Which of the following code fragments gives a random double value between 2 and 5 inclusive? You can assume that the random number seed has been set and any needed definitions and initializations have been made.

Question

a) ```
3.0*rand() + 2
```
b) ```
3.0*(RAND_MAX-rand())/RAND_MAX + 2
```
c) ```
((RAND_MAX-rand())/static_cast(RAND_MAX))*3 + 2```
d) ```
(RAND_MAX-rand())/static_cast(RAND_MAX)*5 -2
```
e) ```
rand()/static_cast(RAND_MAX)*2 + 3
```

Answer 1

d) ```
(RAND_MAX-rand())/static_cast(RAND_MAX)*5 -2
```

Part a) The library function rand() returns an int, not a double, so this is not correct. Part b) is syntactically correct but rand() returns an int between 0 and RAND_MAX. On one compiler available to this writer RAND_MAX is 2,147,483,647 (the value of the largest possible int value, MAX_INT). In 3.0*(RAND_MAX-rand()) the subtraction is done then the multiplication. Integer overflow is assured on this system before the computation is complete. On another compiler, RAND_MAX is 32,767, so b) would work there. This solution is to be avoided. Part c): Divide and multiply associate left to right, so the expression
```
(RAND_MAX-rand())/static_cast(RAND_MAX)
```
gives a double between 0.0 and 1.0. Multiplying by 3 scales this to 0.0 to 3.0, adding 2 gives 2.0 to 5.0 as required, without overflow. Parts d) and e) are wrong:
d) gives a random number between 2.0 and 7.0. e) gives a random double between 3.0 and 5.0.