Return to Answer

3.1

1. Change i-- to n--
2. i<n to -i<n
3. (Unfortunately invalid answer because I wasn't checking with compiler before seeing other answers)

3.2.1

int n = 20 

to

uint n = 20

(Compiler dependent...)

3.2.2

int n = -20L;

for(i = -1; i%n; i--)

prints 19 + signs, same as with int n = 20L;. However, I wouldn't have come up with it if I hadn't seen other answers to 3.2.1

3.1

1. Change i-- to n--
2. i<n to -i<n
3. (Unfortunately invalid answer because I wasn't checking with compiler before seeing other answers)

3.2.1

int n = 20 

to

uint n = 20

(Compiler dependent...)

3.2.2

int n =-20L;

for(i = -1; i%n; i--)

prints 19 + signs, same as with int n = 20L;. However, I wouldn't have come up with it if I hadn't seen other answers to 3.2.1

3.1

1. Change i-- to n--
2. i<n to -i<n
3. (Unfortunately invalid answer because I wasn't checking with compiler before seeing other answers)

3.2.1

int n = 20 

to

uint n = 20

(Compiler dependent...)

3.2.2

int n = -20L;

for(i = -1; i%n; i--)

prints 19 + signs, same as with int n = 20L;

3.1

1. Change i-- to n--
2. i<n to -i<n
3. (Unfortunately invalid answer because I wasn't checking with compiler before seeing other answers)

3.2.1

int n = 20 

to

uint n = 20

(Compiler dependent...)

3.2.2

int n = -20L;

for(i = -1; i%n; i--)

prints 19 + signs, same as with int n = 20L;. However, I wouldn't have come up with it if I hadn't seen other answers to 3.2.1

3.1

1. Change i-- to n--
2. i<n to -i<n
3. (Unfortunately invalid answer because I wasn't checking with compiler before seeing other answers)

3.2.1

int n = 20 

to

uint n = 20

(Compiler dependent...)

3.2.2

int n = -20L;

for(i = -1; i%n; i--)

prints 19 + signs.

3.1

1. Change i-- to n--
2. i<n to -i<n
3. (Unfortunately invalid answer because I wasn't checking with compiler before seeing other answers)

3.2.1

int n = 20 

to

uint n = 20

(Compiler dependent...)

3.2.2

int n = -20L;

for(i = -1; i%n; i--)

prints 19 + signs, same as with int n = 20L;