C++ – How far can memory leaks go

cmemorymemory-leaks

I've run into memory leaks many times. Usually when I'm malloc-ing like there's no tomorrow, or dangling FILE *s like dirty laundry. I generally assume (read: hope desperately) that all memory is cleaned up at least when the program terminates. Are there any situations where leaked memory won't be collected when the program terminates, or crashes?

If the answer varies widely from language-to-language, then let's focus on C(++).

Please note hyperbolic usage of the phrase, 'like there's no tomorrow', and 'dangling … like dirty laundry'. Unsafe* malloc*ing can hurt the ones you love. Also, please use caution with dirty laundry.

Best Answer

No. Operating systems free all resources held by processes when they exit.

This applies to all resources the operating system maintains: memory, open files, network connections, window handles...

That said, if the program is running on an embedded system without an operating system, or with a very simple or buggy operating system, the memory might be unusable until a reboot. But if you were in that situation you probably wouldn't be asking this question.

The operating system may take a long time to free certain resources. For example the TCP port that a network server uses to accept connections may take minutes to become free, even if properly closed by the program. A networked program may also hold remote resources such as database objects. The remote system should free those resources when the network connection is lost, but it may take even longer than the local operating system.

Related Solutions

How to determine the size of the array in C

Executive summary:

int a[17];
size_t n = sizeof(a)/sizeof(a[0]);

Full answer:

To determine the size of your array in bytes, you can use the sizeof operator:

int a[17];
size_t n = sizeof(a);

On my computer, ints are 4 bytes long, so n is 68.

To determine the number of elements in the array, we can divide the total size of the array by the size of the array element. You could do this with the type, like this:

int a[17];
size_t n = sizeof(a) / sizeof(int);

and get the proper answer (68 / 4 = 17), but if the type of a changed you would have a nasty bug if you forgot to change the sizeof(int) as well.

So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.

int a[17];
size_t n = sizeof(a) / sizeof(a[0]);

Another advantage is that you can now easily parameterize the array name in a macro and get:

#define NELEMS(x)  (sizeof(x) / sizeof((x)[0]))

int a[17];
size_t n = NELEMS(a);

C++ – How to set, clear, and toggle a single bit

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.