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What is the return type of bounded-sqrt,bounded-sqrt-iter?
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The problem is in the conditional that has clauses that return values from different types: Number and Boolean. In order to accommodate such conditionals we allow union types in contract specifications. The resulting contracts:
Signature: bounded-sqrt(x,bound)
Purpose: To compute the square root of x, using Newton’s approximations method, if number of iterations does not exceed ’bound’
Type: [Number*Number -> Number union Boolean]
Example:
(sqrt 16. 7) should produce 4.000000636692939(sqrt 16. 4) should produce #f
Pre-conditions: x >= 0, bound >= 0Signature: bounded-sqrt-iter(guess,x,bound)
Purpose: To compute the square root of x, starting with ’guess’ as initial guess, if number of iterations does not exceed ’bound’
Type: [Number*Number*Number -> Number union Boolean]
Example:
(sqrt 1 16. 7) should produce 4.000000636692939(sqrt 1 16. 4) should produce #fPre-conditions: x >= 0, bound >= 0, guess != 0 Your Comment
Signature: bounded-sqrt(x,bound)
Purpose: To compute the square root of x, using Newton’s approximations method, if number of iterations does not exceed ’bound’
Type: [Number*Number -> Number union Boolean]
Example:
(sqrt 16. 7) should produce 4.000000636692939(sqrt 16. 4) should produce #f
Pre-conditions: x >= 0, bound >= 0Signature: bounded-sqrt-iter(guess,x,bound)
Purpose: To compute the square root of x, starting with ’guess’ as initial guess, if number of iterations does not exceed ’bound’
Type: [Number*Number*Number -> Number union Boolean]
Example:
(sqrt 1 16. 7) should produce 4.000000636692939(sqrt 1 16. 4) should produce #fPre-conditions: x >= 0, bound >= 0, guess != 0 Your Comment
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