为什么使用int i的值; 未定义的行为，但使用rand（）的值不是？

Question

If I don't initialise i , I don't know its value, but also I cannot predict the value of rand() .

But on the other hand, I know the value of the uninitialised i is between INT_MIN and INT_MAX , also I know the value of rand() is between 0 and RAND_MAX .

Why is using the value of the uninitialised i undefined behaviour but using the value of rand() is not?

Answer 1

The value of an uninitialized variable is undefined . The return value of rand() is well-defined as being the next number in the pseudo-random sequence for the given seed .

You can rely on rand() returning a pseudorandom number. You cannot rely on any characteristic of the value of an uninitialized int i .

I looked up the C99 standard (ISO/IEC 9899:1999), which is the only standard document I have available, but I seriously doubt these things have changed. In chapter 6.2.6 Representations of types , it is stated that integers are allowed to be stored in memory with padding bits , the value of which is unspecified but may include parity bits, which would be set upon initialization and any arithmetic operation on the integer. Certain representations (like eg a parity mismatch) could be trap representations , the behaviour of which is undefined (but might well terminate your program).

So, no, you cannot even rely on an uninitialized int i to be INT_MIN <= i <= INT_MAX .

Answer 2

The standard says reading from an uninitialized variable is undefined behaviour, so it is. You cannot say that it is between INT_MIN and INT_MAX . In fact, cannot really think of that variable as holding any value, so you couldn't even check that hypothesis ^* .

rand() on the other hand is designed to produce a random number within a range. If reading from the result of rand() were undefined, it wouldn't be part of any library because it couldn't be used.

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^{* Usually "undefined behaviour" provides scope for optimization. The optimizer can do whatever it wants with an uninitialized variable, working under the assumption that it isn't read from.}

Answer 3

Value of rand is defined to be (pseudo-)random. Value of an uninitialized variable is not defined to be anything. That is the difference, whether something is defined to be anything meaningful (while rand() is also meaningful - it gives (pseudo-)random numbers) or undefined.

Answer 4

The short answer, like others have said, is because the standard says so.

The act of accessing the value of a variable (described in the standard as performing an lvalue to rvalue conversion) that is uninitialised gives undefined behaviour.

rand() is specified as giving a value between 0 and RAND_MAX , where RAND_MAX (a macro declared in <cstdlib> for C++, and <stdlib.h> for C) is specified as having a value which is at least 32767 .

Answer 5

rand uses an algorithm to generate your number. So rand is not undefined and if when you use srand to start the random generation you choose a value like 42, you will always get the same value each time you launch your app try to launch this example you will see:

#include <stdio.h>
#include <stdlib.h>

int main()
{
  srand(42);

  printf("%d\n", rand());
  return 0;
}

see also: https://en.wikipedia.org/wiki/List_of_random_number_generators

Answer 6

Consider variable x in the following code:

uint32_t blah(uint32_t q, uint32_t r)
{
  uint16_t x;
  if (q)
    x=foo(q); // Assume return type of foo() is uint16_t
  if (r)
    x=bar(r); // Assume return type of bar() is uint16_t
  return x;
}

Since code never assigns x a value outside the range 0-65535, a compiler for a 32-bit processor could legitimately allocate a 32-bit register for it. Indeed, since the value of q is never used after the first time x is written, a compiler could use the same register to hold x has had been used to hold q . Ensuring that the function would always return a value 0-65535 would require adding some extra instructions compared with simply letting it return whatever happened to be in the register allocated to x .

Note that from the point of view of the Standard authors, if the Standard isn't going to require that the compiler make x hold a value from 0-65535, it may as well not specify anything about what may happen if code tries to use x . Implementations that wish to offer any guarantees about behavior in such cases are free to do so, but the Standard imposes no requirements.