12.5E %12.5E %12.5E %12.5E %12.5E"

julia> @sprintf(af, a...)
ERROR: LoadError: ArgumentError: First argument to `@sprintf` must be a format string.
Stacktrace:
 [1] var"@sprintf"(__source__::LineNumberNode, __module__::Module, fmt::Any, args::Vararg{Any})
   @ Printf C:\Users\jakez\.julia\juliaup\julia-1.12.1+0.x64.w64.mingw32\share\julia\stdlib\v1.12\Printf\src\Printf.jl:1035
in expression starting at REPL[34]:1

Is there any way to get something like a string variable to work in @sprintf?

[Seelengrab]

This is intended behavior. The reason it works this way is because the macro does parse-time validation of the formatting string. Use Printf.Format & Printf.format to use runtime-defined formatting strings (though using a bespoke solution is likely better in the long term).

[engineer-antonio]

Hello,

The Functional Solution: Printf.Format and Printf.format
If you need a dynamic string, you should abandon the macro (@printf) and use the functions from the Printf package. This is the correct way to handle formats generated at runtime.

Julia
using Printf

a = [1, 2, 3, 4, 5, 6]
af = "%12.5E"^length(a)

1. Create a format object

fmt = Printf.Format(af)

2. Use the format function (not the macro)

resultado = Printf.format(fmt, a...)