How Can I Replace All Variables in a Go Sprintf String with the Same Value?

Using Sprintf with Identical Value Replacement

In Sprintf, the ability to replace all variables in the formatted string with the same value is possible but requires modifying the format string.

Explicit Argument Indices

The key is to use explicit argument indices before each formatting verb. This allows you to specify which argument should be used for that particular substitution. In Printf, Sprintf, and Fprintf, the default behavior is to sequentially use the passed arguments. However, explicit argument indices, denoted by [n], allow you to select a specific argument index.

Example

Consider the following example:

val := "foo"
s := fmt.Sprintf("%[1]v in %[1]v is %[1]v", val)
fmt.Println(s)

Here, we use the explicit argument index [1] before each formatting verb. This instructs Sprintf to use the first argument, i.e., val, for all three substitutions. As a result, the output will be:

foo in foo is foo

Single-Line Simplification

The above example can be simplified to a single line:

fmt.Printf("%[1]v in %[1]v is %[1]v", "foo")

Omitting First Index

Moreover, you can omit the first explicit argument index, as it automatically defaults to 1:

fmt.Printf("%v in %[1]v is %[1]v", "foo")

By understanding explicit argument indices, you can specify the same value for multiple replacements in Sprintf, regardless of the number of format specifiers.

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