Zod 4: .default() short-circuits on undefined and must match the OUTPUT type — use .prefault() for v3 behavior
// Zod 3: default 'tuna' is parsed through the transform → 4
z.string().transform(v => v.length).default('tuna').parse(undefined) // 4
// Zod 4: default must be the output (number)
z.string().transform(v => v.length).default(0).parse(undefined) // 0
// Zod 4: keep the v3 semantics
z.string().transform(v => v.length).prefault('tuna').parse(undefined) // 4
Optional + default inside objects:
const S = z.object({ a: z.string().default('tuna').optional() })
S.parse({}) // Zod 3: {} Zod 4: { a: 'tuna' }
If you need the key to stay absent, drop the .default().
Symptom
After upgrading to zod@4, a schema like z.string().transform(s => s.length).default('tuna') fails to type-check, or parsed objects suddenly contain default values for optional fields that used to be absent. Wrong assumption: the default is fed through the schema (and transforms) like a normal input, and .default(...).optional() leaves missing keys missing.
Cause
Zod 4 (zod@4.0.0 at the package root, 2025-07-08) changed .default(): if the input is undefined, ZodDefault short-circuits parsing and returns the default as-is, so the default must be assignable to the output type. Defaults inside object properties are now applied even within optional fields. The old "parse the default" behavior moved to the new .prefault() (pre-parse default).
Verify
Unit-test schema.parse(undefined) / schema.parse({}) and compare with the expected output shape.
Sources
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