Function hasMethod [src]

Returns true if a type has a name method; false otherwise. Result is always comptime-known.

Prototype

pub inline fn hasMethod(comptime T: type, comptime name: []const u8) bool

Parameters

T: typename: []const u8

Example

 test hasMethod {
    try std.testing.expect(!hasMethod(u32, "foo"));
    try std.testing.expect(!hasMethod([]u32, "len"));
    try std.testing.expect(!hasMethod(struct { u32, u64 }, "len"));

    const S1 = struct {
        pub fn foo() void {}
    };

    try std.testing.expect(hasMethod(S1, "foo"));
    try std.testing.expect(hasMethod(*S1, "foo"));

    try std.testing.expect(!hasMethod(S1, "bar"));
    try std.testing.expect(!hasMethod(*[1]S1, "foo"));
    try std.testing.expect(!hasMethod(*[10]S1, "foo"));
    try std.testing.expect(!hasMethod([]S1, "foo"));

    const S2 = struct {
        foo: fn () void,
    };

    try std.testing.expect(!hasMethod(S2, "foo"));

    const U = union {
        pub fn foo() void {}
    };

    try std.testing.expect(hasMethod(U, "foo"));
    try std.testing.expect(hasMethod(*U, "foo"));
    try std.testing.expect(!hasMethod(U, "bar"));
}

Source

  pub inline fn hasMethod(comptime T: type, comptime name: []const u8) bool {
    return switch (@typeInfo(T)) {
        .pointer => |P| switch (P.size) {
            .one => hasFn(P.child, name),
            .many, .slice, .c => false,
        },
        else => hasFn(T, name),
    };
}  