In the following cases, the parameter name is required when calling a swift function
1. when defining a regular function without external parameter name, the caller does not specify the first parameter's name, but need to specify all following parameter's name
func join0(string: String, toString: String, withJoiner: String) -> String {
return string + toString + withJoiner
}
join0("hello", toString: "world", withJoiner: ", ")
2. if the function's first as well as other parameters have external names, (internal name and external name are indicated by a space),
func join1(string s1: String, toString s2: String, withJoiner: String) -> String {
return s1 + withJoiner + s2
}
join1(string: "hello", toString: "world", withJoiner: ", ")
3. if the function's first parameters starts with '#', it indicates internal name and external name are same
func join2(#string: String, toString: String, withJoiner: String) -> String {
return s1 + joiner + s2
}
join1(string: "hello", toString: "world", withJoiner: ", ")
4. if the function's first parameters has default value, the caller must specify the parameter name to call it
func join3(string: String ="abc", toString: String, withJoiner: String) -> String {
return s1 + joiner + s2
}
join3(string: "hello", toString: "world", withJoiner: ", ")
5.In any case, if the parameter's external name is '_', then caller does not need specify the parameter's name,
func join4( _ string: String = "abc", _ toString: String, _ withJoiner: String) -> String {
return string + withJoiner + toString
}
join4("hello", "world", ", ")
Sometimes, if a parameter has a default value, but does not want to have an external name, then, explicitly specify '_' as external parameter name as shown below:
func join(s1: String, s2: String, _ joiner: String = " ") -> String {return s1 + joiner + s2}join("hello", "world", "-")
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